9.1. In the product of three natural numbers, each factor was decreased by 3. Could the product have increased by exactly $2016$?

(N. Agakhanov, I. Bogdanov)
Answer. Yes, it could.

Solution. The product $1 \cdot 1 \cdot 676$ serves as an example. After the specified operation, it becomes $(-2) \cdot(-2) \cdot 673 = 2692 = 676 + 2016$.

Remark. The given example is the only one. Here is how to come up with it. Suppose two of the factors were 1, and the third was $-a$. Their product was $a$, and after the reduction, it became $(-2)^2(a-3) = 4a - 12$. Therefore, for $4a - 12 = a + 2016$, the condition is satisfied. Solving this equation, we get $a = 676$.

Comment. Only the correct answer - 0 points,

A set of numbers that satisfies the condition is presented - 7 points.

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