1. Given a prime number $p$. All natural numbers from 1 to $p$ are written in a row in ascending order. Find all $p$ for which this row can be divided into several blocks of consecutive numbers so that the sums of the numbers in all blocks are equal.
Answer: $p=3$. Let $k-$ be the number of blocks, $S-$ be the sum in each block. Since $p(p+1) / 2=k S$, either $k$ or $S$ is divisible by $p$. Clearly, $k<p$, so $S$ is a multiple of $p$. Let the leftmost group consist of numbers from 1 to $m$. Then $m(m+1) / 2$ is a multiple of $p$, from which $m \geqslant p-1$, i.e., $m=p-1$. But this means that the sum of the numbers from 1 to $p-1$ is equal to $p$, which is only possible when $p=3$.