11-1. Given three numbers. If each of them is decreased by 1, then their product also decreases by 1. If all the original numbers are decreased by 2, then their product also decreases by 2.

a) By how much will the product decrease if all the original numbers are decreased by 3?

b) Provide any three numbers that satisfy the condition of the problem.
Answer: a) by 9; b) $1,1,1$.

Solution. a) Let the initial numbers be $a, b$, and $c$. Denote $ab + bc + ca = x$ and $a + b + c = y$. According to the condition, $(a-1)(b-1)(c-1) = abc - 1$ and $(a-2)(b-2)(c-2) = abc - 2$. Expanding the brackets and combining like terms, we get $x - y = 0$ and $-2x + 4y = 6$, from which $x = y = 3$. Then the required difference is $abc - (a-3)(b-3)(c-3) = 3x - 9y + 27 = 3 \cdot 3 - 9 \cdot 3 + 27 = 9$.

b) For example, the numbers $a = b = c = 1$ work. Remark. It can be proven that the given set of real numbers is the only one.

Criteria. 2 points for each correct answer in parts a) and b). Proving the relation $ab + bc + ca = a + b + c = 3$ - 3 points.