1. In how many ways can the number 1991 be written as a sum of consecutive natural numbers?
Solution. Let

$$
n+(n+1)+\ldots+(n+k)=1991
$$

Then

$$
(k+1) n+\frac{k(k+1)}{2}=1991, \text { i.e. }(k+1)(2 n+k)=2 \cdot 11 \cdot 181
$$

Now, since $(k+1)(2 n+k)=2 \cdot 11 \cdot 181>k^{2}$, we get $k<64$, so $k+1 \in\{2,11,22\}$.

a) If $k+1=2$, then $k=1$ and $995+996=1991$.

b) If $k+1=11$, then $k=10$, so $2 n+10=362$, i.e., $n=176$. In addition, $176+177+\ldots+186=1991$.

c) If $k+1=22$, then $k=21,2 n+21=181$ and $n=80$. In addition, $80+81+\ldots+101=1991$.

Finally, from the above considerations, it follows that the number 1991 can be written in three ways as the sum of consecutive natural numbers.