Combine 3, 3, 5 & 7 to get 24












11












$begingroup$


Using operators plus, minus, multiplied by and divide by, and as many brackets as you want, can you do a formula which uses 3, 3, 5 and 7 to make 24?



Each number must be used and can only be used once (so there will be two 3's).



So, for example, (3x7)+3 makes 24, but this isn't valid because the 5 wasn't used.



I don't know if this is possible, btw! It's in a game I'm playing.






mathematics formation-of-numbers







share|improve this question











$endgroup$








  • 23




    $begingroup$
    Well, I noticed that $${large 57-33=24}.$$ Does this break any rules?
    $endgroup$
    – Mr Pie
    Jul 26 '18 at 10:33












  • $begingroup$
    It's called 24 game for anyone wondering.
    $endgroup$
    – user7393973
    Jul 26 '18 at 15:43










  • $begingroup$
    @user477343 Accepted comment.
    $endgroup$
    – Manoj Kumar
    Jul 26 '18 at 18:13










  • $begingroup$
    Challenge 24. Nostalgia.
    $endgroup$
    – Sevvy325
    Jul 26 '18 at 20:58










  • $begingroup$
    @user477343 Definitely a good answer puzzling.stackexchange.com/questions/5555/…
    $endgroup$
    – Raystafarian
    Jul 27 '18 at 6:26
















11












$begingroup$


Using operators plus, minus, multiplied by and divide by, and as many brackets as you want, can you do a formula which uses 3, 3, 5 and 7 to make 24?



Each number must be used and can only be used once (so there will be two 3's).



So, for example, (3x7)+3 makes 24, but this isn't valid because the 5 wasn't used.



I don't know if this is possible, btw! It's in a game I'm playing.






mathematics formation-of-numbers







share|improve this question











$endgroup$








  • 23




    $begingroup$
    Well, I noticed that $${large 57-33=24}.$$ Does this break any rules?
    $endgroup$
    – Mr Pie
    Jul 26 '18 at 10:33












  • $begingroup$
    It's called 24 game for anyone wondering.
    $endgroup$
    – user7393973
    Jul 26 '18 at 15:43










  • $begingroup$
    @user477343 Accepted comment.
    $endgroup$
    – Manoj Kumar
    Jul 26 '18 at 18:13










  • $begingroup$
    Challenge 24. Nostalgia.
    $endgroup$
    – Sevvy325
    Jul 26 '18 at 20:58










  • $begingroup$
    @user477343 Definitely a good answer puzzling.stackexchange.com/questions/5555/…
    $endgroup$
    – Raystafarian
    Jul 27 '18 at 6:26














11












11








11


1



$begingroup$


Using operators plus, minus, multiplied by and divide by, and as many brackets as you want, can you do a formula which uses 3, 3, 5 and 7 to make 24?



Each number must be used and can only be used once (so there will be two 3's).



So, for example, (3x7)+3 makes 24, but this isn't valid because the 5 wasn't used.



I don't know if this is possible, btw! It's in a game I'm playing.






mathematics formation-of-numbers







share|improve this question











$endgroup$




Using operators plus, minus, multiplied by and divide by, and as many brackets as you want, can you do a formula which uses 3, 3, 5 and 7 to make 24?



Each number must be used and can only be used once (so there will be two 3's).



So, for example, (3x7)+3 makes 24, but this isn't valid because the 5 wasn't used.



I don't know if this is possible, btw! It's in a game I'm playing.






mathematics formation-of-numbers




mathematics formation-of-numbers






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Dec 11 '18 at 9:54









Gamow

34.1k10125365




34.1k10125365










asked Jul 26 '18 at 8:32









Max WilliamsMax Williams

159116




159116








  • 23




    $begingroup$
    Well, I noticed that $${large 57-33=24}.$$ Does this break any rules?
    $endgroup$
    – Mr Pie
    Jul 26 '18 at 10:33












  • $begingroup$
    It's called 24 game for anyone wondering.
    $endgroup$
    – user7393973
    Jul 26 '18 at 15:43










  • $begingroup$
    @user477343 Accepted comment.
    $endgroup$
    – Manoj Kumar
    Jul 26 '18 at 18:13










  • $begingroup$
    Challenge 24. Nostalgia.
    $endgroup$
    – Sevvy325
    Jul 26 '18 at 20:58










  • $begingroup$
    @user477343 Definitely a good answer puzzling.stackexchange.com/questions/5555/…
    $endgroup$
    – Raystafarian
    Jul 27 '18 at 6:26














  • 23




    $begingroup$
    Well, I noticed that $${large 57-33=24}.$$ Does this break any rules?
    $endgroup$
    – Mr Pie
    Jul 26 '18 at 10:33












  • $begingroup$
    It's called 24 game for anyone wondering.
    $endgroup$
    – user7393973
    Jul 26 '18 at 15:43










