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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.

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    $\begingroup$ 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. $\endgroup$ – m4n0 Jul 26 '18 at 18:13
  • $\begingroup$ Challenge 24. Nostalgia. $\endgroup$ – Sevvy325 Jul 26 '18 at 20:58
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    $\begingroup$ @user477343 Definitely a good answer puzzling.stackexchange.com/questions/5555/… $\endgroup$ – Raystafarian Jul 27 '18 at 6:26

11 Answers 11

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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

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

(3 XOR 5) x (3 XOR 7)

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

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    $\begingroup$ Nice twist in this answer! $\endgroup$ – Paul Palmpje Jul 26 '18 at 16:32
  • $\begingroup$ Could someone please explain this answer? I know XOR as a logic gate but I don't understand how can it be used here? $\endgroup$ – defectedWBC yesterday
  • $\begingroup$ @defectedWBC - XOR on numbers treats each bit independently. 3 XOR 5 is: 0011 XOR 0101, which is 0110, or 6. Likewise, 3 XOR 7 is: 0011 XOR 0111 = 0100, or 4. If you're on a windows machine, you can go into Calculator and set it to Programmer mode, and perform XOR's as well. $\endgroup$ – Kevin yesterday
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what about this one?

using factorials of 3
$7+5+3!+3! = 24$

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    $\begingroup$ 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
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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. 🙁

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

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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!

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

$3 * (7 - 5) ^ 3$

Breaks the rule but hey

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please check it

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

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    $\begingroup$ It's got an extra 3, so it's breaking the rules. $\endgroup$ – Bernat Jul 26 '18 at 12:59
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Seems pretty straight forward, sorry I don't know how to hide it:

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

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    $\begingroup$ Sorry, this is wrong: 35 - 9 = 26, not 24. $\endgroup$ – Fabio says Reinstate Monica Jul 26 '18 at 17:00
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    $\begingroup$ Sure, engineer, I can do math! Out of practice it seems. $\endgroup$ – CrossRoads Jul 26 '18 at 17:03
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How about ((3 x 5) - 7) x 3

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    $\begingroup$ Welcome to Puzzling! This answer was already given, there's no need to post it again. $\endgroup$ – Glorfindel Jul 26 '18 at 14:53
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    $\begingroup$ How is this any different from the accepted answer? $\endgroup$ – iBug Jul 26 '18 at 14:53
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check it 3!+3!+5+7 =6+6+12 =24

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    $\begingroup$ 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

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