Make 0 0 0 0 = 8












60












$begingroup$


Can you find a way to make:




$0 0 0 0 = 8$




by adding any operations or symbols? You can use only these symbols:




$+, -, *, !, /, hat, , ()$.




It is limited to this list, and concatenation is also allowed. You cannot add other numbers to the equation.










share|improve this question









$endgroup$





This question has an open bounty worth +100
reputation from user477343 ending in 6 days.


One or more of the answers is exemplary and worthy of an additional bounty.


This bounty will be awarded to @let_the_coding_begin's very first and very elegant answer — it has over 100 upvotes (including my own)! In my opinion, it is just as good the accepted answer, if not, better. So well cone, and congratulations! :D












  • 43




    $begingroup$
    Just put a slash over the equal sign!
    $endgroup$
    – Yout Ried
    Sep 7 '18 at 0:40






  • 13




    $begingroup$
    @YoutRied: standard loophole: neither funny nor creative. It's usually the first answer on any math puzzle.
    $endgroup$
    – Thomas Weller
    Sep 7 '18 at 18:40






  • 2




    $begingroup$
    Are we allowed to move anything, or is it insert only?
    $endgroup$
    – ctrl-alt-delor
    Sep 7 '18 at 19:33






  • 2




    $begingroup$
    Are you allowed to concatenate operators? (e.g !!, **, --, and ++)
    $endgroup$
    – Ole Tange
    Sep 9 '18 at 1:09






  • 1




    $begingroup$
    @OleTange The accepted answer does concatenate operators. Also, the question allows it.
    $endgroup$
    – haykam
    Sep 9 '18 at 13:59
















60












$begingroup$


Can you find a way to make:




$0 0 0 0 = 8$




by adding any operations or symbols? You can use only these symbols:




$+, -, *, !, /, hat, , ()$.




It is limited to this list, and concatenation is also allowed. You cannot add other numbers to the equation.










share|improve this question









$endgroup$





This question has an open bounty worth +100
reputation from user477343 ending in 6 days.


One or more of the answers is exemplary and worthy of an additional bounty.


This bounty will be awarded to @let_the_coding_begin's very first and very elegant answer — it has over 100 upvotes (including my own)! In my opinion, it is just as good the accepted answer, if not, better. So well cone, and congratulations! :D












  • 43




    $begingroup$
    Just put a slash over the equal sign!
    $endgroup$
    – Yout Ried
    Sep 7 '18 at 0:40






  • 13




    $begingroup$
    @YoutRied: standard loophole: neither funny nor creative. It's usually the first answer on any math puzzle.
    $endgroup$
    – Thomas Weller
    Sep 7 '18 at 18:40






  • 2




    $begingroup$
    Are we allowed to move anything, or is it insert only?
    $endgroup$
    – ctrl-alt-delor
    Sep 7 '18 at 19:33






  • 2




    $begingroup$
    Are you allowed to concatenate operators? (e.g !!, **, --, and ++)
    $endgroup$
    – Ole Tange
    Sep 9 '18 at 1:09






  • 1




    $begingroup$
    @OleTange The accepted answer does concatenate operators. Also, the question allows it.
    $endgroup$
    – haykam
    Sep 9 '18 at 13:59














60












60








60


7



$begingroup$


Can you find a way to make:




$0 0 0 0 = 8$




by adding any operations or symbols? You can use only these symbols:




$+, -, *, !, /, hat, , ()$.




It is limited to this list, and concatenation is also allowed. You cannot add other numbers to the equation.










share|improve this question









$endgroup$




Can you find a way to make:




$0 0 0 0 = 8$




by adding any operations or symbols? You can use only these symbols:




$+, -, *, !, /, hat, , ()$.




It is limited to this list, and concatenation is also allowed. You cannot add other numbers to the equation.







mathematics number-theory






share|improve this question













share|improve this question











share|improve this question




share|improve this question










asked Sep 7 '18 at 0:35









kraby15kraby15

2,4093930




2,4093930






This question has an open bounty worth +100
reputation from user477343 ending in 6 days.


One or more of the answers is exemplary and worthy of an additional bounty.


This bounty will be awarded to @let_the_coding_begin's very first and very elegant answer — it has over 100 upvotes (including my own)! In my opinion, it is just as good the accepted answer, if not, better. So well cone, and congratulations! :D








This question has an open bounty worth +100
reputation from user477343 ending in 6 days.


One or more of the answers is exemplary and worthy of an additional bounty.


This bounty will be awarded to @let_the_coding_begin's very first and very elegant answer — it has over 100 upvotes (including my own)! In my opinion, it is just as good the accepted answer, if not, better. So well cone, and congratulations! :D










  • 43




    $begingroup$
    Just put a slash over the equal sign!
    $endgroup$
    – Yout Ried
    Sep 7 '18 at 0:40






  • 13




    $begingroup$
    @YoutRied: standard loophole: neither funny nor creative. It's usually the first answer on any math puzzle.
    $endgroup$
    – Thomas Weller
    Sep 7 '18 at 18:40






  • 2




    $begingroup$
    Are we allowed to move anything, or is it insert only?
    $endgroup$
    – ctrl-alt-delor
    Sep 7 '18 at 19:33






  • 2




    $begingroup$
    Are you allowed to concatenate operators? (e.g !!, **, --, and ++)
    $endgroup$
    – Ole Tange
    Sep 9 '18 at 1:09






  • 1




    $begingroup$
    @OleTange The accepted answer does concatenate operators. Also, the question allows it.
    $endgroup$
    – haykam
    Sep 9 '18 at 13:59














  • 43




    $begingroup$
    Just put a slash over the equal sign!
    $endgroup$
    – Yout Ried
    Sep 7 '18 at 0:40






  • 13




    $begingroup$
    @YoutRied: standard loophole: neither funny nor creative. It's usually the first answer on any math puzzle.
    $endgroup$
    – Thomas Weller
    Sep 7 '18 at 18:40






  • 2




    $begingroup$
    Are we allowed to move anything, or is it insert only?
    $endgroup$
    – ctrl-alt-delor
    Sep 7 '18 at 19:33






  • 2




    $begingroup$
    Are you allowed to concatenate operators? (e.g !!, **, --, and ++)
    $endgroup$
    – Ole Tange
    Sep 9 '18 at 1:09






  • 1




    $begingroup$
    @OleTange The accepted answer does concatenate operators. Also, the question allows it.
    $endgroup$
    – haykam
    Sep 9 '18 at 13:59








43




43




$begingroup$
Just put a slash over the equal sign!
$endgroup$
– Yout Ried
Sep 7 '18 at 0:40




$begingroup$
Just put a slash over the equal sign!
$endgroup$
– Yout Ried
Sep 7 '18 at 0:40




13




13




$begingroup$
@YoutRied: standard loophole: neither funny nor creative. It's usually the first answer on any math puzzle.
$endgroup$
– Thomas Weller
Sep 7 '18 at 18:40




$begingroup$
@YoutRied: standard loophole: neither funny nor creative. It's usually the first answer on any math puzzle.
$endgroup$
– Thomas Weller
Sep 7 '18 at 18:40




2




2




$begingroup$
Are we allowed to move anything, or is it insert only?
$endgroup$
– ctrl-alt-delor
Sep 7 '18 at 19:33




$begingroup$
Are we allowed to move anything, or is it insert only?
$endgroup$
– ctrl-alt-delor
Sep 7 '18 at 19:33




2




2




$begingroup$
Are you allowed to concatenate operators? (e.g !!, **, --, and ++)
$endgroup$
– Ole Tange
Sep 9 '18 at 1:09




$begingroup$
Are you allowed to concatenate operators? (e.g !!, **, --, and ++)
$endgroup$
– Ole Tange
Sep 9 '18 at 1:09




1




1




$begingroup$
@OleTange The accepted answer does concatenate operators. Also, the question allows it.
$endgroup$
– haykam
Sep 9 '18 at 13:59




$begingroup$
@OleTange The accepted answer does concatenate operators. Also, the question allows it.
$endgroup$
– haykam
Sep 9 '18 at 13:59










25 Answers
25






active

oldest

votes


















102












$begingroup$

I think that




$left( 0! + 0! + 0! + 0! right)!! = 8$.




This is because




$0! = 1$ and $4!! = 8$. Note that $left( 0! + 0! + 0! + 0! right)!! = left( 1+1+1+1 right)!! = left (4 right)!! = 8$.




This works and is valid because




The question says I’m allowed to use any of the following symbols in my answer, I am not restricted to using $!$ as an operation only.







share|improve this answer











$endgroup$









  • 67




    $begingroup$
    for those who are confused, !! is a mathematical operator that gives the product of all positive integers upto the argument which have the same parity (odd/even) as the argument, its called a semifactorial. (thus 4!! is 2*4 = 8)
    $endgroup$
    – casualcoder
    Sep 7 '18 at 7:16








  • 13




    $begingroup$
    @casualcoder most people assume $n!!$ for an arbitrary non-negative integer $n$ is the same as $(n!)!$, but it's not. That brings up some confusion, so I usually write $n!_k$ to denote $n$ with $k$ factorials, but not on this site.
    $endgroup$
    – user477343
    Sep 7 '18 at 11:49








  • 4




    $begingroup$
    @casualcoder Google disagrees with Wolfram on this.
    $endgroup$
    – user1717828
    Sep 7 '18 at 14:46








  • 12




    $begingroup$
    Google disagrees with Wolfram on a lot of things. When it comes to math, chances are Wolfram is the correct one.
    $endgroup$
    – AlexanderJ93
    Sep 7 '18 at 23:03






  • 3




    $begingroup$
    @user1717828, user477343’s comment above explains why Google’s interpretation is not correct in this context.
    $endgroup$
    – El-Guest
    Sep 7 '18 at 23:09



















111












$begingroup$

A lateral thinking answer:




0! 0 0 0, because the binary equivalent of 8 is 1000 :)







share|improve this answer











$endgroup$









  • 9




    $begingroup$
    I like this! very direct and minimal.
    $endgroup$
    – Ruadhan2300
    Sep 7 '18 at 14:25






  • 2




    $begingroup$
    My favourite one! I did wonder if someone would go binary.
    $endgroup$
    – oliver-clare
    Sep 7 '18 at 15:07










  • $begingroup$
    I was just checking existing answers to see if anyone else had thought of ! In the sense of negation, and binary. So "yes" :)
    $endgroup$
    – Stilez
    Sep 10 '18 at 8:24












  • $begingroup$
    I have reached my daily voting limit (DVL), but Imma vote this when I can, get it to $99$, and then hope that another user actually puts another upvote :P
    $endgroup$
    – user477343
    Oct 3 '18 at 7:11





















73












$begingroup$


$0 + 0 + 0 + 0 ~~!!=~ 8$




because




$ !!= $ is an alternative way of writing $ ne $.







share|improve this answer









$endgroup$









  • 4




    $begingroup$
    This is the answer!
    $endgroup$
    – user51438
    Sep 8 '18 at 2:20






  • 4




    $begingroup$
    @TheodoreWeld nope, this question is tagged mathematics, and you have to use a slash similar to Yout Ried instead of this
    $endgroup$
    – phuclv
    Sep 8 '18 at 4:24





















58












$begingroup$

Lateral thinking!




$$0+0+substack{0\0}=0+0+8=8$$







share|improve this answer









$endgroup$









  • 9




    $begingroup$
    First I thought "how does picking 0 out of 0 give you 8 combinations", then I got it.
    $endgroup$
    – Fabian Röling
    Sep 7 '18 at 12:01






  • 86




    $begingroup$
    Looks more like vertical thinking to me.
    $endgroup$
    – Evargalo
    Sep 7 '18 at 12:02



















36












$begingroup$

let me try:




$0! Vert 0 - 0!-0! =8$

$10-1-1=8$


$Vert$ is a concatenation operation







share|improve this answer











$endgroup$









  • 1




    $begingroup$
    good solution but in this case you cant use '[' or ']'
    $endgroup$
    – casualcoder
    Sep 7 '18 at 7:13






  • 3




    $begingroup$
    @casualcoder It was explicitly allowed. If this wasn't meant to be a valid solution, why even allow it, as 00 = 0+0 = 0?
    $endgroup$
    – Zizy Archer
    Sep 7 '18 at 7:51






  • 1




    $begingroup$
    I've never used $[ldots ]$ as an operator before, let alone for what you use it as in this answer. I usually use it as a set notation; i.e., $[n]={1,2,ldots n}$ :P
    $endgroup$
    – user477343
    Sep 7 '18 at 11:53








  • 2




    $begingroup$
    I think "concatenation allowed" in these puzzles usually means that, for example, 2 8 7 can be 28 ÷ 7 = 4, but not (2+8)7 = 107
    $endgroup$
    – Chronocidal
    Sep 7 '18 at 14:22






  • 3




    $begingroup$
    @Chronocidal What you are saying is probably true about most of these puzzles. But in this particular puzzle where all the digits are 0, that interpretation does not make a lot of sense. To me this answer feels more legitimate than the accepted answer. That's because concatenation is explicitly allowed, but the !! is not explicitly allowed (though it may be implied).
    $endgroup$
    – kasperd
    Sep 8 '18 at 12:20



















30












$begingroup$


$((0!+0!)^{(0!+0!)})!!$




Evaluation:




$((0!+0!)^{(0!+0!)})!!$

$rightarrow ((1+1)^{(1+1)})!!$

$rightarrow (2^2)!!$

$rightarrow 4!! = 8$







share|improve this answer









$endgroup$













  • $begingroup$
    How did I not think of that?? DVL16 :
    $endgroup$
    – user477343
    Oct 3 '18 at 7:12



















29












$begingroup$

It's different:




$,++$

$0;;;0$

$,++$

$0;;;0$

$,++$


An ASCII art $8$ using only four $0$'s and $+$'s.







share|improve this answer









$endgroup$





















    18












    $begingroup$


    0 + 0 + 0 + 0 = !8




    because




    In C/C++, ! refers to the logical not operator, where all non-zero values become 0, and 0 becomes 1.







    share|improve this answer









    $endgroup$













    • $begingroup$
      I think this should be "the binary not operator".
      $endgroup$
      – Raimund Krämer
      Sep 11 '18 at 8:17






    • 3




      $begingroup$
      @RaimundKrämer ITYM unary not. C99 calls it the "logical negation operator" and files it in the chapter "Unary operators".
      $endgroup$
      – Jens
      Sep 11 '18 at 15:56










    • $begingroup$
      But !8 can be a subfactorial of 8.
      $endgroup$
      – rus9384
      Sep 12 '18 at 11:39










    • $begingroup$
      @Raimund Krämer, In programming, we have logical operators and bitwise operators. Logical operators (!, &&, || in C) operate on boolean values (true and false). Bitwise operators (~, &, |, ^ in C) operate on bits (0 and 1) of numbers. Calling either set binary operators would be confusing, as both set work on binary values.
      $endgroup$
      – ikegami
      Sep 13 '18 at 11:12












    • $begingroup$
      @Raimund Krämer Also, "binary operator" can mean any operator on 2 values (as opposed to a unary operator such as ! or ~)
      $endgroup$
      – l k
      Sep 18 '18 at 6:23



















    15












    $begingroup$

    It's just a matter of perspective ...




    0!/0 + 0!/0 = ∞




    My reasoning....




    0/0 is undefined so we have to first change the 0's into 1's with 0!

    (...and why did you write the infinity symbol sideways in your question?)







    share|improve this answer











    $endgroup$









    • 1




      $begingroup$
      x/0 isn't infinite, though...
      $endgroup$
      – Adam Smith
      Sep 8 '18 at 17:49






    • 1




      $begingroup$
      Lateral thinking was yesterday. Vertical thinking is the new kid in town!
      $endgroup$
      – Jens
      Sep 11 '18 at 15:59






    • 1




      $begingroup$
      x/0 == +infinity, per ieee 754
      $endgroup$
      – j__m
      Sep 15 '18 at 12:40



















    10












    $begingroup$


    $$[+!0]+[0]-!0-!0$$




    Works in JavaScript. Hit F12 and type this into the console (This equation editor uses "−" instead of "-" so copy and paste doesn't quite work). Otherwise, it works in the same way as @malioboro and @Arnaldur's answers.



    In fact, you can make any JavaScript application run just by using a combination of 6 characters, which is what inspired me to make this. I substituted + for 0 when asking JSF**k to do 10-2.






    share|improve this answer









    $endgroup$









    • 3




      $begingroup$
      I like this abuse of javascript a lot. !0 for true, +!0 for 1, [1] for array containing 1, [1]+[0] for concatenate two arrays to get the string "10" ; subtract 1 from that twice to end up as 8. Love it.
      $endgroup$
      – Ross Presser
      Sep 8 '18 at 4:10






    • 2




      $begingroup$
      but the question is tagged mathematics and not programming
      $endgroup$
      – phuclv
      Sep 8 '18 at 4:21








    • 3




      $begingroup$
      Who says you can use square brackets....
      $endgroup$
      – user52269
      Sep 8 '18 at 6:57



















    8












    $begingroup$


    $concat(0!,0) - 0! - 0! = 8$




    becomes:




    $concat(1,0) - 1 - 1 = 8$




    and finally:




    $10 - 2 = 8$




    cool puzzle!






    share|improve this answer









    $endgroup$









    • 2




      $begingroup$
      Great answer! But it has already been answered (check out @malioboro 's answer). Nonetheless, welcome to the Puzzling Stack Exchange (Puzzling.SE)! Since you are new to the Stack Exchange community, let alone this site, I strongly suggest that you visit the Help Center for more info; in particular, I suggest going here, then here for questions (not answers) :D
      $endgroup$
      – user477343
      Sep 7 '18 at 7:44





















    8












    $begingroup$


    $0 + 0 + 0 + 0 equiv 8$



    Adding the symbol $-$ over the equals sign makes it a congruence sign.
    Considering the congruence relation, we must be working mod N, where N divides 8.







    share|improve this answer









    $endgroup$





















      3












      $begingroup$


      Just put the minus symbol over the first zero to give it the look of a figure 8 and use plus to add the zeros.







      share|improve this answer









      $endgroup$













      • $begingroup$
        Great answer, but unfortunately, this is a duplicate of @TheSimpliFire 's answer.
        $endgroup$
        – user477343
        Sep 7 '18 at 11:51








      • 2




        $begingroup$
        @user477343 No, not quite, since this answer puts a line through the 0 instead of putting two 0s on top of each other. In this answer, only one 0 becomes an 8.
        $endgroup$
        – Fabian Röling
        Sep 7 '18 at 12:02










      • $begingroup$
        @FabianRöling As I read it, it uses the minus symbol to be put over the first zero and make it look like an $8$. In my imagination, I assumed the zero is the bottom circle of $8$ since there is something over it, of which is another circle $largecirc$. I did not fully understand what this answer was talking about in saying to use a subtraction symbol, but just assumed it was just like TheSimpliFire's answer due to their similarity; and since that is a great answer, I thought this was great, too. If you understand this answer, may you please explain to me in other words if you can? :
        $endgroup$
        – user477343
        Sep 7 '18 at 12:09








      • 3




        $begingroup$
        @user477343 It uses the horizontal line of the minus sign to cut through the middle of the $0$. It is "over" the zero in the sense of overwritten.
        $endgroup$
        – Jaap Scherphuis
        Sep 7 '18 at 12:55






      • 1




        $begingroup$
        Yeah. 'over' in the sense of "in front of" on a dimension between you and your monitor.
        $endgroup$
        – Sentinel
        Sep 7 '18 at 14:44



















      3












      $begingroup$

      Question limits the symbols, not the operations. So with the symbol + can make the operator ++.