  • $begingroup$
    @user477343 Accepted comment.
    $endgroup$
    – Manoj Kumar
    Jul 26 '18 at 18:13










  • $begingroup$
    Challenge 24. Nostalgia.
    $endgroup$
    – Sevvy325
    Jul 26 '18 at 20:58










  • $begingroup$
    @user477343 Definitely a good answer puzzling.stackexchange.com/questions/5555/…
    $endgroup$
    – Raystafarian
    Jul 27 '18 at 6:26








23




23




$begingroup$
Well, I noticed that $${large 57-33=24}.$$ Does this break any rules?
$endgroup$
– Mr Pie
Jul 26 '18 at 10:33






$begingroup$
Well, I noticed that $${large 57-33=24}.$$ Does this break any rules?
$endgroup$
– Mr Pie
Jul 26 '18 at 10:33














$begingroup$
It's called 24 game for anyone wondering.
$endgroup$
– user7393973
Jul 26 '18 at 15:43




$begingroup$
It's called 24 game for anyone wondering.
$endgroup$
– user7393973
Jul 26 '18 at 15:43












$begingroup$
@user477343 Accepted comment.
$endgroup$
– Manoj Kumar
Jul 26 '18 at 18:13




$begingroup$
@user477343 Accepted comment.
$endgroup$
– Manoj Kumar
Jul 26 '18 at 18:13












$begingroup$
Challenge 24. Nostalgia.
$endgroup$
– Sevvy325
Jul 26 '18 at 20:58




$begingroup$
Challenge 24. Nostalgia.
$endgroup$
– Sevvy325
Jul 26 '18 at 20:58












$begingroup$
@user477343 Definitely a good answer puzzling.stackexchange.com/questions/5555/…
$endgroup$
– Raystafarian
Jul 27 '18 at 6:26




$begingroup$
@user477343 Definitely a good answer puzzling.stackexchange.com/questions/5555/…
$endgroup$
– Raystafarian
Jul 27 '18 at 6:26










12 Answers
12






active

oldest

votes


















26












$begingroup$

Well, I tried to solve it the hard way (using double fractions) but actually it's quite easy.




(3 * 5 - 7) * 3 = (15 - 7) * 3 = 8 * 3 = 24







share|improve this answer









$endgroup$









  • 1




    $begingroup$
    Indeed, by brute-force search, this is the only formula up to commutative transformation.
    $endgroup$
    – LegionMammal978
    Jul 26 '18 at 12:15



















11












$begingroup$

Heh, how about:



(3 XOR 5) x (3 XOR 7)


... yes, yes, I know XOR's not allowed. Poor XOR. Nobody ever invites him.






share|improve this answer











$endgroup$













  • $begingroup$
    Nice twist in this answer!
    $endgroup$
    – Paul Palmpje
    Jul 26 '18 at 16:32



















6












$begingroup$

We can use as many brackets as we want, and Wikipedia states




Square brackets, as in [π] = 3, are sometimes used to denote the floor function, which rounds a real number down to the next integer.




So I will use square brackets to denote the floor function.




$$([5 div 3]+7) times 3 = (1+7) times 3 = 8 times 3 = 24$$




If we massage the rules a little more, and can put numbers in different positions, we can use Falling Factorial notation, where



$$(x)_n = x(x-1)(x-2) cdot cdot cdot (x-n+1) = frac{x!}{(x-n)!}$$




$$[(5)_3 div 7] times 3 = big[frac{5!}{(5-3)!} div 7 big] times 3 = [60 div 7] times 3 = 8 times 3 = 24$$




Unfortunately, there is no way to use combinations to get the given input to 24. 🙁






share|improve this answer











$endgroup$













  • $begingroup$
    argh, it was going so well.
    $endgroup$
    – Max Williams
    Aug 6 '18 at 7:45



















5












$begingroup$

what about this one?




using factorials of 3

$7+5+3!+3! = 24$







share|improve this answer











$endgroup$









  • 1




    $begingroup$
    Nice but the question lists specific operators. The factorial is not one of them. I solved one of these using binomials once. Although this does not use new operators it does rely on the placement of the numbers and should not be accepted.
    $endgroup$
    – Paul Palmpje
    Jul 26 '18 at 12:44



















3












$begingroup$

How about:




3 + 3 + 5 + 7 + 6 = 24




Explanation:




Each number must be used and can only be used once, but the rules don't say other numbers cannot be used!







share|improve this answer









$endgroup$













  • $begingroup$
    Hmm, I guess i did miss an "only" there.
    $endgroup$
    – Max Williams
    Aug 6 '18 at 7:46



















1












$begingroup$

I also thought about :




$3 * (7 - 5) ^ 3$




Breaks the rule but hey






share|improve this answer











$endgroup$





















    1












    $begingroup$

    When we




    combine 3 and 3 it makes 33




    and when we




    combine 5 and 7 it makes 57




    then




    subtract 57-33




    it will give the answer 24






    share|improve this answer











    $endgroup$





















      0












      $begingroup$

      I use 24 cards in my math classroom. Try this solution: (3*5-7)*3






      share|improve this answer








      New contributor




      canadianbell21 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.






