## Problem 2.

Determine the smallest 100 consecutive natural numbers whose sum is divisible by 105.
Solution. Let $n, n+1, \ldots, n+99$ be the 100 consecutive numbers, $n \in \mathbb{N}$. Then their sum is $S=n+(n+1)+\ldots+(n+99)=100 n+\frac{99(99+1)}{2}=100 n+50 \cdot 99=$ $50(2 n+99)$. From $105=3 \cdot 5 \cdot 7 \mid S$ it follows that $21=3 \cdot 7 \mid(2 n+99)$, where $2 n+99 \geq 99$. The minimality condition from the problem's hypothesis leads to the conclusion that $2 n+99=105$, from which $n=3$. Therefore, the numbers sought are: $3,4,5, \ldots 100,101,102$.