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The 24 Game is an arithmetical card game in which the objective is to find a way to manipulate four integers so that the end result is 24.

How do you get $24$ using $5, 5, 5,$ and $1?$

Solution: $\displaystyle5\times\left[5-\left(\frac{1}{5}\right)\right].$

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    $\begingroup$ Define $x,y,z,w\mapsto 24$, then $5,5,5,1\mapsto 24$. $\endgroup$ – YoTengoUnLCD Aug 15 '16 at 21:47
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    $\begingroup$ In the "24 game" I have on my phone, only $+$, $-$, $\cdot$, $/$ and parentheses are allowed - you should clarify if that is also what you intended with this question (it would invalidate all the current answers). $\endgroup$ – Henrik - stop hurting Monica Aug 15 '16 at 21:48
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    $\begingroup$ @Henrik Spoil-sport $\endgroup$ – ÍgjøgnumMeg Aug 15 '16 at 21:55
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    $\begingroup$ @YoTengoUnLCD I loved that comment, especially since I used this game when teaching pre-algebra students about order of operations. I think the mapping concept would have been too much though (well maybe not...looking back, I am quite annoyed I was not introduced to the concept of a mapping until far later than I should have been). Fwiw, this site gives 10 easy solutions right away. $\endgroup$ – Daniel W. Farlow Aug 15 '16 at 22:26
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    $\begingroup$ Can this just be moved to the mathematics puzzle section? $\endgroup$ – Olive Stemforn Aug 17 '16 at 18:22
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Here's a solution using just the $+$, $-$, $\times$, $/$ operations, and parentheses: $$ 5 \times (5 - (1/5)) $$

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With Euler's Totient function:

$$\phi(5 \cdot 5) + \phi(5 \cdot 1)$$

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Here is one for you guys

$$ -1^5 + 5*5 $$

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  • $\begingroup$ Where does the $2$'s come from? And what does $(5)$ mean? $\endgroup$ – Henrik - stop hurting Monica Aug 15 '16 at 21:45
  • $\begingroup$ sorry i did prime factorization. edited $\endgroup$ – Erock Brox Aug 15 '16 at 21:48
  • $\begingroup$ You should add $$ around the expression. $\endgroup$ – Matt Watkins Aug 15 '16 at 21:58
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One possible answer is $$5 \cdot 5 - \lceil \frac{1}{5} \rceil$$

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One easy way:

$$(5-1)!\cdot\frac{5}5$$

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Or $$\sqrt{5\cdot 5} \cdot5 - 1$$

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$1\times\frac{5}{5}\Gamma(5){}{}{}{}{}{}{}{}$

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  • $\begingroup$ Care to elaborate a bit? $\endgroup$ – Namaste Aug 15 '16 at 22:26
  • $\begingroup$ Gamma function $\Gamma(n) = (n-1)!$ $\endgroup$ – John Aug 15 '16 at 22:30
  • $\begingroup$ Was a bit of a silly answer, although it's true. @John for $n \in \Bbb Z^+$ $\endgroup$ – ÍgjøgnumMeg Aug 15 '16 at 23:13
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One possible way:

$$(5-5)+(5-1)!=24$$

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With derangements:

$$!5 - 5(5 - 1)$$

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