2. Find any seven consecutive natural numbers, each of which can be changed (increased or decreased) by 1 in such a way that the product of the seven resulting numbers equals the product of the seven original numbers. (Method Commission of the All-Russian Olympiad)
Solution. Solution. For example, the numbers from 3 to 9 fit: replace 3 with 2, 4 with 5, 5 with 6, and the numbers in each of the pairs $(6,7)$ and $(8,9)$ with each other. In the end, we get $3 \cdot 4 \cdot 5 \cdot 6$ $\cdot 7 \cdot 8 \cdot 9=2 \cdot 5 \cdot 6 \cdot 7 \cdot 6 \cdot 9 \cdot 8$

Remark. If it was possible to change each of $n$ consecutive natural numbers $m, \ldots, m+n-1$ by 1 in such a way that their product remains the same, then the same can be done with $n+2$ consecutive natural numbers $m, \ldots, m+n-1, m+n, m+n+1$: it is sufficient to add the replacements $(m+n) \leftrightarrow(m+n+1)$ to the suitable replacements of the numbers $m, \ldots, m+n-1$. This is exactly how the example in our solution was constructed: first, three suitable numbers $3,4,5$ were found, and then the pairs of numbers 6,7 and 8, 9 were added. By adding the next pairs, we can obtain examples for any odd number of consecutive natural numbers. Examples for any even number of consecutive natural numbers can also be constructed in an obvious way.