9.3. 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?

(A. Kuznetsov)
Answer: $N=3$.

Solution. Note immediately that $d_{s+1-i}=N / d_{i}$ for all $i=$ $=1,2, \ldots, s$.

The number $d_{i+1}-d_{i}$ is divisible by the GCD $\left(d_{i}, d_{i+1}\right)$, so the GCD $\left(d_{i}, d_{i+1}\right) \leqslant d_{i+1}-d_{i}$. For $i=1, \ldots, s-1$, let $r_{i}=\left(d_{i+1}-d_{i}\right)-$ GCD $\left(d_{i}, d_{i+1}\right) \geqslant 0$. According to 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

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

Subtracting the second equation from the first, we get $r_{1}+\ldots+r_{s-1}=$ $=1$. This means that $r_{k}=1$ for some $k$, while all other $r_{i}$ are zero.

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

Since $d_{k}$ and $d_{k+1}$ are two consecutive divisors of the number $N$, then $\frac{N}{d_{k+1}}$ and $\frac{N}{d_{k}}$ are also two consecutive divisors of the number $N$. Therefore, if $\frac{N}{d_{k+1}}=d_{m}$, then $\frac{N}{d_{k}}=d_{m+1}$. In this case,

$$
\begin{aligned}
\operatorname{GCD}\left(d_{m}, d_{m+1}\right)= & \frac{N}{\operatorname{LCM}\left(d_{k}, d_{k+1}\right)}=\frac{N \cdot \operatorname{GCD}\left(d_{k}, d_{k+1}\right)}{d_{k} d_{k+1}}=1$, which is possible only when $k=m$ (and, consequently, $s=2 k)$.

Thus, $d_{k+1}=\frac{N}{d_{k}}$, meaning the number $N=d_{k} d_{k+1}$ is odd. But then $d_{s-1} \leqslant \frac{N}{3}$, from which GCD $\left(d_{s-1}, d_{s}\right) \leqslant d_{s-1} \leqslant \frac{N}{3}$. Therefore, $1 \geqslant r_{s-1} \geqslant \frac{2 N}{3}-\frac{N}{3}=\frac{N}{3}$, i.e., $N \leqslant 3$. Since $N>1$, we get the only possible value $N=3$, which, as is easy to verify, satisfies the condition.