      $endgroup$





















        -1












        $begingroup$

        check it
        3!+3!+5+7
        =6+6+12
        =24






        share|improve this answer









        $endgroup$













        • $begingroup$
          Welcome to Puzzling! This uses an operator (!) which is not allowed according to the rules in the question.
          $endgroup$
          – Glorfindel
          Jul 30 '18 at 6:52



















        -2












        $begingroup$

        Seems pretty straight forward, sorry I don't know how to hide it:




        (5x7) - (3x3) = 35 - 9 = 24







        share|improve this answer











        $endgroup$









        • 2




          $begingroup$
          Sorry, this is wrong: 35 - 9 = 26, not 24.
          $endgroup$
          – Fabio Turati
          Jul 26 '18 at 17:00






        • 1




          $begingroup$
          Sure, engineer, I can do math! Out of practice it seems.
          $endgroup$
          – CrossRoads
          Jul 26 '18 at 17:03



















        -3












        $begingroup$

        please check it




        $(3*3*5)-(7*3) = 45-21 = 24$







        share|improve this answer











        $endgroup$









        • 4




          $begingroup$
          It's got an extra 3, so it's breaking the rules.
          $endgroup$
          – Bernat
          Jul 26 '18 at 12:59



















        -3












        $begingroup$


        How about ((3 x 5) - 7) x 3







        share|improve this answer











        $endgroup$









        • 3




          $begingroup$
          Welcome to Puzzling! This answer was already given, there's no need to post it again.
          $endgroup$
          – Glorfindel
          Jul 26 '18 at 14:53










        • $begingroup$
          How is this any different from the accepted answer?
          $endgroup$
          – iBuͦͦͦg
          Jul 26 '18 at 14:53












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        12 Answers
        12






        active

        oldest

        votes








        12 Answers
        12






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        26












        $begingroup$

        Well, I tried to solve it the hard way (using double fractions) but actually it's quite easy.




        (3 * 5 - 7) * 3 = (15 - 7) * 3 = 8 * 3 = 24







        share|improve this answer









        $endgroup$









        • 1




          $begingroup$
          Indeed, by brute-force search, this is the only formula up to commutative transformation.
          $endgroup$
          – LegionMammal978
          Jul 26 '18 at 12:15
















        26












        $begingroup$

        Well, I tried to solve it the hard way (using double fractions) but actually it's quite easy.




        (3 * 5 - 7) * 3 = (15 - 7) * 3 = 8 * 3 = 24







        share|improve this answer









        $endgroup$









        • 1




          $begingroup$
          Indeed, by brute-force search, this is the only formula up to commutative transformation.
          $endgroup$
          – LegionMammal978
          Jul 26 '18 at 12:15














        26












        26








        26





        $begingroup$

        Well, I tried to solve it the hard way (using double fractions) but actually it's quite easy.




        (3 * 5 - 7) * 3 = (15 - 7) * 3 = 8 * 3 = 24







        share|improve this answer









        $endgroup$



        Well, I tried to solve it the hard way (using double fractions) but actually it's quite easy.




        (3 * 5 - 7) * 3 = (15 - 7) * 3 = 8 * 3 = 24








        share|improve this answer












        share|improve this answer



        share|improve this answer










        answered Jul 26 '18 at 8:55









        Paul PalmpjePaul Palmpje

        40634




        40634








        • 1




          $begingroup$
          Indeed, by brute-force search, this is the only formula up to commutative transformation.
          $endgroup$
          – LegionMammal978
          Jul 26 '18 at 12:15














        • 1




          $begingroup$
          Indeed, by brute-force search, this is the only formula up to commutative transformation.
          $endgroup$
          – LegionMammal978
          Jul 26 '18 at 12:15








        1




        1




        $begingroup$
        Indeed, by brute-force search, this is the only formula up to commutative transformation.
        $endgroup$
        – LegionMammal978
        Jul 26 '18 at 12:15




        $begingroup$
        Indeed, by brute-force search, this is the only formula up to commutative transformation.
        $endgroup$
        – LegionMammal978
        Jul 26 '18 at 12:15











        11












        $begingroup$

        Heh, how about:



        (3 XOR 5) x (3 XOR 7)


        ... yes, yes, I know XOR's not allowed. Poor XOR. Nobody ever invites him.






        share|improve this answer











        $endgroup$













        • $begingroup$
          Nice twist in this answer!
          $endgroup$
          – Paul Palmpje
          Jul 26 '18 at 16:32
















        11












        $begingroup$

        Heh, how about:



        (3 XOR 5) x (3 XOR 7)


        ... yes, yes, I know XOR's not allowed. Poor XOR. Nobody ever invites him.






