Combine 3, 3, 5 & 7 to get 24
$begingroup$
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
edited Dec 11 '18 at 9:54
Gamow
34.1k10125365
asked Jul 26 '18 at 8:32
Max WilliamsMax Williams
159116
$endgroup$
add a comment |
$begingroup$
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
edited Dec 11 '18 at 9:54
Gamow
34.1k10125365
asked Jul 26 '18 at 8:32
Max WilliamsMax Williams
159116
$endgroup$
23
$begingroup$
Well, I noticed that $${large 57-33=24}.$$ Does this break any rules?
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– Mr Pie
Jul 26 '18 at 10:33
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It's called 24 game for anyone wondering.
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– user7393973
Jul 26 '18 at 15:43
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@user477343 Accepted comment.
$endgroup$
– Manoj Kumar
Jul 26 '18 at 18:13
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Challenge 24. Nostalgia.
$endgroup$
– Sevvy325
Jul 26 '18 at 20:58
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@user477343 Definitely a good answer puzzling.stackexchange.com/questions/5555/…
$endgroup$
– Raystafarian
Jul 27 '18 at 6:26
add a comment |
$begingroup$
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
edited Dec 11 '18 at 9:54
Gamow
34.1k10125365
asked Jul 26 '18 at 8:32
Max WilliamsMax Williams
159116
$endgroup$
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
edited Dec 11 '18 at 9:54
Gamow
34.1k10125365
asked Jul 26 '18 at 8:32
Max WilliamsMax Williams
159116
edited Dec 11 '18 at 9:54
Gamow
34.1k10125365
asked Jul 26 '18 at 8:32
Max WilliamsMax Williams
159116
edited Dec 11 '18 at 9:54
Gamow
34.1k10125365
edited Dec 11 '18 at 9:54
Gamow
34.1k10125365
edited Dec 11 '18 at 9:54
Gamow
34.1k10125365
34.1k10125365
asked Jul 26 '18 at 8:32
Max WilliamsMax Williams
159116
asked Jul 26 '18 at 8:32
Max WilliamsMax Williams
159116
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
add a comment |
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
add a comment |
12 Answers
12
active
oldest
votes
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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
answered Jul 26 '18 at 8:55
Paul PalmpjePaul Palmpje
40634
$endgroup$
1
$begingroup$
Indeed, by brute-force search, this is the only formula up to commutative transformation.
$endgroup$
– LegionMammal978
Jul 26 '18 at 12:15
add a comment |
$begingroup$
Heh, how about:
(3 XOR 5) x (3 XOR 7)
... yes, yes, I know XOR's not allowed. Poor XOR. Nobody ever invites him.
answered Jul 26 '18 at 15:50
KevinKevin
2195
$endgroup$
$begingroup$
Nice twist in this answer!
$endgroup$
– Paul Palmpje
Jul 26 '18 at 16:32
add a comment |
$begingroup$
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. 🙁
edited Jul 27 '18 at 5:27
Laurel
818310
answered Jul 26 '18 at 18:34
HaveSpacesuitHaveSpacesuit
1694
$endgroup$
$begingroup$
argh, it was going so well.
$endgroup$
– Max Williams
Aug 6 '18 at 7:45
add a comment |
$begingroup$
what about this one?
using factorials of 3
$7+5+3!+3! = 24$
edited Jul 26 '18 at 13:51
Saeïdryl
3,078828
answered Jul 26 '18 at 12:37
K. DumreK. Dumre
573
$endgroup$
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
add a comment |
$begingroup$
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!
answered Jul 27 '18 at 8:49
Masked ManMasked Man
1
$endgroup$
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Hmm, I guess i did miss an "only" there.
$endgroup$
– Max Williams
Aug 6 '18 at 7:46
add a comment |
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I also thought about :
$3 * (7 - 5) ^ 3$
Breaks the rule but hey
answered Jul 27 '18 at 14:03
animalknoxanimalknox
192
$endgroup$
add a comment |
$begingroup$
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
edited Jul 30 '18 at 7:44
Glorfindel
14.7k45687
answered Jul 30 '18 at 7:19
Nikita gargNikita garg
111
$endgroup$
add a comment |
$begingroup$
I use 24 cards in my math classroom. Try this solution: (3*5-7)*3
answered 16 mins ago
canadianbell21canadianbell21
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.