      (++(++(++(++(++(++(++0000))))))) = 8







      share|improve this answer











      $endgroup$













      • $begingroup$
        I think you'd need the prefix form if you actually wanted that to work.
        $endgroup$
        – LegionMammal978
        Sep 7 '18 at 23:48






      • 1




        $begingroup$
        Doh. Of course you can't increment a literal in the first place.
        $endgroup$
        – David Browne - Microsoft
        Sep 7 '18 at 23:55






      • 1




        $begingroup$
        but that symbol doesn't exist in mathematics
        $endgroup$
        – phuclv
        Sep 8 '18 at 4:20






      • 2




        $begingroup$
        You need to pass an l-value to the ++ and -- operators, and neither their result or a literal is one.
        $endgroup$
        – NieDzejkob
        Sep 8 '18 at 14:35



















      3












      $begingroup$

      Here is an answer that doesn't use the semi-factorial or any concatenation.




      $$0 + 0 - 0! / 0 = (-8)!$$




      The left side is $-1/0$ and the right side is $-infty$.




      Plugging the expression into Wolfram.






      share|improve this answer









      $endgroup$









      • 4




        $begingroup$
        Division by zero is undefined (the limit of 0!/x as x goes to 0 is plus or minus infinity) and so are negative integer factorials (they are not equivalent to negative infinity). Even if they both evaluated to +∞, you can prove nonsense assuming you can equate infinities and treat like a real number. E.g., |1/0| = +∞, adding 1 to positive infinity is still positive infinity, hence |1/0| + 1 = +∞, hence |1/0| = |1/0| +1 by transitivity of equality. Subtract |1/0| from both sides and prove 0 = 1.
        $endgroup$
        – dr jimbob
        Sep 8 '18 at 18:11



















      3












      $begingroup$


      concatenate(0!, 0) - concatenate(0! + 0!) = 8.




      Note that 0! = 1




      (0!, 0) = 10, and (0! + 0!) = 2, so 10 - 2 = 8







      share|improve this answer











      $endgroup$













      • $begingroup$
        Unfortunately concatenate is not a valid operation.
        $endgroup$
        – boboquack
        Sep 11 '18 at 1:37










      • $begingroup$
        Congrats, concatenation is allowed now! Good answer, but people beat you to it. Welcome to Puzzling.SE! Here's a bonus puzzle: What's the word? Hilarious - extremely amusing
        $endgroup$
        – Alto
        Sep 11 '18 at 1:47










      • $begingroup$
        Also, concatenation is combination. Basically, (2, 4) = 24, you get it.
        $endgroup$
        – Alto
        Sep 11 '18 at 1:48



















      3












      $begingroup$

      This could work too:




      (0!+0+0)/0 = ∞




      Explanation




      (0!+0+0)/0 = 1/0 which is infinity (8 but put horizontally)







      share|improve this answer









      $endgroup$













      • $begingroup$
        That is not infinity — it is undefined :
        $endgroup$
        – user477343
        Oct 3 '18 at 11:33





















      3












      $begingroup$


      Add a - above the equals to get $0000 equiv 8$, which is true assuming we are working in the ring $mathbb{Z}/mathbb{2Z}$. (Note I'm trying to avoid writing $[0] = [8]$...)







      share|improve this answer











      $endgroup$





















        0












        $begingroup$


        $00^{00} = 0^0$ which is indeterminate. In some sense, an indeterminate form can be equal to any value, since in Calculus, a function that approaches "$0^0$" can approach any real value, including $8$. So in that sense, $0^0 = 8$.


        If you don't like concatenating two zeroes as "$00$", then $0^0 + 0 + 0$ also works.







        share|improve this answer











        $endgroup$









        • 8




          $begingroup$
          There are no widely accepted definitions under which your equation is considered to be true.
          $endgroup$
          – Tanner Swett
          Sep 7 '18 at 17:34










        • $begingroup$
          @TannerSwett Didn't think it was that unreasonable. My thinking is you can construct a function $f(x)^{g(x)}$ where $f$ and $g$ both limit to zero at some point $c$ such that $f(x)^{g(x)}$ limits to any real number at $c$, including $8$. Admittedly, writing $0^0 = 8$ is basically mathematically false. It is indeterminate for the reason I stated above. But this is Puzzling SE, not Mathematics SE, so I figured it was ok to think outside the box and not be formal with math.
          $endgroup$
          – RothX
          Sep 10 '18 at 13:23












        • $begingroup$
          Yeah, I dunno if I'd call it "that unreasonable". But if I saw a student say that the equation $0^0 = 8$ is true, I'd think that they probably misunderstand what an indeterminate form means. If I saw a math professor say the same thing, I'd ask them if they could please be a little more rigorous.
          $endgroup$
          – Tanner Swett
          Sep 10 '18 at 13:59










        • $begingroup$
          @TannerSwett Yeah, I completely agree. I'd never write $0^0 = 8$ in a mathematical setting, but since this is a puzzle, not a math problem, I thought maybe I could stretch a bit.
          $endgroup$
          – RothX
          Sep 11 '18 at 1:58



















        0












        $begingroup$


        $ ((0! + 0!)$)*(0! + 0!) = 2^2*2 = 8 $.




        Further explanation:




        The $$$ operation denotes the superfactorial defined as : $ n$ = (n!) uparrow uparrow (n!)$.







        share|improve this answer











        $endgroup$









        • 3




          $begingroup$
          Good idea, but make sure to read the question fully. It gives you a list of what operations can be used and $ is not one of them.
          $endgroup$
          – Sensoray
          Sep 7 '18 at 15:26



















        0












        $begingroup$

        If you turn the problem around




        enter image description here







        share|improve this answer









        $endgroup$













        • $begingroup$
          Similar to @rrauenza's answer
          $endgroup$
          – TheSimpliFire
          Sep 8 '18 at 8:11










        • $begingroup$
          Yes. A different way to express the same idea.
          $endgroup$
          – Florian F
          Sep 8 '18 at 9:13



















        0












        $begingroup$

        Similar to @Vaelus




        $0+0+0+0 leq 8$




        Explanation




        You can get the $leq$ by adding a $-$ inclined on top of the $=$







        share|improve this answer









        $endgroup$













        • $begingroup$
          The goal is to strictly make $0,0,0,0=8$ and not to "make the statement true" (therefore allowing this answer to be acceptable) :
          $endgroup$
          – user477343
          Oct 3 '18 at 11:35












        • $begingroup$
          When you say "the goal is to make [math expression]", then you're under the influence of the interpretation of that mathematical expression. IMO, "make the statement true" is the one and only interpretation to be taken, while you could argue more strict rules about not manipulating already existing symbols.
          $endgroup$
          – villasv
          Oct 3 '18 at 13:18












        • $begingroup$
          I see, you have a point. I should've said that my comment is solely based on how I interpreted the question, possibly explaining why your answer was sadly downvoted. I am positive that there exists another answer that manipulates the equation in a similar way that you have. Sorry if I sounded mean. You can get a $(+1)$ but in at least $2$ hours, once my daily voting limit (DVL) is over :)
          $endgroup$
          – user477343
          Oct 3 '18 at 21:00





















        0












        $begingroup$


        As per the list of allowed symbols we are clearly allowed to use "$,$" and "$.$"

        This is doubly evident as otherwise how would we use the $mathbb{concatenation}$ function without a comma to separate the arguments?




        So the solution is:




        $0! - mathbb{concatenation}(., 0 + 0! + 0!!) = .8$

        $1 - mathbb{concatenation}(., 0 + 1 + 1) = .8$

        $1 - mathbb{concatenation}(., 2) = .8$

        $1 - .2 = .8$







        share|improve this answer









        $endgroup$













        • $begingroup$
          You may simplify 0!! as 0!.
          $endgroup$
          – Cœur
          Sep 11 '18 at 15:28










        • $begingroup$
          I felt like $0+0!+0!!$ looked more elegant/purposeful as if it was part of a sequence like $a(n, m) = a(n-1,m)!; a(0,m)=m$
          $endgroup$
          – SamYonnou
          Sep 11 '18 at 16:58



















        0












        $begingroup$


        If the order can be changed:



        (0 0! 8)0 = 0!



        (0 0! 8) is a permutation.







        share|improve this answer








        New contributor




        Artem Lugin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
        Check out our Code of Conduct.






        $endgroup$









        • 2




          $begingroup$
          $0 * 0 * 0 * 8 = 0$ would be much simpler, though no more a solution to this problem than yours. I don't think it's reasonable to think reordering is allowed.
          $endgroup$
          – Rubio
          3 hours ago










        • $begingroup$
          nice try, but seems that this is not the kind of answer the OP was looking for. beware of downvotes, yet keep up your hard work and happy puzzling :)
          $endgroup$
          – Omega Krypton
          1 hour ago





















        0












        $begingroup$


        0 0 0 0 = 8

        0 0 0 0 = 8!

        0 0 0 0 = 40320

        0 0 0 0 = 4! * 0! * 3! * 2! * 0!

        0 0 0 0 = 288

        0 0 0 0 = 2+8/8

        0! 0! 0! 0 = 2+1

        0! + 0! + 0! + 0 = 2+1

        1 + 1 + 1 + 0 = 2 + 1

        3 = 3




        Where's my prize money?



        Just kidding, this and I are awful.






        share|improve this answer








        New contributor




        Samy Bencherif is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
        Check out our Code of Conduct.






        $endgroup$













        • $begingroup$
          I don't see how this focuses on answering the actual question, but... very cool! :D
          $endgroup$
          – user477343
          2 hours ago












        Your Answer





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        25 Answers
        25






        active

        oldest

        votes








        25 Answers
        25






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        102












        $begingroup$

        I think that




        $left( 0! + 0! + 0! + 0! right)!! = 8$.




        This is because




        $0! = 1$ and $4!! = 8$. Note that $left( 0! + 0! + 0! + 0! right)!! = left( 1+1+1+1 right)!! = left (4 right)!! = 8$.




        This works and is valid because




        The question says I’m allowed to use any of the following symbols in my answer, I am not restricted to using $!$ as an operation only.







        share|improve this answer











        $endgroup$









        • 67




          $begingroup$
          for those who are confused, !! is a mathematical operator that gives the product of all positive integers upto the argument which have the same parity (odd/even) as the argument, its called a semifactorial. (thus 4!! is 2*4 = 8)
          $endgroup$
          – casualcoder
          Sep 7 '18 at 7:16








        • 13




          $begingroup$
          @casualcoder most people assume $n!!$ for an arbitrary non-negative integer $n$ is the same as $(n!)!$, but it's not. That brings up some confusion, so I usually write $n!_k$ to denote $n$ with $k$ factorials, but not on this site.
          $endgroup$
          – user477343
          Sep 7 '18 at 11:49








        • 4




          $begingroup$
          @casualcoder Google disagrees with Wolfram on this.
          $endgroup$
          – user1717828
          Sep 7 '18 at 14:46








        • 12




          $begingroup$
          Google disagrees with Wolfram on a lot of things. When it comes to math, chances are Wolfram is the correct one.
          $endgroup$
          – AlexanderJ93
          Sep 7 '18 at 23:03






        • 3




          $begingroup$
          @user1717828, user477343’s comment above explains why Google’s interpretation is not correct in this context.
          $endgroup$
          – El-Guest
          Sep 7 '18 at 23:09
















        102












        $begingroup$

        I think that




        $left( 0! + 0! + 0! + 0! right)!! = 8$.




        This is because




        $0! = 1$ and $4!! = 8$. Note that $left( 0! + 0! + 0! + 0! right)!! = left( 1+1+1+1 right)!! = left (4 right)!! = 8$.




        This works and is valid because




        The question says I’m allowed to use any of the following symbols in my answer, I am not restricted to using $!$ as an operation only.







        share|improve this answer











        $endgroup$









        • 67




          $begingroup$
          for those who are confused, !! is a mathematical operator that gives the product of all positive integers upto the argument which have the same parity (odd/even) as the argument, its called a semifactorial. (thus 4!! is 2*4 = 8)
          $endgroup$
          – casualcoder
          Sep 7 '18 at 7:16








        • 13




          $begingroup$
          @casualcoder most people assume $n!!$ for an arbitrary non-negative integer $n$ is the same as $(n!)!$, but it's not. That brings up some confusion, so I usually write $n!_k$ to denote $n$ with $k$ factorials, but not on this site.
          $endgroup$
          – user477343
          Sep 7 '18 at 11:49








        • 4




          $begingroup$
          @casualcoder Google disagrees with Wolfram on this.
          $endgroup$
          – user1717828
          Sep 7 '18 at 14:46








        • 12




          $begingroup$
          Google disagrees with Wolfram on a lot of things. When it comes to math, chances are Wolfram is the correct one.
          $endgroup$
          – AlexanderJ93
          Sep 7 '18 at 23:03






        • 3




          $begingroup$
          @user1717828, user477343’s comment above explains why Google’s interpretation is not correct in this context.
          $endgroup$
          – El-Guest
          Sep 7 '18 at 23:09














        102












        102








        102





        $begingroup$

        I think that




        $left( 0! + 0! + 0! + 0! right)!! = 8$.




        This is because




        $0! = 1$ and $4!! = 8$. Note that $left( 0! + 0! + 0! + 0! right)!! = left( 1+1+1+1 right)!! = left (4 right)!! = 8$.




        This works and is valid because




        The question says I’m allowed to use any of the following symbols in my answer, I am not restricted to using $!$ as an operation only.







        share|improve this answer











        $endgroup$



        I think that




        $left( 0! + 0! + 0! + 0! right)!! = 8$.




        This is because




        $0! = 1$ and $4!! = 8$. Note that $left( 0! + 0! + 0! + 0! right)!! = left( 1+1+1+1 right)!! = left (4 right)!! = 8$.




        This works and is valid because




        The question says I’m allowed to use any of the following symbols in my answer, I am not restricted to using $!$ as an operation only.








        share|improve this answer














        share|improve this answer



        share|improve this answer








        edited Sep 7 '18 at 6:42









        Laurel

        818310




        818310










        answered Sep 7 '18 at 0:40









        El-GuestEl-Guest

        21.7k35092




        21.7k35092








        • 67




          $begingroup$
          for those who are confused, !! is a mathematical operator that gives the product of all positive integers upto the argument which have the same parity (odd/even) as the argument, its called a semifactorial. (thus 4!! is 2*4 = 8)
          $endgroup$
          – casualcoder
          Sep 7 '18 at 7:16








        • 13




          $begingroup$
          @casualcoder most people assume $n!!$ for an arbitrary non-negative integer $n$ is the same as $(n!)!$, but it's not. That brings up some confusion, so I usually write $n!_k$ to denote $n$ with $k$ factorials, but not on this site.
          $endgroup$
          – user477343
          Sep 7 '18 at 11:49








        • 4




          $begingroup$
          @casualcoder Google disagrees with Wolfram on this.
          $endgroup$
          – user1717828
          Sep 7 '18 at 14:46








        • 12




          $begingroup$
          Google disagrees with Wolfram on a lot of things. When it comes to math, chances are Wolfram is the correct one.
          $endgroup$
          – AlexanderJ93
          Sep 7 '18 at 23:03






        • 3




          $begingroup$
          @user1717828, user477343’s comment above explains why Google’s interpretation is not correct in this context.
          $endgroup$
          – El-Guest
          Sep 7 '18 at 23:09














        • 67




          $begingroup$
          for those who are confused, !! is a mathematical operator that gives the product of all positive integers upto the argument which have the same parity (odd/even) as the argument, its called a semifactorial. (thus 4!! is 2*4 = 8)
          $endgroup$
          – casualcoder
          Sep 7 '18 at 7:16








        • 13




          $begingroup$
          @casualcoder most people assume $n!!$ for an arbitrary non-negative integer $n$ is the same as $(n!)!$, but it's not. That brings up some confusion, so I usually write $n!_k$ to denote $n$ with $k$ factorials, but not on this site.
          $endgroup$
          – user477343
          Sep 7 '18 at 11:49








        • 4




          $begingroup$
          @casualcoder Google disagrees with Wolfram on this.
          $endgroup$
          – user1717828
          Sep 7 '18 at 14:46








        • 12




          $begingroup$
          Google disagrees with Wolfram on a lot of things. When it comes to math, chances are Wolfram is the correct one.
          $endgroup$
          – AlexanderJ93
          Sep 7 '18 at 23:03






        • 3




          $begingroup$
          @user1717828, user477343’s comment above explains why Google’s interpretation is not correct in this context.
          $endgroup$
          – El-Guest
          Sep 7 '18 at 23:09








        67




        67




        $begingroup$
        for those who are confused, !! is a mathematical operator that gives the product of all positive integers upto the argument which have the same parity (odd/even) as the argument, its called a semifactorial. (thus 4!! is 2*4 = 8)
        $endgroup$
        – casualcoder
        Sep 7 '18 at 7:16






        $begingroup$
        for those who are confused, !! is a mathematical operator that gives the product of all positive integers upto the argument which have the same parity (odd/even) as the argument, its called a semifactorial. (thus 4!! is 2*4 = 8)
        $endgroup$
        – casualcoder
        Sep 7 '18 at 7:16






        13




        13




        $begingroup$
        @casualcoder most people assume $n!!$ for an arbitrary non-negative integer $n$ is the same as $(n!)!$, but it's not. That brings up some confusion, so I usually write $n!_k$ to denote $n$ with $k$ factorials, but not on this site.
        $endgroup$
        – user477343
        Sep 7 '18 at 11:49






        $begingroup$
        @casualcoder most people assume $n!!$ for an arbitrary non-negative integer $n$ is the same as $(n!)!$, but it's not. That brings up some confusion, so I usually write $n!_k$ to denote $n$ with $k$ factorials, but not on this site.
        $endgroup$
        – user477343
        Sep 7 '18 at 11:49






        4




        4




        $begingroup$
        @casualcoder Google disagrees with Wolfram on this.
        $endgroup$
        – user1717828
        Sep 7 '18 at 14:46