        share|improve this answer











        $endgroup$













        • $begingroup$
          Nice twist in this answer!
          $endgroup$
          – Paul Palmpje
          Jul 26 '18 at 16:32














        11












        11








        11





        $begingroup$

        Heh, how about:



        (3 XOR 5) x (3 XOR 7)


        ... yes, yes, I know XOR's not allowed. Poor XOR. Nobody ever invites him.






        share|improve this answer











        $endgroup$



        Heh, how about:



        (3 XOR 5) x (3 XOR 7)


        ... yes, yes, I know XOR's not allowed. Poor XOR. Nobody ever invites him.







        share|improve this answer














        share|improve this answer



        share|improve this answer








        edited Jul 26 '18 at 17:49

























        answered Jul 26 '18 at 15:50









        KevinKevin

        2195




        2195












        • $begingroup$
          Nice twist in this answer!
          $endgroup$
          – Paul Palmpje
          Jul 26 '18 at 16:32


















        • $begingroup$
          Nice twist in this answer!
          $endgroup$
          – Paul Palmpje
          Jul 26 '18 at 16:32
















        $begingroup$
        Nice twist in this answer!
        $endgroup$
        – Paul Palmpje
        Jul 26 '18 at 16:32




        $begingroup$
        Nice twist in this answer!
        $endgroup$
        – Paul Palmpje
        Jul 26 '18 at 16:32











        6












        $begingroup$

        We can use as many brackets as we want, and Wikipedia states




        Square brackets, as in [π] = 3, are sometimes used to denote the floor function, which rounds a real number down to the next integer.




        So I will use square brackets to denote the floor function.




        $$([5 div 3]+7) times 3 = (1+7) times 3 = 8 times 3 = 24$$




        If we massage the rules a little more, and can put numbers in different positions, we can use Falling Factorial notation, where



        $$(x)_n = x(x-1)(x-2) cdot cdot cdot (x-n+1) = frac{x!}{(x-n)!}$$




        $$[(5)_3 div 7] times 3 = big[frac{5!}{(5-3)!} div 7 big] times 3 = [60 div 7] times 3 = 8 times 3 = 24$$




        Unfortunately, there is no way to use combinations to get the given input to 24. 🙁






        share|improve this answer











        $endgroup$













        • $begingroup$
          argh, it was going so well.
          $endgroup$
          – Max Williams
          Aug 6 '18 at 7:45
















        6












        $begingroup$

        We can use as many brackets as we want, and Wikipedia states




        Square brackets, as in [π] = 3, are sometimes used to denote the floor function, which rounds a real number down to the next integer.




        So I will use square brackets to denote the floor function.




        $$([5 div 3]+7) times 3 = (1+7) times 3 = 8 times 3 = 24$$




        If we massage the rules a little more, and can put numbers in different positions, we can use Falling Factorial notation, where



        $$(x)_n = x(x-1)(x-2) cdot cdot cdot (x-n+1) = frac{x!}{(x-n)!}$$




        $$[(5)_3 div 7] times 3 = big[frac{5!}{(5-3)!} div 7 big] times 3 = [60 div 7] times 3 = 8 times 3 = 24$$




        Unfortunately, there is no way to use combinations to get the given input to 24. 🙁






        share|improve this answer











        $endgroup$













        • $begingroup$
          argh, it was going so well.
          $endgroup$
          – Max Williams
          Aug 6 '18 at 7:45














        6












        6








        6





        $begingroup$

        We can use as many brackets as we want, and Wikipedia states




        Square brackets, as in [π] = 3, are sometimes used to denote the floor function, which rounds a real number down to the next integer.




        So I will use square brackets to denote the floor function.




        $$([5 div 3]+7) times 3 = (1+7) times 3 = 8 times 3 = 24$$




        If we massage the rules a little more, and can put numbers in different positions, we can use Falling Factorial notation, where



        $$(x)_n = x(x-1)(x-2) cdot cdot cdot (x-n+1) = frac{x!}{(x-n)!}$$




        $$[(5)_3 div 7] times 3 = big[frac{5!}{(5-3)!} div 7 big] times 3 = [60 div 7] times 3 = 8 times 3 = 24$$




        Unfortunately, there is no way to use combinations to get the given input to 24. 🙁






        share|improve this answer











        $endgroup$



        We can use as many brackets as we want, and Wikipedia states




        Square brackets, as in [π] = 3, are sometimes used to denote the floor function, which rounds a real number down to the next integer.




