Problem 683

[Solution] Let the two-digit number be $a=10 x+y$, then its reverse number is $10 y+x$, where $1 \leqslant x \leqslant 9, \quad 0 \leqslant y \leqslant 9, \quad x, y \in Z$.
From the problem, we have
$$10 x+y=(10 y+x) q+q, \quad q \in N .$$

That is
$$(10-q) x-(10 q-1) y=q \text {. }$$
(1) When $q=1$, (1) becomes $9 x-9 y=1$.
This is impossible;
(2) When $q=2$, (1) becomes $8 x-19 y=2$.
Thus, $y$ is even.
When $y=2$, $x=5$, at this time $a=52$.
When $y=4,8$, the right side of the equation cannot be divided by 4, while the left side can be divided by 4, so there is no solution.
When $y=6$, it is easy to verify that there is no solution.
Therefore, when $q=2$, $a=52$.
(3) When $q=3$, (1) becomes
$$7 x-29 y=3 .$$

When $y \leqslant 2$, $x$ is not an integer.
When $y \geqslant 3$,
$$7 x=29 y+3 \geqslant 90, \quad x>10 .$$

So, when $q=3$, there is no solution.
(4) When $q=4$, (1) becomes
$$6 x-39 y=4 \text {. }$$

The left side of this equation can be divided by 3, but the right side cannot be divided by 3, so there is no integer solution.
(5) When $q \geqslant 5$, we have
$$\begin{aligned} 5 x & \geqslant(10-q) x \\ & =(10 q-1) y+q \\ & \geqslant 49 y+q . \end{aligned}$$

If $y \geqslant 1$, then $5 x \geqslant 49 y+q \geqslant 54$, thus $x>10$, which is impossible.
If $y=0$, then from (1) we get
$$q=(10-q) x \geqslant 10-q .$$

This gives $x=1, q=5$ or $x=4, q=8$, which is also impossible.
In summary, the only solution to this problem is $a=52$.