2. Petya found the sum of all odd divisors of some even natural number $n$, and Vasya - the sum of all its even divisors. Can the product of their results be equal to 2016? If so, find all such numbers $n$.
Answer: 88 and 192.

Solution. Let $n=m \cdot 2^{k}$ be the initial even number, and $m$ be its odd factor. The sum of the odd divisors of the number $n$ coincides with the sum $s$ of all divisors of the number $m$, and thus, Petya will get the number $s$. The sum of all even divisors of $n$ consists of the sum of divisors divisible only by 2, the sum of divisors divisible only by 4, and so on, and finally, the sum of divisors divisible by $2^{k}$. Each odd divisor, when multiplied by 2, becomes a divisor divisible only by 2, so the first sum is $2s$, the second sum is $4s$, and so on. Therefore, the sum of all even divisors will be $2s + 4s + \ldots + 2^{k}s = 2(2^{k} - 1)s$. The product of Petya's and Vasya's results is $2s^{2}(2^{k} - 1) = 2016$, and thus, $s^{2}(2^{k} - 1) = 1008$.

The obtained equation means that the number $1008 = 24 \cdot 32 \cdot 7$ is divisible by the square of a natural number $s > 1$ and a number of the form $2^{k} - 1$. Among the factorizations of the number $1008 = 2^{2} \cdot 252 = 3^{2} \cdot 112 = 4^{2} \cdot 63 = 6^{2} \cdot 28 = 12^{2} \cdot 7$, containing the square of a natural number, the factor of the form $2^{k} - 1$ is present only in the factorizations $4^{2} \cdot 63$ and $12^{2} \cdot 7$. Therefore, $s = 4, k = 6$, or $s = 12, k = 3$. In the first case, $m = 3$, and thus, $n = 3 \cdot 2^{6} = 192$; in the second case, $m = 11$, and thus, $n = 11 \cdot 2^{3} = 88$.

Criteria. In the case of a fully justified solution - a score of 7 points.