        So I will use square brackets to denote the floor function.




        $$([5 div 3]+7) times 3 = (1+7) times 3 = 8 times 3 = 24$$




        If we massage the rules a little more, and can put numbers in different positions, we can use Falling Factorial notation, where



        $$(x)_n = x(x-1)(x-2) cdot cdot cdot (x-n+1) = frac{x!}{(x-n)!}$$




        $$[(5)_3 div 7] times 3 = big[frac{5!}{(5-3)!} div 7 big] times 3 = [60 div 7] times 3 = 8 times 3 = 24$$




        Unfortunately, there is no way to use combinations to get the given input to 24. 🙁







        share|improve this answer














        share|improve this answer



        share|improve this answer








        edited Jul 27 '18 at 5:27









        Laurel

        818310




        818310










        answered Jul 26 '18 at 18:34









        HaveSpacesuitHaveSpacesuit

        1694




        1694












        • $begingroup$
          argh, it was going so well.
          $endgroup$
          – Max Williams
          Aug 6 '18 at 7:45


















        • $begingroup$
          argh, it was going so well.
          $endgroup$
          – Max Williams
          Aug 6 '18 at 7:45
















        $begingroup$
        argh, it was going so well.
        $endgroup$
        – Max Williams
        Aug 6 '18 at 7:45




        $begingroup$
        argh, it was going so well.
        $endgroup$
        – Max Williams
        Aug 6 '18 at 7:45











        5












        $begingroup$

        what about this one?




        using factorials of 3

        $7+5+3!+3! = 24$







        share|improve this answer











        $endgroup$









        • 1




          $begingroup$
          Nice but the question lists specific operators. The factorial is not one of them. I solved one of these using binomials once. Although this does not use new operators it does rely on the placement of the numbers and should not be accepted.
          $endgroup$
          – Paul Palmpje
          Jul 26 '18 at 12:44
















        5












        $begingroup$

        what about this one?




        using factorials of 3

        $7+5+3!+3! = 24$







        share|improve this answer











        $endgroup$









        • 1




          $begingroup$
          Nice but the question lists specific operators. The factorial is not one of them. I solved one of these using binomials once. Although this does not use new operators it does rely on the placement of the numbers and should not be accepted.
          $endgroup$
          – Paul Palmpje
          Jul 26 '18 at 12:44














        5












        5








        5





        $begingroup$

        what about this one?




        using factorials of 3

        $7+5+3!+3! = 24$







        share|improve this answer











        $endgroup$



        what about this one?




        using factorials of 3

        $7+5+3!+3! = 24$








        share|improve this answer














        share|improve this answer



        share|improve this answer








        edited Jul 26 '18 at 13:51









        Saeïdryl

        3,078828




        3,078828










        answered Jul 26 '18 at 12:37









        K. DumreK. Dumre

        573




        573








        • 1




          $begingroup$
          Nice but the question lists specific operators. The factorial is not one of them. I solved one of these using binomials once. Although this does not use new operators it does rely on the placement of the numbers and should not be accepted.
          $endgroup$
          – Paul Palmpje
          Jul 26 '18 at 12:44














        • 1




          $begingroup$
          Nice but the question lists specific operators. The factorial is not one of them. I solved one of these using binomials once. Although this does not use new operators it does rely on the placement of the numbers and should not be accepted.
          $endgroup$
          – Paul Palmpje
          Jul 26 '18 at 12:44








        1




        1




        $begingroup$
        Nice but the question lists specific operators. The factorial is not one of them. I solved one of these using binomials once. Although this does not use new operators it does rely on the placement of the numbers and should not be accepted.
        $endgroup$
        – Paul Palmpje
        Jul 26 '18 at 12:44




        $begingroup$
        Nice but the question lists specific operators. The factorial is not one of them. I solved one of these using binomials once. Although this does not use new operators it does rely on the placement of the numbers and should not be accepted.
        $endgroup$
        – Paul Palmpje
        Jul 26 '18 at 12:44











        3












        $begingroup$

        How about:




        3 + 3 + 5 + 7 + 6 = 24




        Explanation:




        Each number must be used and can only be used once, but the rules don't say other numbers cannot be used!







        share|improve this answer









        $endgroup$













        • $begingroup$
          Hmm, I guess i did miss an "only" there.
          $endgroup$
          – Max Williams
          Aug 6 '18 at 7:46
















        3












        $begingroup$

        How about:




        3 + 3 + 5 + 7 + 6 = 24




        Explanation:




        Each number must be used and can only be used once, but the rules don't say other numbers cannot be used!







        share|improve this answer









        $endgroup$













        • $begingroup$
          Hmm, I guess i did miss an "only" there.
          $endgroup$
          – Max Williams
          Aug 6 '18 at 7:46














        3












        3








        3





        $begingroup$

        How about:




        3 + 3 + 5 + 7 + 6 = 24




        Explanation:




        Each number must be used and can only be used once, but the rules don't say other numbers cannot be used!







        share|improve this answer









        $endgroup$



        How about:




        3 + 3 + 5 + 7 + 6 = 24




        Explanation:




        Each number must be used and can only be used once, but the rules don't say other numbers cannot be used!