$endgroup$
add a comment |
$begingroup$
check it
3!+3!+5+7
=6+6+12
=24
answered Jul 30 '18 at 6:27
Nikita gargNikita garg
111
$endgroup$
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Welcome to Puzzling! This uses an operator (!) which is not allowed according to the rules in the question.
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– Glorfindel
Jul 30 '18 at 6:52
add a comment |
$begingroup$
Seems pretty straight forward, sorry I don't know how to hide it:
(5x7) - (3x3) = 35 - 9 = 24
edited Jul 26 '18 at 17:23
Joe-You-Know
6,60821173
answered Jul 26 '18 at 16:46
CrossRoadsCrossRoads
97
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2
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Sorry, this is wrong: 35 - 9 = 26, not 24.
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– Fabio Turati
Jul 26 '18 at 17:00
1
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Sure, engineer, I can do math! Out of practice it seems.
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– CrossRoads
Jul 26 '18 at 17:03
add a comment |
$begingroup$
please check it
$(3*3*5)-(7*3) = 45-21 = 24$
edited Jul 26 '18 at 13:50
Saeïdryl
3,078828
answered Jul 26 '18 at 12:52
Nikita gargNikita garg
111
$endgroup$
4
$begingroup$
It's got an extra 3, so it's breaking the rules.
$endgroup$
– Bernat
Jul 26 '18 at 12:59
add a comment |
$begingroup$
How about ((3 x 5) - 7) x 3
edited Jul 26 '18 at 18:08
rhsquared
8,30031949
answered Jul 26 '18 at 14:43
user51019user51019
1
$endgroup$
3
$begingroup$
Welcome to Puzzling! This answer was already given, there's no need to post it again.
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– Glorfindel
Jul 26 '18 at 14:53
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How is this any different from the accepted answer?
$endgroup$
– iBuͦͦͦg
Jul 26 '18 at 14:53
add a comment |
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12 Answers
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active
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12 Answers
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active
oldest
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$begingroup$
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
answered Jul 26 '18 at 8:55
Paul PalmpjePaul Palmpje
40634
$endgroup$
1
$begingroup$
Indeed, by brute-force search, this is the only formula up to commutative transformation.
$endgroup$
– LegionMammal978
Jul 26 '18 at 12:15
add a comment |
$begingroup$
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
answered Jul 26 '18 at 8:55
Paul PalmpjePaul Palmpje
40634
$endgroup$
1
$begingroup$
Indeed, by brute-force search, this is the only formula up to commutative transformation.
$endgroup$
– LegionMammal978
Jul 26 '18 at 12:15
add a comment |
$begingroup$
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
answered Jul 26 '18 at 8:55
Paul PalmpjePaul Palmpje
40634
$endgroup$
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
answered Jul 26 '18 at 8:55
Paul PalmpjePaul Palmpje
40634
answered Jul 26 '18 at 8:55
Paul PalmpjePaul Palmpje
40634
answered Jul 26 '18 at 8:55
Paul PalmpjePaul Palmpje
40634
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
add a comment |
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
add a comment |
$begingroup$
Heh, how about:
(3 XOR 5) x (3 XOR 7)
... yes, yes, I know XOR's not allowed. Poor XOR. Nobody ever invites him.
answered Jul 26 '18 at 15:50
KevinKevin
2195
$endgroup$
$begingroup$
Nice twist in this answer!
$endgroup$
– Paul Palmpje
Jul 26 '18 at 16:32
add a comment |
$begingroup$
Heh, how about:
(3 XOR 5) x (3 XOR 7)
... yes, yes, I know XOR's not allowed. Poor XOR. Nobody ever invites him.
answered Jul 26 '18 at 15:50
KevinKevin
2195
$endgroup$
$begingroup$
Nice twist in this answer!
$endgroup$
– Paul Palmpje
Jul 26 '18 at 16:32
add a comment |
$begingroup$
Heh, how about:
(3 XOR 5) x (3 XOR 7)
... yes, yes, I know XOR's not allowed. Poor XOR. Nobody ever invites him.
answered Jul 26 '18 at 15:50
KevinKevin
2195
$endgroup$
Heh, how about:
(3 XOR 5) x (3 XOR 7)
... yes, yes, I know XOR's not allowed. Poor XOR. Nobody ever invites him.
answered Jul 26 '18 at 15:50
KevinKevin
2195
edited Jul 26 '18 at 17:49
answered Jul 26 '18 at 15:50
KevinKevin
2195
answered Jul 26 '18 at 15:50
KevinKevin
2195
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
add a comment |
$begingroup$
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
add a comment |
$begingroup$
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. 🙁
edited Jul 27 '18 at 5:27
Laurel
818310
answered Jul 26 '18 at 18:34
HaveSpacesuitHaveSpacesuit
1694
$endgroup$
$begingroup$
argh, it was going so well.