        $begingroup$
        @casualcoder Google disagrees with Wolfram on this.
        $endgroup$
        – user1717828
        Sep 7 '18 at 14:46






        12




        12




        $begingroup$
        Google disagrees with Wolfram on a lot of things. When it comes to math, chances are Wolfram is the correct one.
        $endgroup$
        – AlexanderJ93
        Sep 7 '18 at 23:03




        $begingroup$
        Google disagrees with Wolfram on a lot of things. When it comes to math, chances are Wolfram is the correct one.
        $endgroup$
        – AlexanderJ93
        Sep 7 '18 at 23:03




        3




        3




        $begingroup$
        @user1717828, user477343’s comment above explains why Google’s interpretation is not correct in this context.
        $endgroup$
        – El-Guest
        Sep 7 '18 at 23:09




        $begingroup$
        @user1717828, user477343’s comment above explains why Google’s interpretation is not correct in this context.
        $endgroup$
        – El-Guest
        Sep 7 '18 at 23:09











        111












        $begingroup$

        A lateral thinking answer:




        0! 0 0 0, because the binary equivalent of 8 is 1000 :)







        share|improve this answer











        $endgroup$









        • 9




          $begingroup$
          I like this! very direct and minimal.
          $endgroup$
          – Ruadhan2300
          Sep 7 '18 at 14:25






        • 2




          $begingroup$
          My favourite one! I did wonder if someone would go binary.
          $endgroup$
          – oliver-clare
          Sep 7 '18 at 15:07










        • $begingroup$
          I was just checking existing answers to see if anyone else had thought of ! In the sense of negation, and binary. So "yes" :)
          $endgroup$
          – Stilez
          Sep 10 '18 at 8:24












        • $begingroup$
          I have reached my daily voting limit (DVL), but Imma vote this when I can, get it to $99$, and then hope that another user actually puts another upvote :P
          $endgroup$
          – user477343
          Oct 3 '18 at 7:11


















        111












        $begingroup$

        A lateral thinking answer:




        0! 0 0 0, because the binary equivalent of 8 is 1000 :)







        share|improve this answer











        $endgroup$









        • 9




          $begingroup$
          I like this! very direct and minimal.
          $endgroup$
          – Ruadhan2300
          Sep 7 '18 at 14:25






        • 2




          $begingroup$
          My favourite one! I did wonder if someone would go binary.
          $endgroup$
          – oliver-clare
          Sep 7 '18 at 15:07










        • $begingroup$
          I was just checking existing answers to see if anyone else had thought of ! In the sense of negation, and binary. So "yes" :)
          $endgroup$
          – Stilez
          Sep 10 '18 at 8:24












        • $begingroup$
          I have reached my daily voting limit (DVL), but Imma vote this when I can, get it to $99$, and then hope that another user actually puts another upvote :P
          $endgroup$
          – user477343
          Oct 3 '18 at 7:11
















        111












        111








        111





        $begingroup$

        A lateral thinking answer:




        0! 0 0 0, because the binary equivalent of 8 is 1000 :)







        share|improve this answer











        $endgroup$



        A lateral thinking answer:




        0! 0 0 0, because the binary equivalent of 8 is 1000 :)








        share|improve this answer














        share|improve this answer



        share|improve this answer








        edited Sep 7 '18 at 14:01









        El-Guest

        21.7k35092




        21.7k35092










        answered Sep 7 '18 at 13:44









        let_the_coding_beginlet_the_coding_begin

        1,141114




        1,141114








        • 9




          $begingroup$
          I like this! very direct and minimal.
          $endgroup$
          – Ruadhan2300
          Sep 7 '18 at 14:25






        • 2




          $begingroup$
          My favourite one! I did wonder if someone would go binary.
          $endgroup$
          – oliver-clare
          Sep 7 '18 at 15:07










        • $begingroup$
          I was just checking existing answers to see if anyone else had thought of ! In the sense of negation, and binary. So "yes" :)
          $endgroup$
          – Stilez
          Sep 10 '18 at 8:24












        • $begingroup$
          I have reached my daily voting limit (DVL), but Imma vote this when I can, get it to $99$, and then hope that another user actually puts another upvote :P
          $endgroup$
          – user477343
          Oct 3 '18 at 7:11
















        • 9




          $begingroup$
          I like this! very direct and minimal.
          $endgroup$
          – Ruadhan2300
          Sep 7 '18 at 14:25






        • 2




          $begingroup$
          My favourite one! I did wonder if someone would go binary.
          $endgroup$
          – oliver-clare
          Sep 7 '18 at 15:07










        • $begingroup$
          I was just checking existing answers to see if anyone else had thought of ! In the sense of negation, and binary. So "yes" :)
          $endgroup$
          – Stilez
          Sep 10 '18 at 8:24












        • $begingroup$
          I have reached my daily voting limit (DVL), but Imma vote this when I can, get it to $99$, and then hope that another user actually puts another upvote :P
          $endgroup$
          – user477343
          Oct 3 '18 at 7:11










        9




        9




        $begingroup$
        I like this! very direct and minimal.
        $endgroup$
        – Ruadhan2300
        Sep 7 '18 at 14:25




        $begingroup$
        I like this! very direct and minimal.
        $endgroup$
        – Ruadhan2300
        Sep 7 '18 at 14:25




        2




        2




        $begingroup$
        My favourite one! I did wonder if someone would go binary.
        $endgroup$
        – oliver-clare
        Sep 7 '18 at 15:07




        $begingroup$
        My favourite one! I did wonder if someone would go binary.
        $endgroup$
        – oliver-clare
        Sep 7 '18 at 15:07












        $begingroup$
        I was just checking existing answers to see if anyone else had thought of ! In the sense of negation, and binary. So "yes" :)
        $endgroup$
        – Stilez
        Sep 10 '18 at 8:24






        $begingroup$
        I was just checking existing answers to see if anyone else had thought of ! In the sense of negation, and binary. So "yes" :)
        $endgroup$
        – Stilez
        Sep 10 '18 at 8:24














        $begingroup$
        I have reached my daily voting limit (DVL), but Imma vote this when I can, get it to $99$, and then hope that another user actually puts another upvote :P
        $endgroup$
        – user477343
        Oct 3 '18 at 7:11






        $begingroup$
        I have reached my daily voting limit (DVL), but Imma vote this when I can, get it to $99$, and then hope that another user actually puts another upvote :P
        $endgroup$
        – user477343
        Oct 3 '18 at 7:11













        73












        $begingroup$


        $0 + 0 + 0 + 0 ~~!!=~ 8$




        because




        $ !!= $ is an alternative way of writing $ ne $.







        share|improve this answer









        $endgroup$









        • 4




          $begingroup$
          This is the answer!
          $endgroup$
          – user51438
          Sep 8 '18 at 2:20






        • 4




          $begingroup$
          @TheodoreWeld nope, this question is tagged mathematics, and you have to use a slash similar to Yout Ried instead of this
          $endgroup$
          – phuclv
          Sep 8 '18 at 4:24


















        73












        $begingroup$


        $0 + 0 + 0 + 0 ~~!!=~ 8$




        because




        $ !!= $ is an alternative way of writing $ ne $.







        share|improve this answer









        $endgroup$









        • 4




          $begingroup$
          This is the answer!
          $endgroup$
          – user51438
          Sep 8 '18 at 2:20






        • 4




          $begingroup$
          @TheodoreWeld nope, this question is tagged mathematics, and you have to use a slash similar to Yout Ried instead of this
          $endgroup$
          – phuclv
          Sep 8 '18 at 4:24
















        73












        73








        73





        $begingroup$


        $0 + 0 + 0 + 0 ~~!!=~ 8$




        because




        $ !!= $ is an alternative way of writing $ ne $.







        share|improve this answer









        $endgroup$




        $0 + 0 + 0 + 0 ~~!!=~ 8$




        because




        $ !!= $ is an alternative way of writing $ ne $.








        share|improve this answer












        share|improve this answer



        share|improve this answer










        answered Sep 7 '18 at 8:43









        Teemu PiippoTeemu Piippo

        71724




        71724








        • 4




          $begingroup$
          This is the answer!
          $endgroup$
          – user51438
          Sep 8 '18 at 2:20






        • 4




          $begingroup$
          @TheodoreWeld nope, this question is tagged mathematics, and you have to use a slash similar to Yout Ried instead of this
          $endgroup$
          – phuclv
          Sep 8 '18 at 4:24
















        • 4




          $begingroup$
          This is the answer!
          $endgroup$
          – user51438
          Sep 8 '18 at 2:20






        • 4




          $begingroup$
          @TheodoreWeld nope, this question is tagged mathematics, and you have to use a slash similar to Yout Ried instead of this
          $endgroup$
          – phuclv
          Sep 8 '18 at 4:24










        4




        4




        $begingroup$
        This is the answer!
        $endgroup$
        – user51438
        Sep 8 '18 at 2:20




        $begingroup$
        This is the answer!
        $endgroup$
        – user51438
        Sep 8 '18 at 2:20




        4




        4




        $begingroup$
        @TheodoreWeld nope, this question is tagged mathematics, and you have to use a slash similar to Yout Ried instead of this
        $endgroup$
        – phuclv
        Sep 8 '18 at 4:24






        $begingroup$
        @TheodoreWeld nope, this question is tagged mathematics, and you have to use a slash similar to Yout Ried instead of this
        $endgroup$
        – phuclv
        Sep 8 '18 at 4:24













        58












        $begingroup$

        Lateral thinking!




        $$0+0+substack{0\0}=0+0+8=8$$







        share|improve this answer









        $endgroup$









        • 9




          $begingroup$
          First I thought "how does picking 0 out of 0 give you 8 combinations", then I got it.
          $endgroup$
          – Fabian Röling
          Sep 7 '18 at 12:01






        • 86




          $begingroup$
          Looks more like vertical thinking to me.
          $endgroup$
          – Evargalo
          Sep 7 '18 at 12:02
















        58












        $begingroup$

        Lateral thinking!




        $$0+0+substack{0\0}=0+0+8=8$$







        share|improve this answer









        $endgroup$









        • 9




          $begingroup$
          First I thought "how does picking 0 out of 0 give you 8 combinations", then I got it.
          $endgroup$
          – Fabian Röling
          Sep 7 '18 at 12:01






        • 86




          $begingroup$
          Looks more like vertical thinking to me.
          $endgroup$
          – Evargalo
          Sep 7 '18 at 12:02














        58












        58








        58





        $begingroup$

        Lateral thinking!




        $$0+0+substack{0\0}=0+0+8=8$$







        share|improve this answer









        $endgroup$



        Lateral thinking!




        $$0+0+substack{0\0}=0+0+8=8$$








        share|improve this answer












        share|improve this answer



        share|improve this answer










        answered Sep 7 '18 at 6:25









        TheSimpliFireTheSimpliFire

        2,215532




        2,215532








        • 9




          $begingroup$
          First I thought "how does picking 0 out of 0 give you 8 combinations", then I got it.
          $endgroup$
          – Fabian Röling
          Sep 7 '18 at 12:01






        • 86




          $begingroup$
          Looks more like vertical thinking to me.
          $endgroup$
          – Evargalo
          Sep 7 '18 at 12:02














        • 9




          $begingroup$
          First I thought "how does picking 0 out of 0 give you 8 combinations", then I got it.
          $endgroup$
          – Fabian Röling
          Sep 7 '18 at 12:01






        • 86




          $begingroup$
          Looks more like vertical thinking to me.
          $endgroup$
          – Evargalo
          Sep 7 '18 at 12:02








        9




        9




        $begingroup$
        First I thought "how does picking 0 out of 0 give you 8 combinations", then I got it.
        $endgroup$
        – Fabian Röling
        Sep 7 '18 at 12:01




        $begingroup$
        First I thought "how does picking 0 out of 0 give you 8 combinations", then I got it.
        $endgroup$
        – Fabian Röling
        Sep 7 '18 at 12:01




        86




        86




        $begingroup$
        Looks more like vertical thinking to me.
        $endgroup$
        – Evargalo
        Sep 7 '18 at 12:02




        $begingroup$
        Looks more like vertical thinking to me.
        $endgroup$
        – Evargalo
        Sep 7 '18 at 12:02











        36












        $begingroup$

        let me try:




        $0! Vert 0 - 0!-0! =8$

        $10-1-1=8$


        $Vert$ is a concatenation operation







        share|improve this answer











        $endgroup$









        • 1




          $begingroup$
          good solution but in this case you cant use '[' or ']'
          $endgroup$
          – casualcoder
          Sep 7 '18 at 7:13






        • 3




          $begingroup$
          @casualcoder It was explicitly allowed. If this wasn't meant to be a valid solution, why even allow it, as 00 = 0+0 = 0?
          $endgroup$
          – Zizy Archer
          Sep 7 '18 at 7:51






        • 1




          $begingroup$
          I've never used $[ldots ]$ as an operator before, let alone for what you use it as in this answer. I usually use it as a set notation; i.e., $[n]={1,2,ldots n}$ :P
          $endgroup$
          – user477343
          Sep 7 '18 at 11:53








        • 2




          $begingroup$
          I think "concatenation allowed" in these puzzles usually means that, for example, 2 8 7 can be 28 ÷ 7 = 4, but not (2+8)7 = 107
          $endgroup$
          – Chronocidal
          Sep 7 '18 at 14:22






        • 3




          $begingroup$
          @Chronocidal What you are saying is probably true about most of these puzzles. But in this particular puzzle where all the digits are 0, that interpretation does not make a lot of sense. To me this answer feels more legitimate than the accepted answer. That's because concatenation is explicitly allowed, but the !! is not explicitly allowed (though it may be implied).
          $endgroup$
          – kasperd
          Sep 8 '18 at 12:20
















        36












        $begingroup$

        let me try:




        $0! Vert 0 - 0!-0! =8$

        $10-1-1=8$


        $Vert$ is a concatenation operation







        share|improve this answer











        $endgroup$









        • 1




          $begingroup$
          good solution but in this case you cant use '[' or ']'
          $endgroup$
          – casualcoder
          Sep 7 '18 at 7:13






        • 3




          $begingroup$
          @casualcoder It was explicitly allowed. If this wasn't meant to be a valid solution, why even allow it, as 00 = 0+0 = 0?
          $endgroup$
          – Zizy Archer
          Sep 7 '18 at 7:51






        • 1




          $begingroup$
          I've never used $[ldots ]$ as an operator before, let alone for what you use it as in this answer. I usually use it as a set notation; i.e., $[n]={1,2,ldots n}$ :P
          $endgroup$
          – user477343
          Sep 7 '18 at 11:53








        • 2




          $begingroup$
          I think "concatenation allowed" in these puzzles usually means that, for example, 2 8 7 can be 28 ÷ 7 = 4, but not (2+8)7 = 107
          $endgroup$
          – Chronocidal
          Sep 7 '18 at 14:22






        • 3




          $begingroup$
          @Chronocidal What you are saying is probably true about most of these puzzles. But in this particular puzzle where all the digits are 0, that interpretation does not make a lot of sense. To me this answer feels more legitimate than the accepted answer. That's because concatenation is explicitly allowed, but the !! is not explicitly allowed (though it may be implied).
          $endgroup$
          – kasperd
          Sep 8 '18 at 12:20














        36












        36








        36





        $begingroup$

        let me try:




        $0! Vert 0 - 0!-0! =8$

        $10-1-1=8$


        $Vert$ is a concatenation operation







        share|improve this answer











        $endgroup$



        let me try:




        $0! Vert 0 - 0!-0! =8$

        $10-1-1=8$


        $Vert$ is a concatenation operation








        share|improve this answer














        share|improve this answer



        share|improve this answer








        edited Sep 8 '18 at 15:29

























        answered Sep 7 '18 at 0:43









        malioboromalioboro

        2,79311033




        2,79311033








        • 1




          $begingroup$
          good solution but in this case you cant use '[' or ']'
          $endgroup$
          – casualcoder
          Sep 7 '18 at 7:13






        • 3




          $begingroup$
          @casualcoder It was explicitly allowed. If this wasn't meant to be a valid solution, why even allow it, as 00 = 0+0 = 0?
          $endgroup$
          – Zizy Archer
          Sep 7 '18 at 7:51






        • 1




          $begingroup$
          I've never used $[ldots ]$ as an operator before, let alone for what you use it as in this answer. I usually use it as a set notation; i.e., $[n]={1,2,ldots n}$ :P
          $endgroup$
          – user477343
          Sep 7 '18 at 11:53








        • 2




          $begingroup$
          I think "concatenation allowed" in these puzzles usually means that, for example, 2 8 7 can be 28 ÷ 7 = 4, but not (2+8)7 = 107
          $endgroup$
          – Chronocidal
          Sep 7 '18 at 14:22






        • 3




          $begingroup$
          @Chronocidal What you are saying is probably true about most of these puzzles. But in this particular puzzle where all the digits are 0, that interpretation does not make a lot of sense. To me this answer feels more legitimate than the accepted answer. That's because concatenation is explicitly allowed, but the !! is not explicitly allowed (though it may be implied).
          $endgroup$
          – kasperd
          Sep 8 '18 at 12:20














        • 1




          $begingroup$
          good solution but in this case you cant use '[' or ']'
          $endgroup$
          – casualcoder
          Sep 7 '18 at 7:13






        • 3




          $begingroup$
          @casualcoder It was explicitly allowed. If this wasn't meant to be a valid solution, why even allow it, as 00 = 0+0 = 0?
          $endgroup$
          – Zizy Archer
          Sep 7 '18 at 7:51






        • 1




          $begingroup$
          I've never used $[ldots ]$ as an operator before, let alone for what you use it as in this answer. I usually use it as a set notation; i.e., $[n]={1,2,ldots n}$ :P
          $endgroup$
          – user477343
          Sep 7 '18 at 11:53








        • 2




          $begingroup$
          I think "concatenation allowed" in these puzzles usually means that, for example, 2 8 7 can be 28 ÷ 7 = 4, but not (2+8)7 = 107
          $endgroup$
          – Chronocidal
          Sep 7 '18 at 14:22






        • 3




          $begingroup$
          @Chronocidal What you are saying is probably true about most of these puzzles. But in this particular puzzle where all the digits are 0, that interpretation does not make a lot of sense. To me this answer feels more legitimate than the accepted answer. That's because concatenation is explicitly allowed, but the !! is not explicitly allowed (though it may be implied).
          $endgroup$
          – kasperd
          Sep 8 '18 at 12:20








        1




        1




        $begingroup$
        good solution but in this case you cant use '[' or ']'
        $endgroup$
        – casualcoder
        Sep 7 '18 at 7:13




        $begingroup$
        good solution but in this case you cant use '[' or ']'
        $endgroup$
        – casualcoder
        Sep 7 '18 at 7:13




