3. A natural number $n$ is called pleasant if it satisfies the following properties:

- its decimal representation consists of 4 digits;
- the first and third digits of $n$ are equal;
- the second and fourth digits of $n$ are equal;
- the product of the digits of $n$ divides $n^{2}$.

Determine all pleasant numbers.
SOLUTION: Every pleasant number can be expressed as

$$
n=101 \cdot(10 a+b)
$$

where $a$ and $b$ are integers with $1 \leq a, b \leq 9$ (the digit zero is excluded because otherwise the product of the digits would be zero, which does not divide any positive number). The condition

$$
(a b)^{2} \mid n^{2}
$$

implies $a b \mid n$, so both $a$ and $b$ divide $n$. However, observe that 101 is a prime number, so $\operatorname{MCD}(a, 101)=$ $\operatorname{MCD}(b, 101)=1$. It follows that $a \mid 10 a+b$ and $b \mid 10 a+b$. Since obviously $a \mid 10 a$, we have $a \mid b$, and thus:

$$
a|b| 10 a \text {. }
$$

At this point, there are three possibilities:

- $b=a$;

- $b=2 a$;

- $b=5 a$.

The first implies $a^{2} \mid 11 a$, i.e., $a \mid 11$ and thus $a=b=1$. The second implies $2 a^{2} \mid 12 a$, i.e., $a \mid 6$, with the three solutions $(a, b) \in\{(1,2),(2,4),(3,6)\}$. The third implies $5 a^{2} \mid 15 a$, i.e., $a \mid 3$, with the only solution $(a, b)=(1,5)$. Finally, note that all the solutions found satisfy the conditions of the problem, so the numbers sought are: 1111, 1212, 2424, 3636, 1515.