11.1. In the product of seven natural numbers, each factor was decreased by 3. Could the product have increased exactly 13 times as a result?

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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 incorrectly given - 5 points.