        3




        3




        $begingroup$
        @casualcoder It was explicitly allowed. If this wasn't meant to be a valid solution, why even allow it, as 00 = 0+0 = 0?
        $endgroup$
        – Zizy Archer
        Sep 7 '18 at 7:51




        $begingroup$
        @casualcoder It was explicitly allowed. If this wasn't meant to be a valid solution, why even allow it, as 00 = 0+0 = 0?
        $endgroup$
        – Zizy Archer
        Sep 7 '18 at 7:51




        1




        1




        $begingroup$
        I've never used $[ldots ]$ as an operator before, let alone for what you use it as in this answer. I usually use it as a set notation; i.e., $[n]={1,2,ldots n}$ :P
        $endgroup$
        – user477343
        Sep 7 '18 at 11:53






        $begingroup$
        I've never used $[ldots ]$ as an operator before, let alone for what you use it as in this answer. I usually use it as a set notation; i.e., $[n]={1,2,ldots n}$ :P
        $endgroup$
        – user477343
        Sep 7 '18 at 11:53






        2




        2




        $begingroup$
        I think "concatenation allowed" in these puzzles usually means that, for example, 2 8 7 can be 28 ÷ 7 = 4, but not (2+8)7 = 107
        $endgroup$
        – Chronocidal
        Sep 7 '18 at 14:22




        $begingroup$
        I think "concatenation allowed" in these puzzles usually means that, for example, 2 8 7 can be 28 ÷ 7 = 4, but not (2+8)7 = 107
        $endgroup$
        – Chronocidal
        Sep 7 '18 at 14:22




        3




        3




        $begingroup$
        @Chronocidal What you are saying is probably true about most of these puzzles. But in this particular puzzle where all the digits are 0, that interpretation does not make a lot of sense. To me this answer feels more legitimate than the accepted answer. That's because concatenation is explicitly allowed, but the !! is not explicitly allowed (though it may be implied).
        $endgroup$
        – kasperd
        Sep 8 '18 at 12:20




        $begingroup$
        @Chronocidal What you are saying is probably true about most of these puzzles. But in this particular puzzle where all the digits are 0, that interpretation does not make a lot of sense. To me this answer feels more legitimate than the accepted answer. That's because concatenation is explicitly allowed, but the !! is not explicitly allowed (though it may be implied).
        $endgroup$
        – kasperd
        Sep 8 '18 at 12:20











        30












        $begingroup$


        $((0!+0!)^{(0!+0!)})!!$




        Evaluation:




        $((0!+0!)^{(0!+0!)})!!$

        $rightarrow ((1+1)^{(1+1)})!!$

        $rightarrow (2^2)!!$

        $rightarrow 4!! = 8$







        share|improve this answer









        $endgroup$













        • $begingroup$
          How did I not think of that?? DVL16 :
          $endgroup$
          – user477343
          Oct 3 '18 at 7:12
















        30












        $begingroup$


        $((0!+0!)^{(0!+0!)})!!$




        Evaluation:




        $((0!+0!)^{(0!+0!)})!!$

        $rightarrow ((1+1)^{(1+1)})!!$

        $rightarrow (2^2)!!$

        $rightarrow 4!! = 8$







        share|improve this answer









        $endgroup$













        • $begingroup$
          How did I not think of that?? DVL16 :
          $endgroup$
          – user477343
          Oct 3 '18 at 7:12














        30












        30








        30





        $begingroup$


        $((0!+0!)^{(0!+0!)})!!$




        Evaluation:




        $((0!+0!)^{(0!+0!)})!!$

        $rightarrow ((1+1)^{(1+1)})!!$

        $rightarrow (2^2)!!$

        $rightarrow 4!! = 8$







        share|improve this answer









        $endgroup$




        $((0!+0!)^{(0!+0!)})!!$




        Evaluation:




        $((0!+0!)^{(0!+0!)})!!$

        $rightarrow ((1+1)^{(1+1)})!!$

        $rightarrow (2^2)!!$

        $rightarrow 4!! = 8$








        share|improve this answer












        share|improve this answer



        share|improve this answer










        answered Sep 7 '18 at 14:04









        u_ndefinedu_ndefined

        2,7901437




        2,7901437












        • $begingroup$
          How did I not think of that?? DVL16 :
          $endgroup$
          – user477343
          Oct 3 '18 at 7:12


















        • $begingroup$
          How did I not think of that?? DVL16 :
          $endgroup$
          – user477343
          Oct 3 '18 at 7:12
















        $begingroup$
        How did I not think of that?? DVL16 :
        $endgroup$
        – user477343
        Oct 3 '18 at 7:12




        $begingroup$
        How did I not think of that?? DVL16 :
        $endgroup$
        – user477343
        Oct 3 '18 at 7:12











        29












        $begingroup$

        It's different:




        $,++$

        $0;;;0$

        $,++$

        $0;;;0$

        $,++$


        An ASCII art $8$ using only four $0$'s and $+$'s.







        share|improve this answer









        $endgroup$


















          29












          $begingroup$

          It's different:




          $,++$

          $0;;;0$

          $,++$

          $0;;;0$

          $,++$


          An ASCII art $8$ using only four $0$'s and $+$'s.







          share|improve this answer









          $endgroup$
















            29












            29








            29





            $begingroup$

            It's different:




            $,++$

            $0;;;0$

            $,++$

            $0;;;0$

            $,++$


            An ASCII art $8$ using only four $0$'s and $+$'s.







            share|improve this answer









            $endgroup$



            It's different:




            $,++$

            $0;;;0$

            $,++$

            $0;;;0$

            $,++$


            An ASCII art $8$ using only four $0$'s and $+$'s.








            share|improve this answer












            share|improve this answer



            share|improve this answer










            answered Sep 7 '18 at 14:46









            JonMark PerryJonMark Perry

            20.7k64099




            20.7k64099























                18












                $begingroup$


                0 + 0 + 0 + 0 = !8




                because




                In C/C++, ! refers to the logical not operator, where all non-zero values become 0, and 0 becomes 1.







                share|improve this answer









                $endgroup$













                • $begingroup$
                  I think this should be "the binary not operator".
                  $endgroup$
                  – Raimund Krämer
                  Sep 11 '18 at 8:17






                • 3




                  $begingroup$
                  @RaimundKrämer ITYM unary not. C99 calls it the "logical negation operator" and files it in the chapter "Unary operators".
                  $endgroup$
                  – Jens
                  Sep 11 '18 at 15:56










                • $begingroup$
                  But !8 can be a subfactorial of 8.
                  $endgroup$
                  – rus9384
                  Sep 12 '18 at 11:39










                • $begingroup$
                  @Raimund Krämer, In programming, we have logical operators and bitwise operators. Logical operators (!, &&, || in C) operate on boolean values (true and false). Bitwise operators (~, &, |, ^ in C) operate on bits (0 and 1) of numbers. Calling either set binary operators would be confusing, as both set work on binary values.
                  $endgroup$
                  – ikegami
                  Sep 13 '18 at 11:12












                • $begingroup$
                  @Raimund Krämer Also, "binary operator" can mean any operator on 2 values (as opposed to a unary operator such as ! or ~)
                  $endgroup$
                  – l k
                  Sep 18 '18 at 6:23
















                18












                $begingroup$


                0 + 0 + 0 + 0 = !8




                because




                In C/C++, ! refers to the logical not operator, where all non-zero values become 0, and 0 becomes 1.







                share|improve this answer









                $endgroup$













                • $begingroup$
                  I think this should be "the binary not operator".
                  $endgroup$
                  – Raimund Krämer
                  Sep 11 '18 at 8:17






                • 3




                  $begingroup$
                  @RaimundKrämer ITYM unary not. C99 calls it the "logical negation operator" and files it in the chapter "Unary operators".
                  $endgroup$
                  – Jens
                  Sep 11 '18 at 15:56










                • $begingroup$
                  But !8 can be a subfactorial of 8.
                  $endgroup$
                  – rus9384
                  Sep 12 '18 at 11:39










                • $begingroup$
                  @Raimund Krämer, In programming, we have logical operators and bitwise operators. Logical operators (!, &&, || in C) operate on boolean values (true and false). Bitwise operators (~, &, |, ^ in C) operate on bits (0 and 1) of numbers. Calling either set binary operators would be confusing, as both set work on binary values.
                  $endgroup$
                  – ikegami
                  Sep 13 '18 at 11:12












                • $begingroup$
                  @Raimund Krämer Also, "binary operator" can mean any operator on 2 values (as opposed to a unary operator such as ! or ~)
                  $endgroup$
                  – l k
                  Sep 18 '18 at 6:23














                18












                18








                18





                $begingroup$


                0 + 0 + 0 + 0 = !8




                because




                In C/C++, ! refers to the logical not operator, where all non-zero values become 0, and 0 becomes 1.







                share|improve this answer









                $endgroup$




                0 + 0 + 0 + 0 = !8




                because




                In C/C++, ! refers to the logical not operator, where all non-zero values become 0, and 0 becomes 1.








                share|improve this answer












                share|improve this answer



                share|improve this answer










                answered Sep 7 '18 at 22:48









                l kl k

                1812




                1812












                • $begingroup$
                  I think this should be "the binary not operator".
                  $endgroup$
                  – Raimund Krämer
                  Sep 11 '18 at 8:17






                • 3




                  $begingroup$
                  @RaimundKrämer ITYM unary not. C99 calls it the "logical negation operator" and files it in the chapter "Unary operators".
                  $endgroup$
                  – Jens
                  Sep 11 '18 at 15:56










                • $begingroup$
                  But !8 can be a subfactorial of 8.
                  $endgroup$
                  – rus9384
                  Sep 12 '18 at 11:39










                • $begingroup$
                  @Raimund Krämer, In programming, we have logical operators and bitwise operators. Logical operators (!, &&, || in C) operate on boolean values (true and false). Bitwise operators (~, &, |, ^ in C) operate on bits (0 and 1) of numbers. Calling either set binary operators would be confusing, as both set work on binary values.
                  $endgroup$
                  – ikegami
                  Sep 13 '18 at 11:12












                • $begingroup$
                  @Raimund Krämer Also, "binary operator" can mean any operator on 2 values (as opposed to a unary operator such as ! or ~)
                  $endgroup$
                  – l k
                  Sep 18 '18 at 6:23


















                • $begingroup$
                  I think this should be "the binary not operator".
                  $endgroup$
                  – Raimund Krämer
                  Sep 11 '18 at 8:17






                • 3




                  $begingroup$
                  @RaimundKrämer ITYM unary not. C99 calls it the "logical negation operator" and files it in the chapter "Unary operators".
                  $endgroup$
                  – Jens
                  Sep 11 '18 at 15:56










                • $begingroup$
                  But !8 can be a subfactorial of 8.
                  $endgroup$
                  – rus9384
                  Sep 12 '18 at 11:39










                • $begingroup$
                  @Raimund Krämer, In programming, we have logical operators and bitwise operators. Logical operators (!, &&, || in C) operate on boolean values (true and false). Bitwise operators (~, &, |, ^ in C) operate on bits (0 and 1) of numbers. Calling either set binary operators would be confusing, as both set work on binary values.
                  $endgroup$
                  – ikegami
                  Sep 13 '18 at 11:12












                • $begingroup$
                  @Raimund Krämer Also, "binary operator" can mean any operator on 2 values (as opposed to a unary operator such as ! or ~)
                  $endgroup$
                  – l k
                  Sep 18 '18 at 6:23
















                $begingroup$
                I think this should be "the binary not operator".
                $endgroup$
                – Raimund Krämer
                Sep 11 '18 at 8:17




                $begingroup$
                I think this should be "the binary not operator".
                $endgroup$
                – Raimund Krämer
                Sep 11 '18 at 8:17




                3




                3




                $begingroup$
                @RaimundKrämer ITYM unary not. C99 calls it the "logical negation operator" and files it in the chapter "Unary operators".
                $endgroup$
                – Jens
                Sep 11 '18 at 15:56




                $begingroup$
                @RaimundKrämer ITYM unary not. C99 calls it the "logical negation operator" and files it in the chapter "Unary operators".
                $endgroup$
                – Jens
                Sep 11 '18 at 15:56












                $begingroup$
                But !8 can be a subfactorial of 8.
                $endgroup$
                – rus9384
                Sep 12 '18 at 11:39




                $begingroup$
                But !8 can be a subfactorial of 8.
                $endgroup$
                – rus9384
                Sep 12 '18 at 11:39












                $begingroup$
                @Raimund Krämer, In programming, we have logical operators and bitwise operators. Logical operators (!, &&, || in C) operate on boolean values (true and false). Bitwise operators (~, &, |, ^ in C) operate on bits (0 and 1) of numbers. Calling either set binary operators would be confusing, as both set work on binary values.
                $endgroup$
                – ikegami
                Sep 13 '18 at 11:12






                $begingroup$
                @Raimund Krämer, In programming, we have logical operators and bitwise operators. Logical operators (!, &&, || in C) operate on boolean values (true and false). Bitwise operators (~, &, |, ^ in C) operate on bits (0 and 1) of numbers. Calling either set binary operators would be confusing, as both set work on binary values.
                $endgroup$
                – ikegami
                Sep 13 '18 at 11:12














                $begingroup$
                @Raimund Krämer Also, "binary operator" can mean any operator on 2 values (as opposed to a unary operator such as ! or ~)
                $endgroup$
                – l k
                Sep 18 '18 at 6:23




                $begingroup$
                @Raimund Krämer Also, "binary operator" can mean any operator on 2 values (as opposed to a unary operator such as ! or ~)
                $endgroup$
                – l k
                Sep 18 '18 at 6:23











                15












                $begingroup$

                It's just a matter of perspective ...




                0!/0 + 0!/0 = ∞




                My reasoning....




                0/0 is undefined so we have to first change the 0's into 1's with 0!

                (...and why did you write the infinity symbol sideways in your question?)







                share|improve this answer











                $endgroup$









                • 1




                  $begingroup$
                  x/0 isn't infinite, though...
                  $endgroup$
                  – Adam Smith
                  Sep 8 '18 at 17:49






                • 1




                  $begingroup$
                  Lateral thinking was yesterday. Vertical thinking is the new kid in town!
                  $endgroup$
                  – Jens
                  Sep 11 '18 at 15:59






                • 1




                  $begingroup$
                  x/0 == +infinity, per ieee 754
                  $endgroup$
                  – j__m
                  Sep 15 '18 at 12:40
















                15












                $begingroup$

                It's just a matter of perspective ...




                0!/0 + 0!/0 = ∞




                My reasoning....




                0/0 is undefined so we have to first change the 0's into 1's with 0!

                (...and why did you write the infinity symbol sideways in your question?)







                share|improve this answer











                $endgroup$









                • 1




                  $begingroup$
                  x/0 isn't infinite, though...
                  $endgroup$
                  – Adam Smith
                  Sep 8 '18 at 17:49






                • 1




                  $begingroup$
                  Lateral thinking was yesterday. Vertical thinking is the new kid in town!
                  $endgroup$
                  – Jens
                  Sep 11 '18 at 15:59






                • 1




                  $begingroup$
                  x/0 == +infinity, per ieee 754
                  $endgroup$
                  – j__m
                  Sep 15 '18 at 12:40














                15












                15








                15





                $begingroup$

                It's just a matter of perspective ...




                0!/0 + 0!/0 = ∞




                My reasoning....




                0/0 is undefined so we have to first change the 0's into 1's with 0!

                (...and why did you write the infinity symbol sideways in your question?)







                share|improve this answer











                $endgroup$



                It's just a matter of perspective ...




                0!/0 + 0!/0 = ∞




                My reasoning....




                0/0 is undefined so we have to first change the 0's into 1's with 0!

                (...and why did you write the infinity symbol sideways in your question?)








                share|improve this answer














                share|improve this answer



                share|improve this answer








                edited Sep 8 '18 at 19:31

























                answered Sep 7 '18 at 18:20









                rrauenzarrauenza

                25915




                25915








                • 1




                  $begingroup$
                  x/0 isn't infinite, though...
                  $endgroup$
                  – Adam Smith
                  Sep 8 '18 at 17:49






                • 1




                  $begingroup$
                  Lateral thinking was yesterday. Vertical thinking is the new kid in town!
                  $endgroup$
                  – Jens
                  Sep 11 '18 at 15:59






                • 1




                  $begingroup$
                  x/0 == +infinity, per ieee 754
                  $endgroup$
                  – j__m
                  Sep 15 '18 at 12:40














                • 1




                  $begingroup$
                  x/0 isn't infinite, though...
                  $endgroup$
                  – Adam Smith
                  Sep 8 '18 at 17:49






                • 1




                  $begingroup$
                  Lateral thinking was yesterday. Vertical thinking is the new kid in town!
                  $endgroup$
                  – Jens
                  Sep 11 '18 at 15:59






                • 1




                  $begingroup$
                  x/0 == +infinity, per ieee 754
                  $endgroup$
                  – j__m
                  Sep 15 '18 at 12:40








                1




                1




                $begingroup$
                x/0 isn't infinite, though...
                $endgroup$
                – Adam Smith
                Sep 8 '18 at 17:49




                $begingroup$
                x/0 isn't infinite, though...
                $endgroup$
                – Adam Smith
                Sep 8 '18 at 17:49




                1




                1




                $begingroup$
                Lateral thinking was yesterday. Vertical thinking is the new kid in town!
                $endgroup$
                – Jens
                Sep 11 '18 at 15:59




                $begingroup$
                Lateral thinking was yesterday. Vertical thinking is the new kid in town!
                $endgroup$
                – Jens
                Sep 11 '18 at 15:59




                1




                1




                $begingroup$
                x/0 == +infinity, per ieee 754
                $endgroup$
                – j__m
                Sep 15 '18 at 12:40




                $begingroup$
                x/0 == +infinity, per ieee 754
                $endgroup$
                – j__m
                Sep 15 '18 at 12:40











                10












                $begingroup$


                $$[+!0]+[0]-!0-!0$$




                Works in JavaScript. Hit F12 and type this into the console (This equation editor uses "−" instead of "-" so copy and paste doesn't quite work). Otherwise, it works in the same way as @malioboro and @Arnaldur's answers.



                In fact, you can make any JavaScript application run just by using a combination of 6 characters, which is what inspired me to make this. I substituted + for 0 when asking JSF**k to do 10-2.






                share|improve this answer









                $endgroup$









                • 3




                  $begingroup$
                  I like this abuse of javascript a lot. !0 for true, +!0 for 1, [1] for array containing 1, [1]+[0] for concatenate two arrays to get the string "10" ; subtract 1 from that twice to end up as 8. Love it.
                  $endgroup$
                  – Ross Presser
                  Sep 8 '18 at 4:10






                • 2




                  $begingroup$
                  but the question is tagged mathematics and not programming
                  $endgroup$
                  – phuclv
                  Sep 8 '18 at 4:21








                • 3




                  $begingroup$
                  Who says you can use square brackets....
                  $endgroup$
                  – user52269
                  Sep 8 '18 at 6:57
















                10












                $begingroup$


                $$[+!0]+[0]-!0-!0$$




                Works in JavaScript. Hit F12 and type this into the console (This equation editor uses "−" instead of "-" so copy and paste doesn't quite work). Otherwise, it works in the same way as @malioboro and @Arnaldur's answers.



