Kuzneuov A.

Sasha chose a natural number $N>1$ and wrote down in ascending order all its natural divisors: $d_{1}<\ldots<d_{s}$ (so that $d_{1}=1$ and $d_{s}=N$). Then for each pair of adjacent numbers, he calculated their greatest common divisor; the sum of the resulting $s-1$ numbers turned out to be

$N-2$. What values could $N$ take?
Note that $d_{s+1-i}=N / d_{i}$ for all $i=1,2,3, \ldots, s$.

The number $d_{i+1}-d_{i}$ is divisible by $\left(d_{i}, d_{i+1}\right)$, so $\left(d_{i}, d_{i+1}\right) \leq d_{i+1}-d_{i}$. For $i=1,2, s-1$, let $r_{i}=\left(d_{i+1}-d_{i}\right)$
$-\left(d_{i}, d_{i+1}\right) \geq 0$. By the condition

$\left(d_{2}-d_{1}\right)+\left(d_{3}-d_{2}\right)+\ldots+\left(d_{s}-d_{s-1}\right)=d_{s}-d_{1}=N-1$, and $\left(d_{1}, d_{2}\right)+\left(d_{2}, d_{3}\right)+\ldots+\left(d_{s-1}, d_{s}\right)=N-2$.

Subtracting, we get $r_{1}+r_{2}+\ldots+r_{s-1}=1$. This means that $r_{k}=1$ for some $k$, and all other $r_{i}$ are zero.

The left side of the equation $\left(d_{k+1}-d_{k}\right)-\left(d_{k}, d_{k+1}\right)=1$ is divisible by $\left(d_{k}, d_{k+1}\right)$, so $\left(d_{k}, d_{k+1}\right)=1$ and $d_{k+1}-d_{k}=2$. This is possible only if both $d_{k}$ and $d_{k+1}$ are odd.

Since $d_{k}$ and $d_{k+1}$ are two consecutive divisors of $N$, then ${ }^{N} / d_{k+1}$ and ${ }^{N} / d_{k}$ are also two consecutive divisors of $N$: if ${ }^{N} / d_{k+1}=d_{m}$, then

$N / d_{k}=d_{m+1}$. In this case, $\left(d_{m}, d_{m+1}\right)=N /\left[d_{k}, d_{k+1}\right]=N\left(d_{k}, d_{k+1}\right) / d_{k} d_{k+1}0$, which is possible only when $k=m$ (and, consequently, $s=2 k$ ).

Thus, $d_{k+1}=N / d_{k}$, meaning the number $N=d_{k} d_{k+1}$ is odd. But then $d_{s-1} \leq N / 3$, from which $\left(d_{s-1}, d_{s}\right) \leq d_{s-1} \leq N / 3$.

Therefore, $1 \geq r_{s-1} \geq 2 N / 3-N / 3=N / 3$, so $N \leq 3$. Since $N>1$, we get the only possible value $N=3$, which, as is easy to verify, satisfies the condition.

## Answer

$N=3$.