        share|improve this answer












        share|improve this answer



        share|improve this answer










        answered Jul 27 '18 at 8:49









        Masked ManMasked Man

        1




        1












        • $begingroup$
          Hmm, I guess i did miss an "only" there.
          $endgroup$
          – Max Williams
          Aug 6 '18 at 7:46


















        • $begingroup$
          Hmm, I guess i did miss an "only" there.
          $endgroup$
          – Max Williams
          Aug 6 '18 at 7:46
















        $begingroup$
        Hmm, I guess i did miss an "only" there.
        $endgroup$
        – Max Williams
        Aug 6 '18 at 7:46




        $begingroup$
        Hmm, I guess i did miss an "only" there.
        $endgroup$
        – Max Williams
        Aug 6 '18 at 7:46











        1












        $begingroup$

        I also thought about :




        $3 * (7 - 5) ^ 3$




        Breaks the rule but hey






        share|improve this answer











        $endgroup$


















          1












          $begingroup$

          I also thought about :




          $3 * (7 - 5) ^ 3$




          Breaks the rule but hey






          share|improve this answer











          $endgroup$
















            1












            1








            1





            $begingroup$

            I also thought about :




            $3 * (7 - 5) ^ 3$




            Breaks the rule but hey






            share|improve this answer











            $endgroup$



            I also thought about :




            $3 * (7 - 5) ^ 3$




            Breaks the rule but hey







            share|improve this answer














            share|improve this answer



            share|improve this answer








            edited Jul 27 '18 at 14:11

























            answered Jul 27 '18 at 14:03









            animalknoxanimalknox

            192




            192























                1












                $begingroup$

                When we




                combine 3 and 3 it makes 33




                and when we




                combine 5 and 7 it makes 57




                then




                subtract 57-33




                it will give the answer 24






                share|improve this answer











                $endgroup$


















                  1












                  $begingroup$

                  When we




                  combine 3 and 3 it makes 33




                  and when we




                  combine 5 and 7 it makes 57




                  then




                  subtract 57-33




                  it will give the answer 24






                  share|improve this answer











                  $endgroup$
















                    1












                    1








                    1





                    $begingroup$

                    When we




                    combine 3 and 3 it makes 33




                    and when we




                    combine 5 and 7 it makes 57




                    then




                    subtract 57-33




                    it will give the answer 24






                    share|improve this answer











                    $endgroup$



                    When we




                    combine 3 and 3 it makes 33




                    and when we




                    combine 5 and 7 it makes 57




                    then




                    subtract 57-33




                    it will give the answer 24







                    share|improve this answer














                    share|improve this answer



                    share|improve this answer








                    edited Jul 30 '18 at 7:44









                    Glorfindel

                    14.7k45687




                    14.7k45687










                    answered Jul 30 '18 at 7:19









                    Nikita gargNikita garg

                    111




                    111























                        0












                        $begingroup$

                        I use 24 cards in my math classroom. Try this solution: (3*5-7)*3






                        share|improve this answer








                        New contributor




                        canadianbell21 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                        Check out our Code of Conduct.






                        $endgroup$


















                          0












                          $begingroup$

                          I use 24 cards in my math classroom. Try this solution: (3*5-7)*3






                          share|improve this answer








                          New contributor




                          canadianbell21 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





                            $begingroup$

                            I use 24 cards in my math classroom. Try this solution: (3*5-7)*3






                            share|improve this answer








                            New contributor




                            canadianbell21 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                            Check out our Code of Conduct.






                            $endgroup$



                            I use 24 cards in my math classroom. Try this solution: (3*5-7)*3







                            share|improve this answer








                            New contributor




                            canadianbell21 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




                            canadianbell21 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                            Check out our Code of Conduct.









                            answered 16 mins ago









                            canadianbell21canadianbell21

                            1




                            1




                            New contributor




                            canadianbell21 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                            Check out our Code of Conduct.





                            New contributor





                            canadianbell21 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                            Check out our Code of Conduct.






                            canadianbell21 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                            Check out our Code of Conduct.























                                -1












                                $begingroup$

                                check it
                                3!+3!+5+7
                                =6+6+12
                                =24






                                share|improve this answer









                                $endgroup$













                                • $begingroup$
                                  Welcome to Puzzling! This uses an operator (!) which is not allowed according to the rules in the question.
                                  $endgroup$
                                  – Glorfindel
                                  Jul 30 '18 at 6:52
















                                -1












                                $begingroup$

                                check it
                                3!+3!+5+7
                                =6+6+12
                                =24






                                share|improve this answer









                                $endgroup$













                                • $begingroup$
                                  Welcome to Puzzling! This uses an operator (!) which is not allowed according to the rules in the question.
                                  $endgroup$
                                  – Glorfindel
                                  Jul 30 '18 at 6:52














                                -1












                                -1








                                -1





                                $begingroup$

                                check it
                                3!+3!+5+7
                                =6+6+12
                                =24






