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The make-24 puzzle is an oldie, but a very fun one at that.

Given four different numbers, produce—through a sequence of operations upon only those four numbers—the number twenty-four.

For example, given $2,2,3,8$ you can make $24$ by: $2\times 3\times\frac82$.

Note that each of the given numbers must be used exactly one time in the solution, and no other digits may appear anywhere.


This question is for the robbers to submit solutions to the problems posed by the cops.

The cops' thread should contain posts which specify the

  • four numbers you are allowed to use
  • operations you are allowed to use

Have a look at the problems people have submitted on the cops' thread and see if you can solve any of them. If you do manage to solve any, show it off here!

Make sure that your answer contains a link back to the problem's post, you can get the url by clicking the "share" link at the bottom of an answer.

share a link to this answer

So, your answers should be somewhat modelled like:

I had a lot of fun solving Chris' problem! Here's his post: https://puzzling.stackexchange.com/a/404. I can use plus, minus, divide, multiplication, square roots and factorials to get $24$ from $3,8,12,50$ and I finally did it!

And my solution:

$$\sqrt{50\times8}+\frac{12}3=24$$

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closed as too broad by Deusovi Mar 22 '17 at 16:32

Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

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Answer to Gareth's Clue

$8/(3-8/3) = 24$

I hope this is how it's done.

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  • $\begingroup$ Wow... that's awesome xD $\endgroup$ – theonlygusti Mar 22 '17 at 16:18
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Answer to theonlygusti 's Clue

$6/(1-3/4) = 24$

I hope this is how it's done.


Edit by Rosie F (Rather than posting a comment, I edit the answer, so that I can hide the spoiler)

There is also the solution

$1^3*4*6$

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  • $\begingroup$ Aww... you used the same technique for both XD Maybe give others a chance ;) But yeah, this is how it's done, hope it'll take off now that we have a few examples! $\endgroup$ – theonlygusti Mar 22 '17 at 16:23

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