                In fact, you can make any JavaScript application run just by using a combination of 6 characters, which is what inspired me to make this. I substituted + for 0 when asking JSF**k to do 10-2.






                share|improve this answer









                $endgroup$









                • 3




                  $begingroup$
                  I like this abuse of javascript a lot. !0 for true, +!0 for 1, [1] for array containing 1, [1]+[0] for concatenate two arrays to get the string "10" ; subtract 1 from that twice to end up as 8. Love it.
                  $endgroup$
                  – Ross Presser
                  Sep 8 '18 at 4:10






                • 2




                  $begingroup$
                  but the question is tagged mathematics and not programming
                  $endgroup$
                  – phuclv
                  Sep 8 '18 at 4:21








                • 3




                  $begingroup$
                  Who says you can use square brackets....
                  $endgroup$
                  – user52269
                  Sep 8 '18 at 6:57














                10












                10








                10





                $begingroup$


                $$[+!0]+[0]-!0-!0$$




                Works in JavaScript. Hit F12 and type this into the console (This equation editor uses "−" instead of "-" so copy and paste doesn't quite work). Otherwise, it works in the same way as @malioboro and @Arnaldur's answers.



                In fact, you can make any JavaScript application run just by using a combination of 6 characters, which is what inspired me to make this. I substituted + for 0 when asking JSF**k to do 10-2.






                share|improve this answer









                $endgroup$




                $$[+!0]+[0]-!0-!0$$




                Works in JavaScript. Hit F12 and type this into the console (This equation editor uses "−" instead of "-" so copy and paste doesn't quite work). Otherwise, it works in the same way as @malioboro and @Arnaldur's answers.



                In fact, you can make any JavaScript application run just by using a combination of 6 characters, which is what inspired me to make this. I substituted + for 0 when asking JSF**k to do 10-2.







                share|improve this answer












                share|improve this answer



                share|improve this answer










                answered Sep 7 '18 at 12:35









                NamytsNamyts

                4647




                4647








                • 3




                  $begingroup$
                  I like this abuse of javascript a lot. !0 for true, +!0 for 1, [1] for array containing 1, [1]+[0] for concatenate two arrays to get the string "10" ; subtract 1 from that twice to end up as 8. Love it.
                  $endgroup$
                  – Ross Presser
                  Sep 8 '18 at 4:10






                • 2




                  $begingroup$
                  but the question is tagged mathematics and not programming
                  $endgroup$
                  – phuclv
                  Sep 8 '18 at 4:21








                • 3




                  $begingroup$
                  Who says you can use square brackets....
                  $endgroup$
                  – user52269
                  Sep 8 '18 at 6:57














                • 3




                  $begingroup$
                  I like this abuse of javascript a lot. !0 for true, +!0 for 1, [1] for array containing 1, [1]+[0] for concatenate two arrays to get the string "10" ; subtract 1 from that twice to end up as 8. Love it.
                  $endgroup$
                  – Ross Presser
                  Sep 8 '18 at 4:10






                • 2




                  $begingroup$
                  but the question is tagged mathematics and not programming
                  $endgroup$
                  – phuclv
                  Sep 8 '18 at 4:21








                • 3




                  $begingroup$
                  Who says you can use square brackets....
                  $endgroup$
                  – user52269
                  Sep 8 '18 at 6:57








                3




                3




                $begingroup$
                I like this abuse of javascript a lot. !0 for true, +!0 for 1, [1] for array containing 1, [1]+[0] for concatenate two arrays to get the string "10" ; subtract 1 from that twice to end up as 8. Love it.
                $endgroup$
                – Ross Presser
                Sep 8 '18 at 4:10




                $begingroup$
                I like this abuse of javascript a lot. !0 for true, +!0 for 1, [1] for array containing 1, [1]+[0] for concatenate two arrays to get the string "10" ; subtract 1 from that twice to end up as 8. Love it.
                $endgroup$
                – Ross Presser
                Sep 8 '18 at 4:10




                2




                2




                $begingroup$
                but the question is tagged mathematics and not programming
                $endgroup$
                – phuclv
                Sep 8 '18 at 4:21






                $begingroup$
                but the question is tagged mathematics and not programming
                $endgroup$
                – phuclv
                Sep 8 '18 at 4:21






                3




                3




                $begingroup$
                Who says you can use square brackets....
                $endgroup$
                – user52269
                Sep 8 '18 at 6:57




                $begingroup$
                Who says you can use square brackets....
                $endgroup$
                – user52269
                Sep 8 '18 at 6:57











                8












                $begingroup$


                $concat(0!,0) - 0! - 0! = 8$




                becomes:




                $concat(1,0) - 1 - 1 = 8$




                and finally:




                $10 - 2 = 8$




                cool puzzle!






                share|improve this answer









                $endgroup$









                • 2




                  $begingroup$
                  Great answer! But it has already been answered (check out @malioboro 's answer). Nonetheless, welcome to the Puzzling Stack Exchange (Puzzling.SE)! Since you are new to the Stack Exchange community, let alone this site, I strongly suggest that you visit the Help Center for more info; in particular, I suggest going here, then here for questions (not answers) :D
                  $endgroup$
                  – user477343
                  Sep 7 '18 at 7:44


















                8












                $begingroup$


                $concat(0!,0) - 0! - 0! = 8$




                becomes:




                $concat(1,0) - 1 - 1 = 8$




                and finally:




                $10 - 2 = 8$




                cool puzzle!






                share|improve this answer









                $endgroup$









                • 2




                  $begingroup$
                  Great answer! But it has already been answered (check out @malioboro 's answer). Nonetheless, welcome to the Puzzling Stack Exchange (Puzzling.SE)! Since you are new to the Stack Exchange community, let alone this site, I strongly suggest that you visit the Help Center for more info; in particular, I suggest going here, then here for questions (not answers) :D
                  $endgroup$
                  – user477343
                  Sep 7 '18 at 7:44
















                8












                8








                8





                $begingroup$


                $concat(0!,0) - 0! - 0! = 8$




                becomes:




                $concat(1,0) - 1 - 1 = 8$




                and finally:




                $10 - 2 = 8$




                cool puzzle!






                share|improve this answer









                $endgroup$




                $concat(0!,0) - 0! - 0! = 8$




                becomes:




                $concat(1,0) - 1 - 1 = 8$




                and finally:




                $10 - 2 = 8$




                cool puzzle!







                share|improve this answer












                share|improve this answer



                share|improve this answer










                answered Sep 7 '18 at 5:38









                ArnaldurArnaldur

                971




                971








                • 2




                  $begingroup$
                  Great answer! But it has already been answered (check out @malioboro 's answer). Nonetheless, welcome to the Puzzling Stack Exchange (Puzzling.SE)! Since you are new to the Stack Exchange community, let alone this site, I strongly suggest that you visit the Help Center for more info; in particular, I suggest going here, then here for questions (not answers) :D
                  $endgroup$
                  – user477343
                  Sep 7 '18 at 7:44
















                • 2




                  $begingroup$
                  Great answer! But it has already been answered (check out @malioboro 's answer). Nonetheless, welcome to the Puzzling Stack Exchange (Puzzling.SE)! Since you are new to the Stack Exchange community, let alone this site, I strongly suggest that you visit the Help Center for more info; in particular, I suggest going here, then here for questions (not answers) :D
                  $endgroup$
                  – user477343
                  Sep 7 '18 at 7:44










                2




                2




                $begingroup$
                Great answer! But it has already been answered (check out @malioboro 's answer). Nonetheless, welcome to the Puzzling Stack Exchange (Puzzling.SE)! Since you are new to the Stack Exchange community, let alone this site, I strongly suggest that you visit the Help Center for more info; in particular, I suggest going here, then here for questions (not answers) :D
                $endgroup$
                – user477343
                Sep 7 '18 at 7:44






                $begingroup$
                Great answer! But it has already been answered (check out @malioboro 's answer). Nonetheless, welcome to the Puzzling Stack Exchange (Puzzling.SE)! Since you are new to the Stack Exchange community, let alone this site, I strongly suggest that you visit the Help Center for more info; in particular, I suggest going here, then here for questions (not answers) :D
                $endgroup$
                – user477343
                Sep 7 '18 at 7:44













                8












                $begingroup$


                $0 + 0 + 0 + 0 equiv 8$



                Adding the symbol $-$ over the equals sign makes it a congruence sign.
                Considering the congruence relation, we must be working mod N, where N divides 8.







                share|improve this answer









                $endgroup$


















                  8












                  $begingroup$


                  $0 + 0 + 0 + 0 equiv 8$



                  Adding the symbol $-$ over the equals sign makes it a congruence sign.
                  Considering the congruence relation, we must be working mod N, where N divides 8.







                  share|improve this answer









                  $endgroup$
















                    8












                    8








                    8





                    $begingroup$


                    $0 + 0 + 0 + 0 equiv 8$



                    Adding the symbol $-$ over the equals sign makes it a congruence sign.
                    Considering the congruence relation, we must be working mod N, where N divides 8.







                    share|improve this answer









                    $endgroup$




                    $0 + 0 + 0 + 0 equiv 8$



                    Adding the symbol $-$ over the equals sign makes it a congruence sign.
                    Considering the congruence relation, we must be working mod N, where N divides 8.








                    share|improve this answer












                    share|improve this answer



                    share|improve this answer










                    answered Sep 8 '18 at 6:39









                    VaelusVaelus

                    41124




                    41124























                        3












                        $begingroup$


                        Just put the minus symbol over the first zero to give it the look of a figure 8 and use plus to add the zeros.







                        share|improve this answer









                        $endgroup$













                        • $begingroup$
                          Great answer, but unfortunately, this is a duplicate of @TheSimpliFire 's answer.
                          $endgroup$
                          – user477343
                          Sep 7 '18 at 11:51








                        • 2




                          $begingroup$
                          @user477343 No, not quite, since this answer puts a line through the 0 instead of putting two 0s on top of each other. In this answer, only one 0 becomes an 8.
                          $endgroup$
                          – Fabian Röling
                          Sep 7 '18 at 12:02










                        • $begingroup$
                          @FabianRöling As I read it, it uses the minus symbol to be put over the first zero and make it look like an $8$. In my imagination, I assumed the zero is the bottom circle of $8$ since there is something over it, of which is another circle $largecirc$. I did not fully understand what this answer was talking about in saying to use a subtraction symbol, but just assumed it was just like TheSimpliFire's answer due to their similarity; and since that is a great answer, I thought this was great, too. If you understand this answer, may you please explain to me in other words if you can? :
                          $endgroup$
                          – user477343
                          Sep 7 '18 at 12:09








                        • 3




                          $begingroup$
                          @user477343 It uses the horizontal line of the minus sign to cut through the middle of the $0$. It is "over" the zero in the sense of overwritten.
                          $endgroup$
                          – Jaap Scherphuis
                          Sep 7 '18 at 12:55






                        • 1




                          $begingroup$
                          Yeah. 'over' in the sense of "in front of" on a dimension between you and your monitor.
                          $endgroup$
                          – Sentinel
                          Sep 7 '18 at 14:44
















                        3












                        $begingroup$


                        Just put the minus symbol over the first zero to give it the look of a figure 8 and use plus to add the zeros.







                        share|improve this answer









                        $endgroup$













                        • $begingroup$
                          Great answer, but unfortunately, this is a duplicate of @TheSimpliFire 's answer.
                          $endgroup$
                          – user477343
                          Sep 7 '18 at 11:51








                        • 2




                          $begingroup$
                          @user477343 No, not quite, since this answer puts a line through the 0 instead of putting two 0s on top of each other. In this answer, only one 0 becomes an 8.
                          $endgroup$
                          – Fabian Röling
                          Sep 7 '18 at 12:02










                        • $begingroup$
                          @FabianRöling As I read it, it uses the minus symbol to be put over the first zero and make it look like an $8$. In my imagination, I assumed the zero is the bottom circle of $8$ since there is something over it, of which is another circle $largecirc$. I did not fully understand what this answer was talking about in saying to use a subtraction symbol, but just assumed it was just like TheSimpliFire's answer due to their similarity; and since that is a great answer, I thought this was great, too. If you understand this answer, may you please explain to me in other words if you can? :
                          $endgroup$
                          – user477343
                          Sep 7 '18 at 12:09








                        • 3




                          $begingroup$
                          @user477343 It uses the horizontal line of the minus sign to cut through the middle of the $0$. It is "over" the zero in the sense of overwritten.
                          $endgroup$
                          – Jaap Scherphuis
                          Sep 7 '18 at 12:55






                        • 1




                          $begingroup$
                          Yeah. 'over' in the sense of "in front of" on a dimension between you and your monitor.
                          $endgroup$
                          – Sentinel
                          Sep 7 '18 at 14:44














                        3












                        3








                        3





                        $begingroup$


                        Just put the minus symbol over the first zero to give it the look of a figure 8 and use plus to add the zeros.







                        share|improve this answer









                        $endgroup$




                        Just put the minus symbol over the first zero to give it the look of a figure 8 and use plus to add the zeros.








                        share|improve this answer












                        share|improve this answer



                        share|improve this answer










                        answered Sep 7 '18 at 11:37









                        SentinelSentinel

                        1,041212




                        1,041212












                        • $begingroup$
                          Great answer, but unfortunately, this is a duplicate of @TheSimpliFire 's answer.
                          $endgroup$
                          – user477343
                          Sep 7 '18 at 11:51








                        • 2




                          $begingroup$
                          @user477343 No, not quite, since this answer puts a line through the 0 instead of putting two 0s on top of each other. In this answer, only one 0 becomes an 8.
                          $endgroup$
                          – Fabian Röling
                          Sep 7 '18 at 12:02










                        • $begingroup$
                          @FabianRöling As I read it, it uses the minus symbol to be put over the first zero and make it look like an $8$. In my imagination, I assumed the zero is the bottom circle of $8$ since there is something over it, of which is another circle $largecirc$. I did not fully understand what this answer was talking about in saying to use a subtraction symbol, but just assumed it was just like TheSimpliFire's answer due to their similarity; and since that is a great answer, I thought this was great, too. If you understand this answer, may you please explain to me in other words if you can? :
                          $endgroup$
                          – user477343
                          Sep 7 '18 at 12:09








                        • 3




                          $begingroup$
                          @user477343 It uses the horizontal line of the minus sign to cut through the middle of the $0$. It is "over" the zero in the sense of overwritten.
                          $endgroup$
                          – Jaap Scherphuis
                          Sep 7 '18 at 12:55






                        • 1




                          $begingroup$
                          Yeah. 'over' in the sense of "in front of" on a dimension between you and your monitor.
                          $endgroup$
                          – Sentinel
                          Sep 7 '18 at 14:44


















                        • $begingroup$
                          Great answer, but unfortunately, this is a duplicate of @TheSimpliFire 's answer.
                          $endgroup$
                          – user477343
                          Sep 7 '18 at 11:51








                        • 2




                          $begingroup$
                          @user477343 No, not quite, since this answer puts a line through the 0 instead of putting two 0s on top of each other. In this answer, only one 0 becomes an 8.
                          $endgroup$
                          – Fabian Röling
                          Sep 7 '18 at 12:02










                        • $begingroup$
                          @FabianRöling As I read it, it uses the minus symbol to be put over the first zero and make it look like an $8$. In my imagination, I assumed the zero is the bottom circle of $8$ since there is something over it, of which is another circle $largecirc$. I did not fully understand what this answer was talking about in saying to use a subtraction symbol, but just assumed it was just like TheSimpliFire's answer due to their similarity; and since that is a great answer, I thought this was great, too. If you understand this answer, may you please explain to me in other words if you can? :
                          $endgroup$
                          – user477343
                          Sep 7 '18 at 12:09








                        • 3




                          $begingroup$
                          @user477343 It uses the horizontal line of the minus sign to cut through the middle of the $0$. It is "over" the zero in the sense of overwritten.
                          $endgroup$
                          – Jaap Scherphuis
                          Sep 7 '18 at 12:55






                        • 1




                          $begingroup$
                          Yeah. 'over' in the sense of "in front of" on a dimension between you and your monitor.
                          $endgroup$
                          – Sentinel
                          Sep 7 '18 at 14:44
















                        $begingroup$
                        Great answer, but unfortunately, this is a duplicate of @TheSimpliFire 's answer.
                        $endgroup$
                        – user477343
                        Sep 7 '18 at 11:51






                        $begingroup$
                        Great answer, but unfortunately, this is a duplicate of @TheSimpliFire 's answer.
                        $endgroup$
                        – user477343
                        Sep 7 '18 at 11:51






                        2




                        2




                        $begingroup$
                        @user477343 No, not quite, since this answer puts a line through the 0 instead of putting two 0s on top of each other. In this answer, only one 0 becomes an 8.
                        $endgroup$
                        – Fabian Röling
                        Sep 7 '18 at 12:02




                        $begingroup$
                        @user477343 No, not quite, since this answer puts a line through the 0 instead of putting two 0s on top of each other. In this answer, only one 0 becomes an 8.
                        $endgroup$
                        – Fabian Röling
                        Sep 7 '18 at 12:02












                        $begingroup$
                        @FabianRöling As I read it, it uses the minus symbol to be put over the first zero and make it look like an $8$. In my imagination, I assumed the zero is the bottom circle of $8$ since there is something over it, of which is another circle $largecirc$. I did not fully understand what this answer was talking about in saying to use a subtraction symbol, but just assumed it was just like TheSimpliFire's answer due to their similarity; and since that is a great answer, I thought this was great, too. If you understand this answer, may you please explain to me in other words if you can? :
                        $endgroup$
                        – user477343
                        Sep 7 '18 at 12:09






                        $begingroup$
                        @FabianRöling As I read it, it uses the minus symbol to be put over the first zero and make it look like an $8$. In my imagination, I assumed the zero is the bottom circle of $8$ since there is something over it, of which is another circle $largecirc$. I did not fully understand what this answer was talking about in saying to use a subtraction symbol, but just assumed it was just like TheSimpliFire's answer due to their similarity; and since that is a great answer, I thought this was great, too. If you understand this answer, may you please explain to me in other words if you can? :
                        $endgroup$
                        – user477343
                        Sep 7 '18 at 12:09






                        3




                        3




                        $begingroup$
                        @user477343 It uses the horizontal line of the minus sign to cut through the middle of the $0$. It is "over" the zero in the sense of overwritten.
                        $endgroup$
                        – Jaap Scherphuis
                        Sep 7 '18 at 12:55




                        $begingroup$
                        @user477343 It uses the horizontal line of the minus sign to cut through the middle of the $0$. It is "over" the zero in the sense of overwritten.
                        $endgroup$
                        – Jaap Scherphuis
                        Sep 7 '18 at 12:55




                        1




                        1




                        $begingroup$
                        Yeah. 'over' in the sense of "in front of" on a dimension between you and your monitor.
                        $endgroup$
                        – Sentinel
                        Sep 7 '18 at 14:44




                        $begingroup$
                        Yeah. 'over' in the sense of "in front of" on a dimension between you and your monitor.
                        $endgroup$
                        – Sentinel
                        Sep 7 '18 at 14:44











                        3












                        $begingroup$

                        Question limits the symbols, not the operations. So with the symbol + can make the operator ++.