                                share|improve this answer









                                $endgroup$



                                check it
                                3!+3!+5+7
                                =6+6+12
                                =24







                                share|improve this answer












                                share|improve this answer



                                share|improve this answer










                                answered Jul 30 '18 at 6:27









                                Nikita gargNikita garg

                                111




                                111












                                • $begingroup$
                                  Welcome to Puzzling! This uses an operator (!) which is not allowed according to the rules in the question.
                                  $endgroup$
                                  – Glorfindel
                                  Jul 30 '18 at 6:52


















                                • $begingroup$
                                  Welcome to Puzzling! This uses an operator (!) which is not allowed according to the rules in the question.
                                  $endgroup$
                                  – Glorfindel
                                  Jul 30 '18 at 6:52
















                                $begingroup$
                                Welcome to Puzzling! This uses an operator (!) which is not allowed according to the rules in the question.
                                $endgroup$
                                – Glorfindel
                                Jul 30 '18 at 6:52




                                $begingroup$
                                Welcome to Puzzling! This uses an operator (!) which is not allowed according to the rules in the question.
                                $endgroup$
                                – Glorfindel
                                Jul 30 '18 at 6:52











                                -2












                                $begingroup$

                                Seems pretty straight forward, sorry I don't know how to hide it:




                                (5x7) - (3x3) = 35 - 9 = 24







                                share|improve this answer











                                $endgroup$









                                • 2




                                  $begingroup$
                                  Sorry, this is wrong: 35 - 9 = 26, not 24.
                                  $endgroup$
                                  – Fabio Turati
                                  Jul 26 '18 at 17:00






                                • 1




                                  $begingroup$
                                  Sure, engineer, I can do math! Out of practice it seems.
                                  $endgroup$
                                  – CrossRoads
                                  Jul 26 '18 at 17:03
















                                -2












                                $begingroup$

                                Seems pretty straight forward, sorry I don't know how to hide it:




                                (5x7) - (3x3) = 35 - 9 = 24







                                share|improve this answer











                                $endgroup$









                                • 2




                                  $begingroup$
                                  Sorry, this is wrong: 35 - 9 = 26, not 24.
                                  $endgroup$
                                  – Fabio Turati
                                  Jul 26 '18 at 17:00






                                • 1




                                  $begingroup$
                                  Sure, engineer, I can do math! Out of practice it seems.
                                  $endgroup$
                                  – CrossRoads
                                  Jul 26 '18 at 17:03














                                -2












                                -2








                                -2





                                $begingroup$

                                Seems pretty straight forward, sorry I don't know how to hide it:




                                (5x7) - (3x3) = 35 - 9 = 24







                                share|improve this answer











                                $endgroup$



                                Seems pretty straight forward, sorry I don't know how to hide it:




                                (5x7) - (3x3) = 35 - 9 = 24








                                share|improve this answer














                                share|improve this answer



                                share|improve this answer








                                edited Jul 26 '18 at 17:23









                                Joe-You-Know

                                6,60821173




                                6,60821173










                                answered Jul 26 '18 at 16:46









                                CrossRoadsCrossRoads

                                97




                                97








                                • 2




                                  $begingroup$
                                  Sorry, this is wrong: 35 - 9 = 26, not 24.
                                  $endgroup$
                                  – Fabio Turati
                                  Jul 26 '18 at 17:00






                                • 1




                                  $begingroup$
                                  Sure, engineer, I can do math! Out of practice it seems.
                                  $endgroup$
                                  – CrossRoads
                                  Jul 26 '18 at 17:03














                                • 2




                                  $begingroup$
                                  Sorry, this is wrong: 35 - 9 = 26, not 24.
                                  $endgroup$
                                  – Fabio Turati
                                  Jul 26 '18 at 17:00






                                • 1




                                  $begingroup$
                                  Sure, engineer, I can do math! Out of practice it seems.
                                  $endgroup$
                                  – CrossRoads
                                  Jul 26 '18 at 17:03








                                2




                                2




                                $begingroup$
                                Sorry, this is wrong: 35 - 9 = 26, not 24.
                                $endgroup$
                                – Fabio Turati
                                Jul 26 '18 at 17:00




                                $begingroup$
                                Sorry, this is wrong: 35 - 9 = 26, not 24.
                                $endgroup$
                                – Fabio Turati
                                Jul 26 '18 at 17:00




                                1




                                1




                                $begingroup$
                                Sure, engineer, I can do math! Out of practice it seems.
                                $endgroup$
                                – CrossRoads
                                Jul 26 '18 at 17:03




                                $begingroup$
                                Sure, engineer, I can do math! Out of practice it seems.
                                $endgroup$
                                – CrossRoads
                                Jul 26 '18 at 17:03











                                -3












                                $begingroup$

                                please check it




                                $(3*3*5)-(7*3) = 45-21 = 24$







                                share|improve this answer











                                $endgroup$









                                • 4




                                  $begingroup$
                                  It's got an extra 3, so it's breaking the rules.
                                  $endgroup$
                                  – Bernat
                                  Jul 26 '18 at 12:59
















