Makkin M.I.

Can all natural divisors of the number 100! (including 1 and the number itself) be divided into two groups such that both groups have the same number of numbers and the product of the numbers in the first group equals the product of the numbers in the second group?

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100 ! is divisible by $31^{3}$, but not by $31^{4}$. Therefore, all factors of the number 100 ! can be divided into quartets of the form $\left\{n, 31 n, 31^{2} n, 31^{3} n\right\}$, where $n-$ is an arbitrary divisor of the number $\frac{100!}{31^{3}}$. From each such quartet, the numbers $n$ and $31^{3} n$ will be placed in one group, and the numbers $31 n$ and $31^{2} n-$ in another.

## Otvet

It is possible.