                        (++(++(++(++(++(++(++0000))))))) = 8







                        share|improve this answer











                        $endgroup$













                        • $begingroup$
                          I think you'd need the prefix form if you actually wanted that to work.
                          $endgroup$
                          – LegionMammal978
                          Sep 7 '18 at 23:48






                        • 1




                          $begingroup$
                          Doh. Of course you can't increment a literal in the first place.
                          $endgroup$
                          – David Browne - Microsoft
                          Sep 7 '18 at 23:55






                        • 1




                          $begingroup$
                          but that symbol doesn't exist in mathematics
                          $endgroup$
                          – phuclv
                          Sep 8 '18 at 4:20






                        • 2




                          $begingroup$
                          You need to pass an l-value to the ++ and -- operators, and neither their result or a literal is one.
                          $endgroup$
                          – NieDzejkob
                          Sep 8 '18 at 14:35
















                        3












                        $begingroup$

                        Question limits the symbols, not the operations. So with the symbol + can make the operator ++.




                        (++(++(++(++(++(++(++0000))))))) = 8







                        share|improve this answer











                        $endgroup$













                        • $begingroup$
                          I think you'd need the prefix form if you actually wanted that to work.
                          $endgroup$
                          – LegionMammal978
                          Sep 7 '18 at 23:48






                        • 1




                          $begingroup$
                          Doh. Of course you can't increment a literal in the first place.
                          $endgroup$
                          – David Browne - Microsoft
                          Sep 7 '18 at 23:55






                        • 1




                          $begingroup$
                          but that symbol doesn't exist in mathematics
                          $endgroup$
                          – phuclv
                          Sep 8 '18 at 4:20






                        • 2




                          $begingroup$
                          You need to pass an l-value to the ++ and -- operators, and neither their result or a literal is one.
                          $endgroup$
                          – NieDzejkob
                          Sep 8 '18 at 14:35














                        3












                        3








                        3





                        $begingroup$

                        Question limits the symbols, not the operations. So with the symbol + can make the operator ++.




                        (++(++(++(++(++(++(++0000))))))) = 8







                        share|improve this answer











                        $endgroup$



                        Question limits the symbols, not the operations. So with the symbol + can make the operator ++.




                        (++(++(++(++(++(++(++0000))))))) = 8








                        share|improve this answer














                        share|improve this answer



                        share|improve this answer








                        edited Sep 7 '18 at 23:53

























                        answered Sep 7 '18 at 21:23









                        David Browne - MicrosoftDavid Browne - Microsoft

                        1394




                        1394












                        • $begingroup$
                          I think you'd need the prefix form if you actually wanted that to work.
                          $endgroup$
                          – LegionMammal978
                          Sep 7 '18 at 23:48






                        • 1




                          $begingroup$
                          Doh. Of course you can't increment a literal in the first place.
                          $endgroup$
                          – David Browne - Microsoft
                          Sep 7 '18 at 23:55






                        • 1




                          $begingroup$
                          but that symbol doesn't exist in mathematics
                          $endgroup$
                          – phuclv
                          Sep 8 '18 at 4:20






                        • 2




                          $begingroup$
                          You need to pass an l-value to the ++ and -- operators, and neither their result or a literal is one.
                          $endgroup$
                          – NieDzejkob
                          Sep 8 '18 at 14:35


















                        • $begingroup$
                          I think you'd need the prefix form if you actually wanted that to work.
                          $endgroup$
                          – LegionMammal978
                          Sep 7 '18 at 23:48






                        • 1




                          $begingroup$
                          Doh. Of course you can't increment a literal in the first place.
                          $endgroup$
                          – David Browne - Microsoft
                          Sep 7 '18 at 23:55






                        • 1




                          $begingroup$
                          but that symbol doesn't exist in mathematics
                          $endgroup$
                          – phuclv
                          Sep 8 '18 at 4:20






                        • 2




                          $begingroup$
                          You need to pass an l-value to the ++ and -- operators, and neither their result or a literal is one.
                          $endgroup$
                          – NieDzejkob
                          Sep 8 '18 at 14:35
















                        $begingroup$
                        I think you'd need the prefix form if you actually wanted that to work.
                        $endgroup$
                        – LegionMammal978
                        Sep 7 '18 at 23:48




                        $begingroup$
                        I think you'd need the prefix form if you actually wanted that to work.
                        $endgroup$
                        – LegionMammal978
                        Sep 7 '18 at 23:48




                        1




                        1




                        $begingroup$
                        Doh. Of course you can't increment a literal in the first place.
                        $endgroup$
                        – David Browne - Microsoft
                        Sep 7 '18 at 23:55




                        $begingroup$
                        Doh. Of course you can't increment a literal in the first place.
                        $endgroup$
                        – David Browne - Microsoft
                        Sep 7 '18 at 23:55




                        1




                        1




                        $begingroup$
                        but that symbol doesn't exist in mathematics
                        $endgroup$
                        – phuclv
                        Sep 8 '18 at 4:20




                        $begingroup$
                        but that symbol doesn't exist in mathematics
                        $endgroup$
                        – phuclv
                        Sep 8 '18 at 4:20




                        2




                        2




                        $begingroup$
                        You need to pass an l-value to the ++ and -- operators, and neither their result or a literal is one.
                        $endgroup$
                        – NieDzejkob
                        Sep 8 '18 at 14:35




                        $begingroup$
                        You need to pass an l-value to the ++ and -- operators, and neither their result or a literal is one.
                        $endgroup$
                        – NieDzejkob
                        Sep 8 '18 at 14:35











                        3












                        $begingroup$

                        Here is an answer that doesn't use the semi-factorial or any concatenation.




                        $$0 + 0 - 0! / 0 = (-8)!$$




                        The left side is $-1/0$ and the right side is $-infty$.




                        Plugging the expression into Wolfram.






                        share|improve this answer









                        $endgroup$









                        • 4




                          $begingroup$
                          Division by zero is undefined (the limit of 0!/x as x goes to 0 is plus or minus infinity) and so are negative integer factorials (they are not equivalent to negative infinity). Even if they both evaluated to +∞, you can prove nonsense assuming you can equate infinities and treat like a real number. E.g., |1/0| = +∞, adding 1 to positive infinity is still positive infinity, hence |1/0| + 1 = +∞, hence |1/0| = |1/0| +1 by transitivity of equality. Subtract |1/0| from both sides and prove 0 = 1.
                          $endgroup$
                          – dr jimbob
                          Sep 8 '18 at 18:11
















                        3












                        $begingroup$

                        Here is an answer that doesn't use the semi-factorial or any concatenation.




                        $$0 + 0 - 0! / 0 = (-8)!$$




                        The left side is $-1/0$ and the right side is $-infty$.




                        Plugging the expression into Wolfram.






                        share|improve this answer









                        $endgroup$









                        • 4




                          $begingroup$
                          Division by zero is undefined (the limit of 0!/x as x goes to 0 is plus or minus infinity) and so are negative integer factorials (they are not equivalent to negative infinity). Even if they both evaluated to +∞, you can prove nonsense assuming you can equate infinities and treat like a real number. E.g., |1/0| = +∞, adding 1 to positive infinity is still positive infinity, hence |1/0| + 1 = +∞, hence |1/0| = |1/0| +1 by transitivity of equality. Subtract |1/0| from both sides and prove 0 = 1.
                          $endgroup$
                          – dr jimbob
                          Sep 8 '18 at 18:11














                        3












                        3








                        3





                        $begingroup$

                        Here is an answer that doesn't use the semi-factorial or any concatenation.




                        $$0 + 0 - 0! / 0 = (-8)!$$




                        The left side is $-1/0$ and the right side is $-infty$.




                        Plugging the expression into Wolfram.






                        share|improve this answer









                        $endgroup$



                        Here is an answer that doesn't use the semi-factorial or any concatenation.




                        $$0 + 0 - 0! / 0 = (-8)!$$




                        The left side is $-1/0$ and the right side is $-infty$.




                        Plugging the expression into Wolfram.







                        share|improve this answer












                        share|improve this answer



                        share|improve this answer










                        answered Sep 8 '18 at 0:20









                        user1717828user1717828

                        1,514617




                        1,514617








                        • 4




                          $begingroup$
                          Division by zero is undefined (the limit of 0!/x as x goes to 0 is plus or minus infinity) and so are negative integer factorials (they are not equivalent to negative infinity). Even if they both evaluated to +∞, you can prove nonsense assuming you can equate infinities and treat like a real number. E.g., |1/0| = +∞, adding 1 to positive infinity is still positive infinity, hence |1/0| + 1 = +∞, hence |1/0| = |1/0| +1 by transitivity of equality. Subtract |1/0| from both sides and prove 0 = 1.
                          $endgroup$
                          – dr jimbob
                          Sep 8 '18 at 18:11














                        • 4




                          $begingroup$
                          Division by zero is undefined (the limit of 0!/x as x goes to 0 is plus or minus infinity) and so are negative integer factorials (they are not equivalent to negative infinity). Even if they both evaluated to +∞, you can prove nonsense assuming you can equate infinities and treat like a real number. E.g., |1/0| = +∞, adding 1 to positive infinity is still positive infinity, hence |1/0| + 1 = +∞, hence |1/0| = |1/0| +1 by transitivity of equality. Subtract |1/0| from both sides and prove 0 = 1.
                          $endgroup$
                          – dr jimbob
                          Sep 8 '18 at 18:11








                        4




                        4




                        $begingroup$
                        Division by zero is undefined (the limit of 0!/x as x goes to 0 is plus or minus infinity) and so are negative integer factorials (they are not equivalent to negative infinity). Even if they both evaluated to +∞, you can prove nonsense assuming you can equate infinities and treat like a real number. E.g., |1/0| = +∞, adding 1 to positive infinity is still positive infinity, hence |1/0| + 1 = +∞, hence |1/0| = |1/0| +1 by transitivity of equality. Subtract |1/0| from both sides and prove 0 = 1.
                        $endgroup$
                        – dr jimbob
                        Sep 8 '18 at 18:11




                        $begingroup$
                        Division by zero is undefined (the limit of 0!/x as x goes to 0 is plus or minus infinity) and so are negative integer factorials (they are not equivalent to negative infinity). Even if they both evaluated to +∞, you can prove nonsense assuming you can equate infinities and treat like a real number. E.g., |1/0| = +∞, adding 1 to positive infinity is still positive infinity, hence |1/0| + 1 = +∞, hence |1/0| = |1/0| +1 by transitivity of equality. Subtract |1/0| from both sides and prove 0 = 1.
                        $endgroup$
                        – dr jimbob
                        Sep 8 '18 at 18:11











                        3












                        $begingroup$


                        concatenate(0!, 0) - concatenate(0! + 0!) = 8.




                        Note that 0! = 1




                        (0!, 0) = 10, and (0! + 0!) = 2, so 10 - 2 = 8







                        share|improve this answer











                        $endgroup$













                        • $begingroup$
                          Unfortunately concatenate is not a valid operation.
                          $endgroup$
                          – boboquack
                          Sep 11 '18 at 1:37










                        • $begingroup$
                          Congrats, concatenation is allowed now! Good answer, but people beat you to it. Welcome to Puzzling.SE! Here's a bonus puzzle: What's the word? Hilarious - extremely amusing
                          $endgroup$
                          – Alto
                          Sep 11 '18 at 1:47










                        • $begingroup$
                          Also, concatenation is combination. Basically, (2, 4) = 24, you get it.
                          $endgroup$
                          – Alto
                          Sep 11 '18 at 1:48
















                        3












                        $begingroup$


                        concatenate(0!, 0) - concatenate(0! + 0!) = 8.




                        Note that 0! = 1




                        (0!, 0) = 10, and (0! + 0!) = 2, so 10 - 2 = 8







                        share|improve this answer











                        $endgroup$













                        • $begingroup$
                          Unfortunately concatenate is not a valid operation.
                          $endgroup$
                          – boboquack
                          Sep 11 '18 at 1:37










                        • $begingroup$
                          Congrats, concatenation is allowed now! Good answer, but people beat you to it. Welcome to Puzzling.SE! Here's a bonus puzzle: What's the word? Hilarious - extremely amusing
                          $endgroup$
                          – Alto
                          Sep 11 '18 at 1:47










                        • $begingroup$
                          Also, concatenation is combination. Basically, (2, 4) = 24, you get it.
                          $endgroup$
                          – Alto
                          Sep 11 '18 at 1:48














                        3












                        3








                        3





                        $begingroup$


                        concatenate(0!, 0) - concatenate(0! + 0!) = 8.




                        Note that 0! = 1




                        (0!, 0) = 10, and (0! + 0!) = 2, so 10 - 2 = 8







                        share|improve this answer











                        $endgroup$




                        concatenate(0!, 0) - concatenate(0! + 0!) = 8.




                        Note that 0! = 1




                        (0!, 0) = 10, and (0! + 0!) = 2, so 10 - 2 = 8








                        share|improve this answer














                        share|improve this answer



                        share|improve this answer








                        edited Sep 11 '18 at 1:59









                        Alto

                        1,232320




                        1,232320










                        answered Sep 11 '18 at 0:57









                        NickNick

                        291




                        291












                        • $begingroup$
                          Unfortunately concatenate is not a valid operation.
                          $endgroup$
                          – boboquack
                          Sep 11 '18 at 1:37










                        • $begingroup$
                          Congrats, concatenation is allowed now! Good answer, but people beat you to it. Welcome to Puzzling.SE! Here's a bonus puzzle: What's the word? Hilarious - extremely amusing
                          $endgroup$
                          – Alto
                          Sep 11 '18 at 1:47










                        • $begingroup$
                          Also, concatenation is combination. Basically, (2, 4) = 24, you get it.
                          $endgroup$
                          – Alto
                          Sep 11 '18 at 1:48


















                        • $begingroup$
                          Unfortunately concatenate is not a valid operation.
                          $endgroup$
                          – boboquack
                          Sep 11 '18 at 1:37










                        • $begingroup$
                          Congrats, concatenation is allowed now! Good answer, but people beat you to it. Welcome to Puzzling.SE! Here's a bonus puzzle: What's the word? Hilarious - extremely amusing
                          $endgroup$
                          – Alto
                          Sep 11 '18 at 1:47










                        • $begingroup$
                          Also, concatenation is combination. Basically, (2, 4) = 24, you get it.
                          $endgroup$
                          – Alto
                          Sep 11 '18 at 1:48
















                        $begingroup$
                        Unfortunately concatenate is not a valid operation.
                        $endgroup$
                        – boboquack
                        Sep 11 '18 at 1:37




                        $begingroup$
                        Unfortunately concatenate is not a valid operation.
                        $endgroup$
                        – boboquack
                        Sep 11 '18 at 1:37












                        $begingroup$
                        Congrats, concatenation is allowed now! Good answer, but people beat you to it. Welcome to Puzzling.SE! Here's a bonus puzzle: What's the word? Hilarious - extremely amusing
                        $endgroup$
                        – Alto
                        Sep 11 '18 at 1:47




                        $begingroup$
                        Congrats, concatenation is allowed now! Good answer, but people beat you to it. Welcome to Puzzling.SE! Here's a bonus puzzle: What's the word? Hilarious - extremely amusing
                        $endgroup$
                        – Alto
                        Sep 11 '18 at 1:47












                        $begingroup$
                        Also, concatenation is combination. Basically, (2, 4) = 24, you get it.
                        $endgroup$
                        – Alto
                        Sep 11 '18 at 1:48




                        $begingroup$
                        Also, concatenation is combination. Basically, (2, 4) = 24, you get it.
                        $endgroup$
                        – Alto
                        Sep 11 '18 at 1:48











                        3












                        $begingroup$

                        This could work too:




                        (0!+0+0)/0 = ∞




                        Explanation




                        (0!+0+0)/0 = 1/0 which is infinity (8 but put horizontally)







                        share|improve this answer









                        $endgroup$













                        • $begingroup$
                          That is not infinity — it is undefined :
                          $endgroup$
                          – user477343
                          Oct 3 '18 at 11:33


















                        3












                        $begingroup$

                        This could work too:




                        (0!+0+0)/0 = ∞




                        Explanation




                        (0!+0+0)/0 = 1/0 which is infinity (8 but put horizontally)







                        share|improve this answer









                        $endgroup$













                        • $begingroup$
                          That is not infinity — it is undefined :
                          $endgroup$
                          – user477343
                          Oct 3 '18 at 11:33
















                        3












                        3








                        3





                        $begingroup$

                        This could work too:




                        (0!+0+0)/0 = ∞




                        Explanation




                        (0!+0+0)/0 = 1/0 which is infinity (8 but put horizontally)







                        share|improve this answer









                        $endgroup$



                        This could work too:




                        (0!+0+0)/0 = ∞




                        Explanation




                        (0!+0+0)/0 = 1/0 which is infinity (8 but put horizontally)








                        share|improve this answer












                        share|improve this answer



                        share|improve this answer










                        answered Sep 13 '18 at 9:10









                        Michal B.Michal B.

                        2665




                        2665












                        • $begingroup$
                          That is not infinity — it is undefined :
                          $endgroup$
                          – user477343
                          Oct 3 '18 at 11:33




















                        • $begingroup$
                          That is not infinity — it is undefined :
                          $endgroup$
                          – user477343
                          Oct 3 '18 at 11:33


















                        $begingroup$
                        That is not infinity — it is undefined :
                        $endgroup$
                        – user477343
                        Oct 3 '18 at 11:33






                        $begingroup$
                        That is not infinity — it is undefined :
                        $endgroup$
                        – user477343
                        Oct 3 '18 at 11:33













                        3












                        $begingroup$


                        Add a - above the equals to get $0000 equiv 8$, which is true assuming we are working in the ring $mathbb{Z}/mathbb{2Z}$. (Note I'm trying to avoid writing $[0] = [8]$...)







