## Task $8 / 85$

Determine the smallest natural number $n$ that can be represented both as the sum of 10 and as the sum of 794 consecutive (not necessarily natural) integers!
For the sum $n=s_{k}$ of $k$ consecutive integers, it is known that $2 s_{k}=k(2 a+k-1)$, where $a$ is the smallest of the $k$ integers. Therefore, the double sum is divisible by $k$ without a remainder. In our case, it is divisible by both 10 and 794, thus it is a common multiple of 10 and 794; due to the minimality of $n=s_{k}$, it is even the least common multiple:

$$
2 n=2 s_{k}=2 \cdot 5 \cdot 397=3970 \quad ; \quad n=s_{k}=5 \cdot 397=1985
$$

Indeed,

$$
1985=\sum_{i=194}^{203} i=\sum_{i=-394}^{399} i
$$