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You are given the numbers 1,3,3,5.
Use any operation of +,−,∗,/,() to build 81 with those four numbers. You must use all four numbers exactly once.
No concatenation, exponential or other symbols like factorial are allowed.

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    $\begingroup$ (3+5)*3*1 = 24 = Kobe. Kobe scored 81 points in '06. Done $\endgroup$
    – anodyne
    Mar 5, 2020 at 17:03
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    $\begingroup$ Can I confirm that the rules are the same as the 24 game, except you're producing 81 instead of 24? It really doesn't look possible to me to even build any number as big as 81 here. With addition and multiplication, the most you can get is (1 + 3) * 3 * 5 = 60. Nor can I see a way to get anything bigger with division, like with something of form a / (b - c/d), because one can't produce a small denominator. $\endgroup$
    – xnor
    Mar 5, 2020 at 17:12
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    $\begingroup$ The () can also be used as binomial coefficient - have a look at puzzling.stackexchange.com/questions/92773/… $\endgroup$
    – ThomasL
    Mar 5, 2020 at 17:35

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I think the answer is

$$ \frac{5}{\binom{\frac{1}{3}}{3}} = 81$$
where we've used the generalised binomial coefficient $$\binom{\frac{1}{3}}{3} = \frac{\frac{1}{3}\left(\frac{1}{3}-1\right)\left(\frac{1}{3}-2\right)}{3!} = \frac{5}{81}$$

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  • $\begingroup$ Well done hexomino! $\endgroup$
    – ThomasL
    Mar 5, 2020 at 18:32

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