                        share|improve this answer











                        $endgroup$


















                          3












                          $begingroup$


                          Add a - above the equals to get $0000 equiv 8$, which is true assuming we are working in the ring $mathbb{Z}/mathbb{2Z}$. (Note I'm trying to avoid writing $[0] = [8]$...)







                          share|improve this answer











                          $endgroup$
















                            3












                            3








                            3





                            $begingroup$


                            Add a - above the equals to get $0000 equiv 8$, which is true assuming we are working in the ring $mathbb{Z}/mathbb{2Z}$. (Note I'm trying to avoid writing $[0] = [8]$...)







                            share|improve this answer











                            $endgroup$




                            Add a - above the equals to get $0000 equiv 8$, which is true assuming we are working in the ring $mathbb{Z}/mathbb{2Z}$. (Note I'm trying to avoid writing $[0] = [8]$...)








                            share|improve this answer














                            share|improve this answer



                            share|improve this answer








                            edited Sep 14 '18 at 2:45









                            a stone arachnid

                            1697




                            1697










                            answered Sep 11 '18 at 15:52









                            Gus314Gus314

                            312




                            312























                                0












                                $begingroup$


                                $00^{00} = 0^0$ which is indeterminate. In some sense, an indeterminate form can be equal to any value, since in Calculus, a function that approaches "$0^0$" can approach any real value, including $8$. So in that sense, $0^0 = 8$.


                                If you don't like concatenating two zeroes as "$00$", then $0^0 + 0 + 0$ also works.







                                share|improve this answer











                                $endgroup$









                                • 8




                                  $begingroup$
                                  There are no widely accepted definitions under which your equation is considered to be true.
                                  $endgroup$
                                  – Tanner Swett
                                  Sep 7 '18 at 17:34










                                • $begingroup$
                                  @TannerSwett Didn't think it was that unreasonable. My thinking is you can construct a function $f(x)^{g(x)}$ where $f$ and $g$ both limit to zero at some point $c$ such that $f(x)^{g(x)}$ limits to any real number at $c$, including $8$. Admittedly, writing $0^0 = 8$ is basically mathematically false. It is indeterminate for the reason I stated above. But this is Puzzling SE, not Mathematics SE, so I figured it was ok to think outside the box and not be formal with math.
                                  $endgroup$
                                  – RothX
                                  Sep 10 '18 at 13:23












                                • $begingroup$
                                  Yeah, I dunno if I'd call it "that unreasonable". But if I saw a student say that the equation $0^0 = 8$ is true, I'd think that they probably misunderstand what an indeterminate form means. If I saw a math professor say the same thing, I'd ask them if they could please be a little more rigorous.
                                  $endgroup$
                                  – Tanner Swett
                                  Sep 10 '18 at 13:59










                                • $begingroup$
                                  @TannerSwett Yeah, I completely agree. I'd never write $0^0 = 8$ in a mathematical setting, but since this is a puzzle, not a math problem, I thought maybe I could stretch a bit.
                                  $endgroup$
                                  – RothX
                                  Sep 11 '18 at 1:58
















                                0












                                $begingroup$


                                $00^{00} = 0^0$ which is indeterminate. In some sense, an indeterminate form can be equal to any value, since in Calculus, a function that approaches "$0^0$" can approach any real value, including $8$. So in that sense, $0^0 = 8$.


                                If you don't like concatenating two zeroes as "$00$", then $0^0 + 0 + 0$ also works.







                                share|improve this answer











                                $endgroup$









                                • 8




                                  $begingroup$
                                  There are no widely accepted definitions under which your equation is considered to be true.
                                  $endgroup$
                                  – Tanner Swett
                                  Sep 7 '18 at 17:34










                                • $begingroup$
                                  @TannerSwett Didn't think it was that unreasonable. My thinking is you can construct a function $f(x)^{g(x)}$ where $f$ and $g$ both limit to zero at some point $c$ such that $f(x)^{g(x)}$ limits to any real number at $c$, including $8$. Admittedly, writing $0^0 = 8$ is basically mathematically false. It is indeterminate for the reason I stated above. But this is Puzzling SE, not Mathematics SE, so I figured it was ok to think outside the box and not be formal with math.
                                  $endgroup$
                                  – RothX
                                  Sep 10 '18 at 13:23












                                • $begingroup$
                                  Yeah, I dunno if I'd call it "that unreasonable". But if I saw a student say that the equation $0^0 = 8$ is true, I'd think that they probably misunderstand what an indeterminate form means. If I saw a math professor say the same thing, I'd ask them if they could please be a little more rigorous.
                                  $endgroup$
                                  – Tanner Swett
                                  Sep 10 '18 at 13:59










                                • $begingroup$
                                  @TannerSwett Yeah, I completely agree. I'd never write $0^0 = 8$ in a mathematical setting, but since this is a puzzle, not a math problem, I thought maybe I could stretch a bit.
                                  $endgroup$
                                  – RothX
                                  Sep 11 '18 at 1:58














                                0












                                0








                                0





                                $begingroup$


                                $00^{00} = 0^0$ which is indeterminate. In some sense, an indeterminate form can be equal to any value, since in Calculus, a function that approaches "$0^0$" can approach any real value, including $8$. So in that sense, $0^0 = 8$.


                                If you don't like concatenating two zeroes as "$00$", then $0^0 + 0 + 0$ also works.







                                share|improve this answer











                                $endgroup$




                                $00^{00} = 0^0$ which is indeterminate. In some sense, an indeterminate form can be equal to any value, since in Calculus, a function that approaches "$0^0$" can approach any real value, including $8$. So in that sense, $0^0 = 8$.


                                If you don't like concatenating two zeroes as "$00$", then $0^0 + 0 + 0$ also works.








                                share|improve this answer














                                share|improve this answer



                                share|improve this answer








                                edited Sep 7 '18 at 13:30

























                                answered Sep 7 '18 at 13:22









                                RothXRothX

                                1273




                                1273








                                • 8




                                  $begingroup$
                                  There are no widely accepted definitions under which your equation is considered to be true.
                                  $endgroup$
                                  – Tanner Swett
                                  Sep 7 '18 at 17:34










                                • $begingroup$
                                  @TannerSwett Didn't think it was that unreasonable. My thinking is you can construct a function $f(x)^{g(x)}$ where $f$ and $g$ both limit to zero at some point $c$ such that $f(x)^{g(x)}$ limits to any real number at $c$, including $8$. Admittedly, writing $0^0 = 8$ is basically mathematically false. It is indeterminate for the reason I stated above. But this is Puzzling SE, not Mathematics SE, so I figured it was ok to think outside the box and not be formal with math.
                                  $endgroup$
                                  – RothX
                                  Sep 10 '18 at 13:23












                                • $begingroup$
                                  Yeah, I dunno if I'd call it "that unreasonable". But if I saw a student say that the equation $0^0 = 8$ is true, I'd think that they probably misunderstand what an indeterminate form means. If I saw a math professor say the same thing, I'd ask them if they could please be a little more rigorous.
                                  $endgroup$
                                  – Tanner Swett
                                  Sep 10 '18 at 13:59










                                • $begingroup$
                                  @TannerSwett Yeah, I completely agree. I'd never write $0^0 = 8$ in a mathematical setting, but since this is a puzzle, not a math problem, I thought maybe I could stretch a bit.
                                  $endgroup$
                                  – RothX
                                  Sep 11 '18 at 1:58














                                • 8




                                  $begingroup$
                                  There are no widely accepted definitions under which your equation is considered to be true.
                                  $endgroup$
                                  – Tanner Swett
                                  Sep 7 '18 at 17:34










                                • $begingroup$
                                  @TannerSwett Didn't think it was that unreasonable. My thinking is you can construct a function $f(x)^{g(x)}$ where $f$ and $g$ both limit to zero at some point $c$ such that $f(x)^{g(x)}$ limits to any real number at $c$, including $8$. Admittedly, writing $0^0 = 8$ is basically mathematically false. It is indeterminate for the reason I stated above. But this is Puzzling SE, not Mathematics SE, so I figured it was ok to think outside the box and not be formal with math.
                                  $endgroup$
                                  – RothX
                                  Sep 10 '18 at 13:23












                                • $begingroup$
                                  Yeah, I dunno if I'd call it "that unreasonable". But if I saw a student say that the equation $0^0 = 8$ is true, I'd think that they probably misunderstand what an indeterminate form means. If I saw a math professor say the same thing, I'd ask them if they could please be a little more rigorous.
                                  $endgroup$
                                  – Tanner Swett
                                  Sep 10 '18 at 13:59










                                • $begingroup$
                                  @TannerSwett Yeah, I completely agree. I'd never write $0^0 = 8$ in a mathematical setting, but since this is a puzzle, not a math problem, I thought maybe I could stretch a bit.
                                  $endgroup$
                                  – RothX
                                  Sep 11 '18 at 1:58








                                8




                                8




                                $begingroup$
                                There are no widely accepted definitions under which your equation is considered to be true.
                                $endgroup$
                                – Tanner Swett
                                Sep 7 '18 at 17:34




                                $begingroup$
                                There are no widely accepted definitions under which your equation is considered to be true.
                                $endgroup$
                                – Tanner Swett
                                Sep 7 '18 at 17:34












                                $begingroup$
                                @TannerSwett Didn't think it was that unreasonable. My thinking is you can construct a function $f(x)^{g(x)}$ where $f$ and $g$ both limit to zero at some point $c$ such that $f(x)^{g(x)}$ limits to any real number at $c$, including $8$. Admittedly, writing $0^0 = 8$ is basically mathematically false. It is indeterminate for the reason I stated above. But this is Puzzling SE, not Mathematics SE, so I figured it was ok to think outside the box and not be formal with math.
                                $endgroup$
                                – RothX
                                Sep 10 '18 at 13:23






                                $begingroup$
                                @TannerSwett Didn't think it was that unreasonable. My thinking is you can construct a function $f(x)^{g(x)}$ where $f$ and $g$ both limit to zero at some point $c$ such that $f(x)^{g(x)}$ limits to any real number at $c$, including $8$. Admittedly, writing $0^0 = 8$ is basically mathematically false. It is indeterminate for the reason I stated above. But this is Puzzling SE, not Mathematics SE, so I figured it was ok to think outside the box and not be formal with math.
                                $endgroup$
                                – RothX
                                Sep 10 '18 at 13:23














                                $begingroup$
                                Yeah, I dunno if I'd call it "that unreasonable". But if I saw a student say that the equation $0^0 = 8$ is true, I'd think that they probably misunderstand what an indeterminate form means. If I saw a math professor say the same thing, I'd ask them if they could please be a little more rigorous.
                                $endgroup$
                                – Tanner Swett
                                Sep 10 '18 at 13:59




                                $begingroup$
                                Yeah, I dunno if I'd call it "that unreasonable". But if I saw a student say that the equation $0^0 = 8$ is true, I'd think that they probably misunderstand what an indeterminate form means. If I saw a math professor say the same thing, I'd ask them if they could please be a little more rigorous.
                                $endgroup$
                                – Tanner Swett
                                Sep 10 '18 at 13:59












                                $begingroup$
                                @TannerSwett Yeah, I completely agree. I'd never write $0^0 = 8$ in a mathematical setting, but since this is a puzzle, not a math problem, I thought maybe I could stretch a bit.
                                $endgroup$
                                – RothX
                                Sep 11 '18 at 1:58




                                $begingroup$
                                @TannerSwett Yeah, I completely agree. I'd never write $0^0 = 8$ in a mathematical setting, but since this is a puzzle, not a math problem, I thought maybe I could stretch a bit.
                                $endgroup$
                                – RothX
                                Sep 11 '18 at 1:58











                                0












                                $begingroup$


                                $ ((0! + 0!)$)*(0! + 0!) = 2^2*2 = 8 $.




                                Further explanation:




                                The $$$ operation denotes the superfactorial defined as : $ n$ = (n!) uparrow uparrow (n!)$.







                                share|improve this answer











                                $endgroup$









                                • 3




                                  $begingroup$
                                  Good idea, but make sure to read the question fully. It gives you a list of what operations can be used and $ is not one of them.
                                  $endgroup$
                                  – Sensoray
                                  Sep 7 '18 at 15:26
















                                0












                                $begingroup$


                                $ ((0! + 0!)$)*(0! + 0!) = 2^2*2 = 8 $.




                                Further explanation:




                                The $$$ operation denotes the superfactorial defined as : $ n$ = (n!) uparrow uparrow (n!)$.







                                share|improve this answer











                                $endgroup$









                                • 3




                                  $begingroup$
                                  Good idea, but make sure to read the question fully. It gives you a list of what operations can be used and $ is not one of them.
                                  $endgroup$
                                  – Sensoray
                                  Sep 7 '18 at 15:26














                                0












                                0








                                0





                                $begingroup$


                                $ ((0! + 0!)$)*(0! + 0!) = 2^2*2 = 8 $.




                                Further explanation:




                                The $$$ operation denotes the superfactorial defined as : $ n$ = (n!) uparrow uparrow (n!)$.







                                share|improve this answer











                                $endgroup$




                                $ ((0! + 0!)$)*(0! + 0!) = 2^2*2 = 8 $.




                                Further explanation:




                                The $$$ operation denotes the superfactorial defined as : $ n$ = (n!) uparrow uparrow (n!)$.








                                share|improve this answer














                                share|improve this answer



                                share|improve this answer








                                edited Sep 7 '18 at 14:41

























                                answered Sep 7 '18 at 14:28









                                RiaRia

                                354111




                                354111








                                • 3




                                  $begingroup$
                                  Good idea, but make sure to read the question fully. It gives you a list of what operations can be used and $ is not one of them.
                                  $endgroup$
                                  – Sensoray
                                  Sep 7 '18 at 15:26














                                • 3




                                  $begingroup$
                                  Good idea, but make sure to read the question fully. It gives you a list of what operations can be used and $ is not one of them.
                                  $endgroup$
                                  – Sensoray
                                  Sep 7 '18 at 15:26








                                3




                                3




                                $begingroup$
                                Good idea, but make sure to read the question fully. It gives you a list of what operations can be used and $ is not one of them.
                                $endgroup$
                                – Sensoray
                                Sep 7 '18 at 15:26




                                $begingroup$
                                Good idea, but make sure to read the question fully. It gives you a list of what operations can be used and $ is not one of them.
                                $endgroup$
                                – Sensoray
                                Sep 7 '18 at 15:26











                                0












                                $begingroup$

                                If you turn the problem around




                                enter image description here







                                share|improve this answer









                                $endgroup$













                                • $begingroup$
                                  Similar to @rrauenza's answer
                                  $endgroup$
                                  – TheSimpliFire
                                  Sep 8 '18 at 8:11










                                • $begingroup$
                                  Yes. A different way to express the same idea.
                                  $endgroup$
                                  – Florian F
                                  Sep 8 '18 at 9:13
















                                0












                                $begingroup$

                                If you turn the problem around




                                enter image description here







                                share|improve this answer









                                $endgroup$













                                • $begingroup$
                                  Similar to @rrauenza's answer
                                  $endgroup$
                                  – TheSimpliFire
                                  Sep 8 '18 at 8:11










                                • $begingroup$
                                  Yes. A different way to express the same idea.
                                  $endgroup$
                                  – Florian F
                                  Sep 8 '18 at 9:13














                                0












                                0








                                0





                                $begingroup$

                                If you turn the problem around




                                enter image description here







                                share|improve this answer









                                $endgroup$



                                If you turn the problem around




                                enter image description here








                                share|improve this answer












                                share|improve this answer



                                share|improve this answer










                                answered Sep 8 '18 at 8:09









                                Florian FFlorian F

                                9,22612260




                                9,22612260












                                • $begingroup$
                                  Similar to @rrauenza's answer
                                  $endgroup$
                                  – TheSimpliFire
                                  Sep 8 '18 at 8:11










                                • $begingroup$
                                  Yes. A different way to express the same idea.
                                  $endgroup$
                                  – Florian F
                                  Sep 8 '18 at 9:13


















                                • $begingroup$
                                  Similar to @rrauenza's answer
                                  $endgroup$
                                  – TheSimpliFire
                                  Sep 8 '18 at 8:11










                                • $begingroup$
                                  Yes. A different way to express the same idea.
                                  $endgroup$
                                  – Florian F
                                  Sep 8 '18 at 9:13
















                                $begingroup$
                                Similar to @rrauenza's answer
                                $endgroup$
                                – TheSimpliFire
                                Sep 8 '18 at 8:11




                                $begingroup$
                                Similar to @rrauenza's answer
                                $endgroup$
                                – TheSimpliFire
                                Sep 8 '18 at 8:11












                                $begingroup$
                                Yes. A different way to express the same idea.
                                $endgroup$
                                – Florian F
                                Sep 8 '18 at 9:13




                                $begingroup$
                                Yes. A different way to express the same idea.
                                $endgroup$
                                – Florian F
                                Sep 8 '18 at 9:13











                                0












                                $begingroup$

                                Similar to @Vaelus




                                $0+0+0+0 leq 8$




                                Explanation




                                You can get the $leq$ by adding a $-$ inclined on top of the $=$







                                share|improve this answer









                                $endgroup$













                                • $begingroup$
                                  The goal is to strictly make $0,0,0,0=8$ and not to "make the statement true" (therefore allowing this answer to be acceptable) :
                                  $endgroup$
                                  – user477343
                                  Oct 3 '18 at 11:35












                                • $begingroup$
                                  When you say "the goal is to make [math expression]", then you're under the influence of the interpretation of that mathematical expression. IMO, "make the statement true" is the one and only interpretation to be taken, while you could argue more strict rules about not manipulating already existing symbols.
                                  $endgroup$
                                  – villasv
                                  Oct 3 '18 at 13:18












                                • $begingroup$
                                  I see, you have a point. I should've said that my comment is solely based on how I interpreted the question, possibly explaining why your answer was sadly downvoted. I am positive that there exists another answer that manipulates the equation in a similar way that you have. Sorry if I sounded mean. You can get a $(+1)$ but in at least $2$ hours, once my daily voting limit (DVL) is over :)
                                  $endgroup$
                                  – user477343
                                  Oct 3 '18 at 21:00


















                                0












                                $begingroup$

                                Similar to @Vaelus




                                $0+0+0+0 leq 8$




                                Explanation




                                You can get the $leq$ by adding a $-$ inclined on top of the $=$







                                share|improve this answer









                                $endgroup$













                                • $begingroup$
                                  The goal is to strictly make $0,0,0,0=8$ and not to "make the statement true" (therefore allowing this answer to be acceptable) :
                                  $endgroup$
                                  – user477343
                                  Oct 3 '18 at 11:35












