[ $\left.\quad \begin{array}{lc}{\left[\begin{array}{l}\text { Processes and operations }\end{array}\right]} \\ {[} & \text { Semivariants }\end{array}\right]$

Authors: Fadin M. Kovalenko K.

Initially, a natural number $N$ is written on the board. At any moment, Misha can choose a number $a>1$ on the board, erase it, and write down all natural divisors of $a$, except for $a$ itself (the same numbers can appear on the board). After some time, it turned out that there are $N^{2}$ numbers on the board. For which $N$ could this have happened?

#
Let $N>1$, and $1=d_{1}<d_{2}<\ldots<d_{k}<d_{k+1}=N-$ be all the divisors of $N$.

Notice that $d_{i} d_{k+2-i}=N$. Therefore,

$d_{1}^{2}+d_{2}^{2}+\ldots+d_{k}^{2}=\frac{N^{2}}{d_{k+1}^{2}}+\frac{N^{2}}{d_{k}^{2}}+\ldots+\frac{N^{2}}{d_{2}^{2}} \leq N^{2}\left(\frac{1}{2^{2}}+\frac{1}{3^{2}}+\ldots+\frac{1}{N^{2}}\right)<N^{2}$ (the last inequality follows from problem $\underline{30898})$.

From this, it follows that

1) on the first step, the sum of the squares of the numbers written on the board decreases;
2) in the subsequent steps, it does not increase.

Thus, the sum of the squares of the numbers written on the board at all moments, except the initial one, is less than $N^{2}$. Since all numbers are natural, their quantity is also less than $N^{2}$.

## Answer

Only for $N=1$.