10.2. In the product of five natural numbers, each factor was decreased by 3. Could the product have increased exactly 15 times as a result? (N. Agakhanov, I. Bogdanov) Answer. Yes, it could.
Solution. As an example, the product $1 \cdot 1 \cdot 1 \cdot 1 \cdot 48$ fits. After the specified operation, it becomes $(-2) \cdot(-2) \cdot(-2) \cdot(-2) \cdot 45=720=15 \cdot 48$.

Remark. The given example is the only one. We will indicate how to come up with it. Suppose that four of the factors were equal to 1, and the fifth was $a$. Their product was $a$, and after the reduction, it became $(-2)^{4}(a-3)=16 a-48$. Therefore, for $16 a-48=15 a$ the condition is satisfied. Solving this equation, we get $a=48$.

Comment. Only the correct answer - 0 points.

The set of numbers satisfying the condition is presented - 7 points.

In an otherwise correct solution, due to an arithmetic error, the answer is the set $1,1,1,1, a$, where the value of $a$ is incorrectly -5 points.