59th Ukrainian National Mathematical Olympiad

It is clear that if $n\ge 3$, $n!\nmid 3$, that is $n!\ne 3k$, for some natural number $k$. But then $2^{n!}=2^{3k}=8^k\equiv 1(\mathrm{mod}7)$. Analogously modulo 9: if $n\ge 3$ we have that $n!\nmid 6$, that is $n!\ne 6l$, for some natural $l$. But then $2^{n!}=2^{6l}=64^l\equiv 1(\mathrm{mod}9)$.

It's easy to see, that the cubes of integer numbers are equal $0$ or $\pm 1$ modulo $7$ and modulo $9$. Without loss of generality, we may assume that $a\ge b\ge c$.

Case 1. $a\ge 3$ and $b\ge 3$. If $c\ge 3$, then $2^{a!}+2^{b!}+2^{c!}\equiv 3(\mathrm{mod}7)$ – is not a cube of an integer. If $c=2$ then $2^{a!}+2^{b!}+2^{c!}\equiv 1+1+4\equiv -1(\mathrm{mod}7)$ – can be a cube of an integer. Check modulo 9. $2^{a!}+2^{b!}+2^{c!}\equiv 1+1+4\equiv 6(\mathrm{mod}9)$ – is not a cube of an integer. If $c=1$ then $2^{a!}+2^{b!}+2^{c!}\equiv 1+1+2\equiv 4(\mathrm{mod}7)$ – is not a cube of an integer.

Case 2. $a\ge 3\Rightarrow b\le 2$. Hence we have three cases. If $b=c=2$ $2^{a!}+2^{b!}+2^{c!}\equiv 1+4+4\equiv 2(\mathrm{mod}7)$ – is not a cube of an integer. If $b=2,c=1$ $2^{a!}+2^{b!}+2^{c!}\equiv 1+4+2\equiv 0(\mathrm{mod}7)$ – can be a cube of an integer. Then analogously modulo 9 $2^{a!}+2^{b!}+2^{c!}\equiv 1+4+2\equiv 7(\mathrm{mod}9)$ – is not a cube of an integer. If $b=c=1$ $2^{a!}+2^{b!}+2^{c!}\equiv 1+2+2\equiv 5(\mathrm{mod}7)$ – is not a cube of an integer.

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Case 3. $2\ge a\ge b\ge c$, here we have 4 cases overall, from which we simply find a single answer in a simple overview.