4. On the board, there is a four-digit number that has exactly six positive divisors, of which exactly two are single-digit and exactly two are two-digit. The larger of the two-digit divisors is a square of a natural number. Determine all numbers that could be written on the board.
SOLUTION. The number of positive divisors of a number whose prime factorization has the form $n=$ $=p_{1}^{k_{1}} p_{2}^{k_{2}} \ldots p_{r}^{k_{r}}$, is $\left(k_{1}+1\right)\left(k_{2}+1\right) \ldots\left(k_{r}+1\right)$. Therefore, a number that has exactly $6=3 \cdot 2$ positive divisors must have one of the forms $p^{5}$ or $p^{2} q$, where $p$ and $q$ are prime numbers.

First, consider the possibility $p^{5}$. This number has divisors $1, p, p^{2}, p^{3}, p^{4}, p^{5}$; clearly $1q$, then $199$, $99999$ and $99999$ and $999<49 q<10000$; these conditions are satisfied by $q \in\{23,29,31,37,41,43,47\}$.

Thus, one of seven numbers is written on the board: $49 \cdot 23=1127,49 \cdot 29=1421$, $49 \cdot 31=1519,49 \cdot 37=1813,49 \cdot 41=2009,49 \cdot 43=2107,49 \cdot 47=2303$.

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