7
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Four numbers are available: $1$, $3$, $4$ and $6$. Every number must be used once and only once with (some of) the operations $+$, $-$, $\times$, $\div$ to form the number $24$.

It's from the book "Art of Exploitation", 2nd edition. Give it a try!

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  • $\begingroup$ Do we have to use every number? Otherwise 6 * 4... $\endgroup$ – Lynch Mar 8 '16 at 12:38
  • $\begingroup$ Yes, every number need to be used only once. $\endgroup$ – user19985 Mar 8 '16 at 12:39
  • $\begingroup$ can you confirm if writing ** or ++ is ok? $\endgroup$ – Ben Mar 8 '16 at 13:08
  • $\begingroup$ no it is not okay, you can only use +, -, * and / once at time. $\endgroup$ – user19985 Mar 8 '16 at 13:09
20
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An answer to this problem is:

$6 \div (1 - \frac34)$

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  • 2
    $\begingroup$ Great first answer and welcome to puzzling.stackexchange. It is customary to hide answers by using the spoiler tag >! to give others a chance to solve it on their own. $\endgroup$ – Hugh Meyers Mar 8 '16 at 12:53
1
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A possible interpretation of the rules:

$1+3+4+6=24_{5}$

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-2
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Another possible answer could be...

$(14-6) \times 3 = 24$

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    $\begingroup$ the problems says the NUMBERS 1,3,4, and 6, not the DIGITS 1,3,4, and 6 $\endgroup$ – Kevin Mar 8 '16 at 17:31
-3
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Bit of an off the wall answer

(6-4)(3+1) forms 24 if you calculate the values within the brackets, and leave the results as they are

Not mathematically correct at all I know, it's an answer that loosely fits the wording of the puzzle

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-4
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Answer could also b:

$6 \times 4 \times 1^3$

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  • 4
    $\begingroup$ Exponents weren't allowed in the puzzle $\endgroup$ – Lynch Mar 8 '16 at 12:51
  • $\begingroup$ where does it say that? I would write a little 3 without the '^' but don't know how to here.... $\endgroup$ – Ben Mar 8 '16 at 12:55
  • 1
    $\begingroup$ The question states the operators you can use are +, -, * and / $\endgroup$ – Lynch Mar 8 '16 at 12:55
  • 1
    $\begingroup$ Doesn't matter if you write multiplication as 3 x 4, or 3 * 4, they both equal 12. The operation is the same, and that's what's limited in the puzzle $\endgroup$ – Lynch Mar 8 '16 at 12:59
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    $\begingroup$ @user902383 I would argue even though both operators ** and * use the same characters, they are very different operators, and it is not at all clear you may combine the characters to a new operator. $\endgroup$ – Block Mar 8 '16 at 13:16