2. When dividing a three-digit number by a two-digit number, we get the sum of the digits of the divisor as the quotient, and the remainder is the number formed by swapping the digits of the divisor. If we multiply this remainder by the quotient and add the divisor, the resulting number consists of the digits of the dividend, but in reverse order. What is the dividend and the divisor?
Let the digits of the three-digit dividend and the two-digit divisor be denoted by $x, y, z$ and $u, v$ respectively. The conditions can be written in the following form:

$$
\begin{aligned}
& 100 x + 10 y + z = (10 u + v)(u + v) + 10 v + u \\
& 100 z + 10 y + x = (10 v + u)(u + v) + 10 u + v
\end{aligned}
$$

and since the remainder is smaller than the divisor, it must be that

$$
u > v
$$

Subtracting (2) from (1) yields

$$
99(x - z) = 9(u - v)(u + v) + 9(v - u) = 9(u - v)(u + v - 1)
$$

which simplifies to

$$
11(x - z) = (u - v)(u + v - 1)
$$

The right-hand side is divisible by 11. However, $u - v$ cannot be divisible by 11 because, as the difference of digits, it is less than 10 and, according to (3), positive. Since 11 is a prime number, if it divides a product, it must divide one of the factors. Thus, $u + v - 1$ is divisible by 11, but it is less than $2 \cdot 11$ because $u$ and $v$ are digits, and it cannot be 0 because then, considering (3), $-10 u + v = 10$ would be true, which is not a solution to the problem. Therefore, $u + v - 1 = 11$, which means

$$
u + v = 12
$$

From (1),

$$
100 x + 10 y + z = (10 u + v) \cdot 12 + 10 v + u = (9 u + 12) \cdot 12 + 120 - 10 u + u = 264 + 99 u
$$

The left-hand side is less than 1000, so

$$
99 u \leq 999 - 264, \quad u \leq \frac{725}{99} = 7 \frac{32}{99}
$$

Considering (3) and (5), we get $u = 7, v = 5$. Substituting these into the right-hand sides of (1) and (2),

$$
\begin{aligned}
& (10 u + v)(u + v) + 10 v + u = 75 \cdot 12 + 57 = 957 \\
& (10 v + u)(u + v) + 10 u + v = 57 \cdot 12 + 75 = 759
\end{aligned}
$$

Thus, we have found that the problem has one solution where the dividend is 957 and the divisor is 75.

Remark: With the relationship (5) at hand, we can also directly determine $x, y$, and $z$. Adding the equalities (1) and (2) and using (5),

$$
\begin{aligned}
& 101(x + z) + 20 y = 11(u + v)(u + v) + 11(u + v) = \\
& = 11(u + v)(u + v + 1) = 1716
\end{aligned}
$$

Dividing the left-hand side by 20, $(x + z)$ leaves a remainder of 16, so these two numbers can only differ by a multiple of 20, but since $x$ and $z$ are digits,

$$
x + z = 16
$$

must hold, and substituting this into (6),

$$
y = 5
$$

follows. From (4) and (3), it follows that necessarily

$$
x > z
$$

16 can only be decomposed into two different single-digit addends as $9 + 7$, so $x = 9, y = 5, z = 7$, and from (1) or (2) and (5), $u$ and $v$ can also be determined. However, the line of thought followed above leads more quickly to the solution.