$endgroup$
– Max Williams
Aug 6 '18 at 7:45
add a comment |
$begingroup$
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. 🙁
edited Jul 27 '18 at 5:27
Laurel
818310
answered Jul 26 '18 at 18:34
HaveSpacesuitHaveSpacesuit
1694
$endgroup$
$begingroup$
argh, it was going so well.
$endgroup$
– Max Williams
Aug 6 '18 at 7:45
add a comment |
$begingroup$
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. 🙁
edited Jul 27 '18 at 5:27
Laurel
818310
answered Jul 26 '18 at 18:34
HaveSpacesuitHaveSpacesuit
1694
$endgroup$
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. 🙁
edited Jul 27 '18 at 5:27
Laurel
818310
answered Jul 26 '18 at 18:34
HaveSpacesuitHaveSpacesuit
1694
edited Jul 27 '18 at 5:27
Laurel
818310
edited Jul 27 '18 at 5:27
Laurel
818310
edited Jul 27 '18 at 5:27
Laurel
818310
818310
answered Jul 26 '18 at 18:34
HaveSpacesuitHaveSpacesuit
1694
answered Jul 26 '18 at 18:34
HaveSpacesuitHaveSpacesuit
1694
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
add a comment |
$begingroup$
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
add a comment |
$begingroup$
what about this one?
using factorials of 3
$7+5+3!+3! = 24$
edited Jul 26 '18 at 13:51
Saeïdryl
3,078828
answered Jul 26 '18 at 12:37
K. DumreK. Dumre
573
$endgroup$
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
add a comment |
$begingroup$
what about this one?
using factorials of 3
$7+5+3!+3! = 24$
edited Jul 26 '18 at 13:51
Saeïdryl
3,078828
answered Jul 26 '18 at 12:37
K. DumreK. Dumre
573
$endgroup$
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
add a comment |
$begingroup$
what about this one?
using factorials of 3
$7+5+3!+3! = 24$
edited Jul 26 '18 at 13:51
Saeïdryl
3,078828
answered Jul 26 '18 at 12:37
K. DumreK. Dumre
573
$endgroup$
what about this one?
using factorials of 3
$7+5+3!+3! = 24$
edited Jul 26 '18 at 13:51
Saeïdryl
3,078828
answered Jul 26 '18 at 12:37
K. DumreK. Dumre
573
edited Jul 26 '18 at 13:51
Saeïdryl
3,078828
edited Jul 26 '18 at 13:51
Saeïdryl
3,078828
edited Jul 26 '18 at 13:51
Saeïdryl
3,078828
3,078828
answered Jul 26 '18 at 12:37
K. DumreK. Dumre
573
answered Jul 26 '18 at 12:37
K. DumreK. Dumre
573
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
add a comment |
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
add a comment |
$begingroup$
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!
answered Jul 27 '18 at 8:49
Masked ManMasked Man
1
$endgroup$
$begingroup$
Hmm, I guess i did miss an "only" there.
$endgroup$
– Max Williams
Aug 6 '18 at 7:46
add a comment |
$begingroup$
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!
answered Jul 27 '18 at 8:49
Masked ManMasked Man
1
$endgroup$
$begingroup$
Hmm, I guess i did miss an "only" there.
$endgroup$
– Max Williams
Aug 6 '18 at 7:46
add a comment |
$begingroup$
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!
answered Jul 27 '18 at 8:49
Masked ManMasked Man
1
$endgroup$
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!