                                • $begingroup$
                                  When you say "the goal is to make [math expression]", then you're under the influence of the interpretation of that mathematical expression. IMO, "make the statement true" is the one and only interpretation to be taken, while you could argue more strict rules about not manipulating already existing symbols.
                                  $endgroup$
                                  – villasv
                                  Oct 3 '18 at 13:18












                                • $begingroup$
                                  I see, you have a point. I should've said that my comment is solely based on how I interpreted the question, possibly explaining why your answer was sadly downvoted. I am positive that there exists another answer that manipulates the equation in a similar way that you have. Sorry if I sounded mean. You can get a $(+1)$ but in at least $2$ hours, once my daily voting limit (DVL) is over :)
                                  $endgroup$
                                  – user477343
                                  Oct 3 '18 at 21:00
















                                0












                                0








                                0





                                $begingroup$

                                Similar to @Vaelus




                                $0+0+0+0 leq 8$




                                Explanation




                                You can get the $leq$ by adding a $-$ inclined on top of the $=$







                                share|improve this answer









                                $endgroup$



                                Similar to @Vaelus




                                $0+0+0+0 leq 8$




                                Explanation




                                You can get the $leq$ by adding a $-$ inclined on top of the $=$








                                share|improve this answer












                                share|improve this answer



                                share|improve this answer










                                answered Sep 8 '18 at 19:44









                                villasvvillasv

                                1487




                                1487












                                • $begingroup$
                                  The goal is to strictly make $0,0,0,0=8$ and not to "make the statement true" (therefore allowing this answer to be acceptable) :
                                  $endgroup$
                                  – user477343
                                  Oct 3 '18 at 11:35












                                • $begingroup$
                                  When you say "the goal is to make [math expression]", then you're under the influence of the interpretation of that mathematical expression. IMO, "make the statement true" is the one and only interpretation to be taken, while you could argue more strict rules about not manipulating already existing symbols.
                                  $endgroup$
                                  – villasv
                                  Oct 3 '18 at 13:18












                                • $begingroup$
                                  I see, you have a point. I should've said that my comment is solely based on how I interpreted the question, possibly explaining why your answer was sadly downvoted. I am positive that there exists another answer that manipulates the equation in a similar way that you have. Sorry if I sounded mean. You can get a $(+1)$ but in at least $2$ hours, once my daily voting limit (DVL) is over :)
                                  $endgroup$
                                  – user477343
                                  Oct 3 '18 at 21:00




















                                • $begingroup$
                                  The goal is to strictly make $0,0,0,0=8$ and not to "make the statement true" (therefore allowing this answer to be acceptable) :
                                  $endgroup$
                                  – user477343
                                  Oct 3 '18 at 11:35












                                • $begingroup$
                                  When you say "the goal is to make [math expression]", then you're under the influence of the interpretation of that mathematical expression. IMO, "make the statement true" is the one and only interpretation to be taken, while you could argue more strict rules about not manipulating already existing symbols.
                                  $endgroup$
                                  – villasv
                                  Oct 3 '18 at 13:18












                                • $begingroup$
                                  I see, you have a point. I should've said that my comment is solely based on how I interpreted the question, possibly explaining why your answer was sadly downvoted. I am positive that there exists another answer that manipulates the equation in a similar way that you have. Sorry if I sounded mean. You can get a $(+1)$ but in at least $2$ hours, once my daily voting limit (DVL) is over :)
                                  $endgroup$
                                  – user477343
                                  Oct 3 '18 at 21:00


















                                $begingroup$
                                The goal is to strictly make $0,0,0,0=8$ and not to "make the statement true" (therefore allowing this answer to be acceptable) :
                                $endgroup$
                                – user477343
                                Oct 3 '18 at 11:35






                                $begingroup$
                                The goal is to strictly make $0,0,0,0=8$ and not to "make the statement true" (therefore allowing this answer to be acceptable) :
                                $endgroup$
                                – user477343
                                Oct 3 '18 at 11:35














                                $begingroup$
                                When you say "the goal is to make [math expression]", then you're under the influence of the interpretation of that mathematical expression. IMO, "make the statement true" is the one and only interpretation to be taken, while you could argue more strict rules about not manipulating already existing symbols.
                                $endgroup$
                                – villasv
                                Oct 3 '18 at 13:18






                                $begingroup$
                                When you say "the goal is to make [math expression]", then you're under the influence of the interpretation of that mathematical expression. IMO, "make the statement true" is the one and only interpretation to be taken, while you could argue more strict rules about not manipulating already existing symbols.
                                $endgroup$
                                – villasv
                                Oct 3 '18 at 13:18














                                $begingroup$
                                I see, you have a point. I should've said that my comment is solely based on how I interpreted the question, possibly explaining why your answer was sadly downvoted. I am positive that there exists another answer that manipulates the equation in a similar way that you have. Sorry if I sounded mean. You can get a $(+1)$ but in at least $2$ hours, once my daily voting limit (DVL) is over :)
                                $endgroup$
                                – user477343
                                Oct 3 '18 at 21:00






                                $begingroup$
                                I see, you have a point. I should've said that my comment is solely based on how I interpreted the question, possibly explaining why your answer was sadly downvoted. I am positive that there exists another answer that manipulates the equation in a similar way that you have. Sorry if I sounded mean. You can get a $(+1)$ but in at least $2$ hours, once my daily voting limit (DVL) is over :)
                                $endgroup$
                                – user477343
                                Oct 3 '18 at 21:00













                                0












                                $begingroup$


                                As per the list of allowed symbols we are clearly allowed to use "$,$" and "$.$"

                                This is doubly evident as otherwise how would we use the $mathbb{concatenation}$ function without a comma to separate the arguments?




                                So the solution is:




                                $0! - mathbb{concatenation}(., 0 + 0! + 0!!) = .8$

                                $1 - mathbb{concatenation}(., 0 + 1 + 1) = .8$

                                $1 - mathbb{concatenation}(., 2) = .8$

                                $1 - .2 = .8$







                                share|improve this answer









                                $endgroup$













                                • $begingroup$
                                  You may simplify 0!! as 0!.
                                  $endgroup$
                                  – Cœur
                                  Sep 11 '18 at 15:28










                                • $begingroup$
                                  I felt like $0+0!+0!!$ looked more elegant/purposeful as if it was part of a sequence like $a(n, m) = a(n-1,m)!; a(0,m)=m$
                                  $endgroup$
                                  – SamYonnou
                                  Sep 11 '18 at 16:58
















                                0












                                $begingroup$


                                As per the list of allowed symbols we are clearly allowed to use "$,$" and "$.$"

                                This is doubly evident as otherwise how would we use the $mathbb{concatenation}$ function without a comma to separate the arguments?




                                So the solution is:




                                $0! - mathbb{concatenation}(., 0 + 0! + 0!!) = .8$

                                $1 - mathbb{concatenation}(., 0 + 1 + 1) = .8$

                                $1 - mathbb{concatenation}(., 2) = .8$

                                $1 - .2 = .8$







                                share|improve this answer









                                $endgroup$













                                • $begingroup$
                                  You may simplify 0!! as 0!.
                                  $endgroup$
                                  – Cœur
                                  Sep 11 '18 at 15:28










                                • $begingroup$
                                  I felt like $0+0!+0!!$ looked more elegant/purposeful as if it was part of a sequence like $a(n, m) = a(n-1,m)!; a(0,m)=m$
                                  $endgroup$
                                  – SamYonnou
                                  Sep 11 '18 at 16:58














                                0












                                0








                                0





                                $begingroup$


                                As per the list of allowed symbols we are clearly allowed to use "$,$" and "$.$"

                                This is doubly evident as otherwise how would we use the $mathbb{concatenation}$ function without a comma to separate the arguments?




                                So the solution is:




                                $0! - mathbb{concatenation}(., 0 + 0! + 0!!) = .8$

                                $1 - mathbb{concatenation}(., 0 + 1 + 1) = .8$

                                $1 - mathbb{concatenation}(., 2) = .8$

                                $1 - .2 = .8$







                                share|improve this answer









                                $endgroup$




                                As per the list of allowed symbols we are clearly allowed to use "$,$" and "$.$"

                                This is doubly evident as otherwise how would we use the $mathbb{concatenation}$ function without a comma to separate the arguments?




                                So the solution is:




                                $0! - mathbb{concatenation}(., 0 + 0! + 0!!) = .8$

                                $1 - mathbb{concatenation}(., 0 + 1 + 1) = .8$

                                $1 - mathbb{concatenation}(., 2) = .8$

                                $1 - .2 = .8$








                                share|improve this answer












                                share|improve this answer



                                share|improve this answer










                                answered Sep 10 '18 at 16:55









                                SamYonnouSamYonnou

                                69067




                                69067












                                • $begingroup$
                                  You may simplify 0!! as 0!.
                                  $endgroup$
                                  – Cœur
                                  Sep 11 '18 at 15:28










                                • $begingroup$
                                  I felt like $0+0!+0!!$ looked more elegant/purposeful as if it was part of a sequence like $a(n, m) = a(n-1,m)!; a(0,m)=m$
                                  $endgroup$
                                  – SamYonnou
                                  Sep 11 '18 at 16:58


















                                • $begingroup$
                                  You may simplify 0!! as 0!.
                                  $endgroup$
                                  – Cœur
                                  Sep 11 '18 at 15:28










                                • $begingroup$
                                  I felt like $0+0!+0!!$ looked more elegant/purposeful as if it was part of a sequence like $a(n, m) = a(n-1,m)!; a(0,m)=m$
                                  $endgroup$
                                  – SamYonnou
                                  Sep 11 '18 at 16:58
















                                $begingroup$
                                You may simplify 0!! as 0!.
                                $endgroup$
                                – Cœur
                                Sep 11 '18 at 15:28




                                $begingroup$
                                You may simplify 0!! as 0!.
                                $endgroup$
                                – Cœur
                                Sep 11 '18 at 15:28












                                $begingroup$
                                I felt like $0+0!+0!!$ looked more elegant/purposeful as if it was part of a sequence like $a(n, m) = a(n-1,m)!; a(0,m)=m$
                                $endgroup$
                                – SamYonnou
                                Sep 11 '18 at 16:58




                                $begingroup$
                                I felt like $0+0!+0!!$ looked more elegant/purposeful as if it was part of a sequence like $a(n, m) = a(n-1,m)!; a(0,m)=m$
                                $endgroup$
                                – SamYonnou
                                Sep 11 '18 at 16:58











                                0












                                $begingroup$


                                If the order can be changed:



                                (0 0! 8)0 = 0!



                                (0 0! 8) is a permutation.







                                share|improve this answer








                                New contributor




                                Artem Lugin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                Check out our Code of Conduct.






                                $endgroup$









                                • 2




                                  $begingroup$
                                  $0 * 0 * 0 * 8 = 0$ would be much simpler, though no more a solution to this problem than yours. I don't think it's reasonable to think reordering is allowed.
                                  $endgroup$
                                  – Rubio
                                  3 hours ago










                                • $begingroup$
                                  nice try, but seems that this is not the kind of answer the OP was looking for. beware of downvotes, yet keep up your hard work and happy puzzling :)
                                  $endgroup$
                                  – Omega Krypton
                                  1 hour ago


















                                0












                                $begingroup$


                                If the order can be changed:



                                (0 0! 8)0 = 0!



                                (0 0! 8) is a permutation.







                                share|improve this answer








                                New contributor




                                Artem Lugin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                Check out our Code of Conduct.






                                $endgroup$









                                • 2




                                  $begingroup$
                                  $0 * 0 * 0 * 8 = 0$ would be much simpler, though no more a solution to this problem than yours. I don't think it's reasonable to think reordering is allowed.
                                  $endgroup$
                                  – Rubio
                                  3 hours ago










                                • $begingroup$
                                  nice try, but seems that this is not the kind of answer the OP was looking for. beware of downvotes, yet keep up your hard work and happy puzzling :)
                                  $endgroup$
                                  – Omega Krypton
                                  1 hour ago
















                                0












                                0








                                0





                                $begingroup$


                                If the order can be changed:



                                (0 0! 8)0 = 0!



                                (0 0! 8) is a permutation.







                                share|improve this answer








                                New contributor




                                Artem Lugin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                Check out our Code of Conduct.






                                $endgroup$




                                If the order can be changed:



                                (0 0! 8)0 = 0!



                                (0 0! 8) is a permutation.








                                share|improve this answer








                                New contributor




                                Artem Lugin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                Check out our Code of Conduct.









                                share|improve this answer



                                share|improve this answer






                                New contributor




                                Artem Lugin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                Check out our Code of Conduct.









                                answered 3 hours ago









                                Artem LuginArtem Lugin

                                856




                                856




                                New contributor




                                Artem Lugin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                Check out our Code of Conduct.





                                New contributor





                                Artem Lugin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                Check out our Code of Conduct.






                                Artem Lugin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                Check out our Code of Conduct.








                                • 2




                                  $begingroup$
                                  $0 * 0 * 0 * 8 = 0$ would be much simpler, though no more a solution to this problem than yours. I don't think it's reasonable to think reordering is allowed.
                                  $endgroup$
                                  – Rubio
                                  3 hours ago










                                • $begingroup$
                                  nice try, but seems that this is not the kind of answer the OP was looking for. beware of downvotes, yet keep up your hard work and happy puzzling :)
                                  $endgroup$
                                  – Omega Krypton
                                  1 hour ago
















                                • 2




                                  $begingroup$
                                  $0 * 0 * 0 * 8 = 0$ would be much simpler, though no more a solution to this problem than yours. I don't think it's reasonable to think reordering is allowed.
                                  $endgroup$
                                  – Rubio
                                  3 hours ago










                                • $begingroup$
                                  nice try, but seems that this is not the kind of answer the OP was looking for. beware of downvotes, yet keep up your hard work and happy puzzling :)
                                  $endgroup$
                                  – Omega Krypton
                                  1 hour ago










                                2




                                2




                                $begingroup$
                                $0 * 0 * 0 * 8 = 0$ would be much simpler, though no more a solution to this problem than yours. I don't think it's reasonable to think reordering is allowed.
                                $endgroup$
                                – Rubio
                                3 hours ago




                                $begingroup$
                                $0 * 0 * 0 * 8 = 0$ would be much simpler, though no more a solution to this problem than yours. I don't think it's reasonable to think reordering is allowed.
                                $endgroup$
                                – Rubio
                                3 hours ago












                                $begingroup$
                                nice try, but seems that this is not the kind of answer the OP was looking for. beware of downvotes, yet keep up your hard work and happy puzzling :)
                                $endgroup$
                                – Omega Krypton
                                1 hour ago






                                $begingroup$
                                nice try, but seems that this is not the kind of answer the OP was looking for. beware of downvotes, yet keep up your hard work and happy puzzling :)
                                $endgroup$
                                – Omega Krypton
                                1 hour ago













                                0












                                $begingroup$


                                0 0 0 0 = 8

                                0 0 0 0 = 8!

                                0 0 0 0 = 40320

                                0 0 0 0 = 4! * 0! * 3! * 2! * 0!

                                0 0 0 0 = 288

                                0 0 0 0 = 2+8/8

                                0! 0! 0! 0 = 2+1

                                0! + 0! + 0! + 0 = 2+1

                                1 + 1 + 1 + 0 = 2 + 1

                                3 = 3




                                Where's my prize money?



                                Just kidding, this and I are awful.






                                share|improve this answer








                                New contributor




                                Samy Bencherif is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                Check out our Code of Conduct.






                                $endgroup$













                                • $begingroup$
                                  I don't see how this focuses on answering the actual question, but... very cool! :D
                                  $endgroup$
                                  – user477343
                                  2 hours ago
















                                0












                                $begingroup$


                                0 0 0 0 = 8

                                0 0 0 0 = 8!

                                0 0 0 0 = 40320

                                0 0 0 0 = 4! * 0! * 3! * 2! * 0!

                                0 0 0 0 = 288

                                0 0 0 0 = 2+8/8

                                0! 0! 0! 0 = 2+1

                                0! + 0! + 0! + 0 = 2+1

                                1 + 1 + 1 + 0 = 2 + 1

                                3 = 3




                                Where's my prize money?



                                Just kidding, this and I are awful.






                                share|improve this answer








                                New contributor




                                Samy Bencherif is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                Check out our Code of Conduct.






                                $endgroup$













                                • $begingroup$
                                  I don't see how this focuses on answering the actual question, but... very cool! :D
                                  $endgroup$
                                  – user477343
                                  2 hours ago














                                0












                                0








                                0





                                $begingroup$


                                0 0 0 0 = 8

                                0 0 0 0 = 8!

                                0 0 0 0 = 40320

                                0 0 0 0 = 4! * 0! * 3! * 2! * 0!

                                0 0 0 0 = 288

                                0 0 0 0 = 2+8/8

                                0! 0! 0! 0 = 2+1

                                0! + 0! + 0! + 0 = 2+1

                                1 + 1 + 1 + 0 = 2 + 1

                                3 = 3




                                Where's my prize money?



                                Just kidding, this and I are awful.






                                share|improve this answer








                                New contributor




                                Samy Bencherif is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                Check out our Code of Conduct.






                                $endgroup$




                                0 0 0 0 = 8

                                0 0 0 0 = 8!

                                0 0 0 0 = 40320

                                0 0 0 0 = 4! * 0! * 3! * 2! * 0!

                                0 0 0 0 = 288

                                0 0 0 0 = 2+8/8

                                0! 0! 0! 0 = 2+1

                                0! + 0! + 0! + 0 = 2+1

                                1 + 1 + 1 + 0 = 2 + 1

                                3 = 3




                                Where's my prize money?



                                Just kidding, this and I are awful.







                                share|improve this answer








                                New contributor




                                Samy Bencherif is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                Check out our Code of Conduct.









                                share|improve this answer



                                share|improve this answer






                                New contributor




                                Samy Bencherif is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                Check out our Code of Conduct.









                                answered 2 hours ago









                                Samy BencherifSamy Bencherif

                                1213




                                1213




                                New contributor




                                Samy Bencherif is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                Check out our Code of Conduct.





                                New contributor





                                Samy Bencherif is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                Check out our Code of Conduct.






                                Samy Bencherif is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                Check out our Code of Conduct.












                                • $begingroup$
                                  I don't see how this focuses on answering the actual question, but... very cool! :D
                                  $endgroup$
                                  – user477343
                                  2 hours ago


















                                • $begingroup$
                                  I don't see how this focuses on answering the actual question, but... very cool! :D
                                  $endgroup$
                                  – user477343
                                  2 hours ago
















                                $begingroup$
                                I don't see how this focuses on answering the actual question, but... very cool! :D
                                $endgroup$
                                – user477343
                                2 hours ago




                                $begingroup$
                                I don't see how this focuses on answering the actual question, but... very cool! :D
                                $endgroup$
                                – user477343
                                2 hours ago


















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