                                -3












                                $begingroup$

                                please check it




                                $(3*3*5)-(7*3) = 45-21 = 24$







                                share|improve this answer











                                $endgroup$









                                • 4




                                  $begingroup$
                                  It's got an extra 3, so it's breaking the rules.
                                  $endgroup$
                                  – Bernat
                                  Jul 26 '18 at 12:59














                                -3












                                -3








                                -3





                                $begingroup$

                                please check it




                                $(3*3*5)-(7*3) = 45-21 = 24$







                                share|improve this answer











                                $endgroup$



                                please check it




                                $(3*3*5)-(7*3) = 45-21 = 24$








                                share|improve this answer














                                share|improve this answer



                                share|improve this answer








                                edited Jul 26 '18 at 13:50









                                Saeïdryl

                                3,078828




                                3,078828










                                answered Jul 26 '18 at 12:52









                                Nikita gargNikita garg

                                111




                                111








                                • 4




                                  $begingroup$
                                  It's got an extra 3, so it's breaking the rules.
                                  $endgroup$
                                  – Bernat
                                  Jul 26 '18 at 12:59














                                • 4




                                  $begingroup$
                                  It's got an extra 3, so it's breaking the rules.
                                  $endgroup$
                                  – Bernat
                                  Jul 26 '18 at 12:59








                                4




                                4




                                $begingroup$
                                It's got an extra 3, so it's breaking the rules.
                                $endgroup$
                                – Bernat
                                Jul 26 '18 at 12:59




                                $begingroup$
                                It's got an extra 3, so it's breaking the rules.
                                $endgroup$
                                – Bernat
                                Jul 26 '18 at 12:59











                                -3












                                $begingroup$


                                How about ((3 x 5) - 7) x 3







                                share|improve this answer











                                $endgroup$









                                • 3




                                  $begingroup$
                                  Welcome to Puzzling! This answer was already given, there's no need to post it again.
                                  $endgroup$
                                  – Glorfindel
                                  Jul 26 '18 at 14:53










                                • $begingroup$
                                  How is this any different from the accepted answer?
                                  $endgroup$
                                  – iBuͦͦͦg
                                  Jul 26 '18 at 14:53
















                                -3












                                $begingroup$


                                How about ((3 x 5) - 7) x 3







                                share|improve this answer











                                $endgroup$









                                • 3




                                  $begingroup$
                                  Welcome to Puzzling! This answer was already given, there's no need to post it again.
                                  $endgroup$
                                  – Glorfindel
                                  Jul 26 '18 at 14:53










                                • $begingroup$
                                  How is this any different from the accepted answer?
                                  $endgroup$
                                  – iBuͦͦͦg
                                  Jul 26 '18 at 14:53














                                -3












                                -3








                                -3





                                $begingroup$


                                How about ((3 x 5) - 7) x 3







                                share|improve this answer











                                $endgroup$




                                How about ((3 x 5) - 7) x 3








                                share|improve this answer














                                share|improve this answer



                                share|improve this answer








                                edited Jul 26 '18 at 18:08









                                rhsquared

                                8,30031949




                                8,30031949










                                answered Jul 26 '18 at 14:43









                                user51019user51019

                                1




                                1








                                • 3




                                  $begingroup$
                                  Welcome to Puzzling! This answer was already given, there's no need to post it again.
                                  $endgroup$
                                  – Glorfindel
                                  Jul 26 '18 at 14:53










                                • $begingroup$
                                  How is this any different from the accepted answer?
                                  $endgroup$
                                  – iBuͦͦͦg
                                  Jul 26 '18 at 14:53














                                • 3




                                  $begingroup$
                                  Welcome to Puzzling! This answer was already given, there's no need to post it again.
                                  $endgroup$
                                  – Glorfindel
                                  Jul 26 '18 at 14:53










                                • $begingroup$
                                  How is this any different from the accepted answer?
                                  $endgroup$
                                  – iBuͦͦͦg
                                  Jul 26 '18 at 14:53








                                3




                                3




                                $begingroup$
                                Welcome to Puzzling! This answer was already given, there's no need to post it again.
                                $endgroup$
                                – Glorfindel
                                Jul 26 '18 at 14:53




                                $begingroup$
                                Welcome to Puzzling! This answer was already given, there's no need to post it again.
                                $endgroup$
                                – Glorfindel
                                Jul 26 '18 at 14:53












                                $begingroup$
                                How is this any different from the accepted answer?
                                $endgroup$
                                – iBuͦͦͦg
                                Jul 26 '18 at 14:53




                                $begingroup$
                                How is this any different from the accepted answer?
                                $endgroup$
                                – iBuͦͦͦg
                                Jul 26 '18 at 14:53


















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