answered Jul 27 '18 at 8:49
Masked ManMasked Man
1
answered Jul 27 '18 at 8:49
Masked ManMasked Man
1
answered Jul 27 '18 at 8:49
Masked ManMasked Man
1
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
add a comment |
$begingroup$
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
add a comment |
$begingroup$
I also thought about :
$3 * (7 - 5) ^ 3$
Breaks the rule but hey
answered Jul 27 '18 at 14:03
animalknoxanimalknox
192
$endgroup$
add a comment |
$begingroup$
I also thought about :
$3 * (7 - 5) ^ 3$
Breaks the rule but hey
answered Jul 27 '18 at 14:03
animalknoxanimalknox
192
$endgroup$
add a comment |
$begingroup$
I also thought about :
$3 * (7 - 5) ^ 3$
Breaks the rule but hey
answered Jul 27 '18 at 14:03
animalknoxanimalknox
192
$endgroup$
I also thought about :
$3 * (7 - 5) ^ 3$
Breaks the rule but hey
answered Jul 27 '18 at 14:03
animalknoxanimalknox
192
edited Jul 27 '18 at 14:11
answered Jul 27 '18 at 14:03
animalknoxanimalknox
192
answered Jul 27 '18 at 14:03
animalknoxanimalknox
192
answered Jul 27 '18 at 14:03
animalknoxanimalknox
192
192
add a comment |
add a comment |
$begingroup$
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
edited Jul 30 '18 at 7:44
Glorfindel
14.7k45687
answered Jul 30 '18 at 7:19
Nikita gargNikita garg
111
$endgroup$
add a comment |
$begingroup$
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
edited Jul 30 '18 at 7:44
Glorfindel
14.7k45687
answered Jul 30 '18 at 7:19
Nikita gargNikita garg
111
$endgroup$
add a comment |
$begingroup$
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
edited Jul 30 '18 at 7:44
Glorfindel
14.7k45687
answered Jul 30 '18 at 7:19
Nikita gargNikita garg
111
$endgroup$
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
edited Jul 30 '18 at 7:44
Glorfindel
14.7k45687
answered Jul 30 '18 at 7:19
Nikita gargNikita garg
111
edited Jul 30 '18 at 7:44
Glorfindel
14.7k45687
edited Jul 30 '18 at 7:44
Glorfindel
14.7k45687
edited Jul 30 '18 at 7:44
Glorfindel
14.7k45687
14.7k45687
answered Jul 30 '18 at 7:19
Nikita gargNikita garg
111
answered Jul 30 '18 at 7:19
Nikita gargNikita garg
111
answered Jul 30 '18 at 7:19
Nikita gargNikita garg
111
111
add a comment |
add a comment |
$begingroup$
I use 24 cards in my math classroom. Try this solution: (3*5-7)*3
answered 16 mins ago
canadianbell21canadianbell21
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.
$endgroup$
add a comment |
$begingroup$
I use 24 cards in my math classroom. Try this solution: (3*5-7)*3
answered 16 mins ago
canadianbell21canadianbell21
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.
$endgroup$
add a comment |
$begingroup$
I use 24 cards in my math classroom. Try this solution: (3*5-7)*3
answered 16 mins ago
canadianbell21canadianbell21
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.
$endgroup$
I use 24 cards in my math classroom. Try this solution: (3*5-7)*3
answered 16 mins ago
canadianbell21canadianbell21
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.
answered 16 mins ago
canadianbell21canadianbell21
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.
answered 16 mins ago
canadianbell21canadianbell21
1
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.
