11.4. In the product of seven natural numbers, each factor was decreased by 3. Could the product have increased exactly 13 times? (N. Agakhanov, I. Bogdanov)
Answer. Yes, it could.

Solution. As an example, the product $1 \cdot 1 \cdot 1 \cdot 1 \cdot 1 \cdot 2 \cdot 16=32$ fits. After the specified operation, it becomes $(-2) \cdot(-2) \cdot(-2) \cdot(-2) \cdot(-2) \cdot(-1) \cdot 13=13 \cdot 32$.

Remark 1. Let's explain how to come up with this example. Suppose five of the factors were 1, the sixth was 2, and the seventh was $a$. Their product was $2a$, and after the reduction, it became $(-2)^{5}(-1)(a-3)=32a-96$. Therefore, for $32a-96=26a$, the condition is satisfied. Solving this equation, we get $a=16$. Thus, the numbers $1,1,1,1,1,2,16$ meet the requirements.

Remark 2. The given example is not the only one.

For instance, the product $1^{4} \cdot 29 \cdot 61 \cdot 64$ also works, transitioning after subtraction to $(-2)^{4} \cdot 26 \cdot 58 \cdot 61$.

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 set $1,1,1,1,1,2, a$ is obtained, where the value of $a$ is incorrect - 5 points.