add a comment |
add a comment |
$begingroup$
check it
3!+3!+5+7
=6+6+12
=24
answered Jul 30 '18 at 6:27
Nikita gargNikita garg
111
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
check it
3!+3!+5+7
=6+6+12
=24
answered Jul 30 '18 at 6:27
Nikita gargNikita garg
111
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
check it
3!+3!+5+7
=6+6+12
=24
answered Jul 30 '18 at 6:27
Nikita gargNikita garg
111
$endgroup$
check it
3!+3!+5+7
=6+6+12
=24
answered Jul 30 '18 at 6:27
Nikita gargNikita garg
111
answered Jul 30 '18 at 6:27
Nikita gargNikita garg
111
answered Jul 30 '18 at 6:27
Nikita gargNikita garg
111
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
add a comment |
$begingroup$
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
add a comment |
$begingroup$
Seems pretty straight forward, sorry I don't know how to hide it:
(5x7) - (3x3) = 35 - 9 = 24
edited Jul 26 '18 at 17:23
Joe-You-Know
6,60821173
answered Jul 26 '18 at 16:46
CrossRoadsCrossRoads
97
$endgroup$
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
add a comment |
$begingroup$
Seems pretty straight forward, sorry I don't know how to hide it:
(5x7) - (3x3) = 35 - 9 = 24
edited Jul 26 '18 at 17:23
Joe-You-Know
6,60821173
answered Jul 26 '18 at 16:46
CrossRoadsCrossRoads
97
$endgroup$
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
add a comment |
$begingroup$
Seems pretty straight forward, sorry I don't know how to hide it:
(5x7) - (3x3) = 35 - 9 = 24
edited Jul 26 '18 at 17:23
Joe-You-Know
6,60821173
answered Jul 26 '18 at 16:46
CrossRoadsCrossRoads
97
$endgroup$
Seems pretty straight forward, sorry I don't know how to hide it:
(5x7) - (3x3) = 35 - 9 = 24
edited Jul 26 '18 at 17:23
Joe-You-Know
6,60821173
answered Jul 26 '18 at 16:46
CrossRoadsCrossRoads
97
edited Jul 26 '18 at 17:23
Joe-You-Know
6,60821173
edited Jul 26 '18 at 17:23
Joe-You-Know
6,60821173
edited Jul 26 '18 at 17:23
Joe-You-Know
6,60821173
6,60821173
answered Jul 26 '18 at 16:46
CrossRoadsCrossRoads
97
answered Jul 26 '18 at 16:46
CrossRoadsCrossRoads
97
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
add a comment |
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
add a comment |
$begingroup$
please check it
$(3*3*5)-(7*3) = 45-21 = 24$
edited Jul 26 '18 at 13:50
Saeïdryl
3,078828
answered Jul 26 '18 at 12:52
Nikita gargNikita garg
111
$endgroup$
4
$begingroup$
It's got an extra 3, so it's breaking the rules.
$endgroup$
– Bernat
Jul 26 '18 at 12:59
add a comment |
$begingroup$
please check it
$(3*3*5)-(7*3) = 45-21 = 24$
edited Jul 26 '18 at 13:50
Saeïdryl
3,078828
answered Jul 26 '18 at 12:52
Nikita gargNikita garg
111
$endgroup$
4
$begingroup$
It's got an extra 3, so it's breaking the rules.
$endgroup$
– Bernat
Jul 26 '18 at 12:59
add a comment |
$begingroup$
please check it
$(3*3*5)-(7*3) = 45-21 = 24$
edited Jul 26 '18 at 13:50
Saeïdryl
3,078828
answered Jul 26 '18 at 12:52
Nikita gargNikita garg
111
$endgroup$
please check it
$(3*3*5)-(7*3) = 45-21 = 24$
edited Jul 26 '18 at 13:50
Saeïdryl
3,078828
answered Jul 26 '18 at 12:52
Nikita gargNikita garg
111
edited Jul 26 '18 at 13:50
Saeïdryl
3,078828
edited Jul 26 '18 at 13:50
Saeïdryl
3,078828
edited Jul 26 '18 at 13:50
Saeïdryl
3,078828
3,078828
answered Jul 26 '18 at 12:52
Nikita gargNikita garg
111
answered Jul 26 '18 at 12:52
Nikita gargNikita garg
111
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
add a comment |
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
add a comment |
$begingroup$
How about ((3 x 5) - 7) x 3
edited Jul 26 '18 at 18:08
rhsquared
8,30031949
answered Jul 26 '18 at 14:43
user51019user51019
1
$endgroup$
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
add a comment |
$begingroup$
How about ((3 x 5) - 7) x 3
edited Jul 26 '18 at 18:08
rhsquared
8,30031949
answered Jul 26 '18 at 14:43
user51019user51019
1
$endgroup$
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
add a comment |
$begingroup$
How about ((3 x 5) - 7) x 3
edited Jul 26 '18 at 18:08
rhsquared
8,30031949
answered Jul 26 '18 at 14:43
user51019user51019
1
$endgroup$
How about ((3 x 5) - 7) x 3
edited Jul 26 '18 at 18:08
rhsquared
8,30031949
answered Jul 26 '18 at 14:43
user51019user51019
1
edited Jul 26 '18 at 18:08
rhsquared
8,30031949
edited Jul 26 '18 at 18:08
rhsquared
8,30031949
edited Jul 26 '18 at 18:08
rhsquared
8,30031949
8,30031949
answered Jul 26 '18 at 14:43
user51019user51019
1
answered Jul 26 '18 at 14:43
user51019user51019
1
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
add a comment |
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
add a comment |
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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