Problem 1295

1. Let $\overline{AB}$ represent the two-digit number formed by the digits $A$ and $B$. Thus, $\overline{AB} = 10A + B$.
2. Similarly, let $\overline{BA}$ represent the two-digit number formed by the digits $B$ and $A$. Thus, $\overline{BA} = 10B + A$.
3. The product of these two numbers is given by:
$$k = (10A + B)(10B + A)$$
4. Expanding the product, we get:
$$k = 10A \cdot 10B + 10A \cdot A + B \cdot 10B + B \cdot A$$
$$k = 100AB + 10A^2 + 10B^2 + AB$$
$$k = 101AB + 10A^2 + 10B^2$$
5. We are given that $k + 1$ is a multiple of $101$. Therefore:
$$k + 1 = 101AB + 10A^2 + 10B^2 + 1$$
must be divisible by $101$.
6. Since $101AB$ is already divisible by $101$, we need:
$$10A^2 + 10B^2 + 1 \equiv 0 \pmod{101}$$
7. Simplifying, we get:
$$10A^2 + 10B^2 \equiv -1 \pmod{101}$$
$$10(A^2 + B^2) \equiv -1 \pmod{101}$$
8. Dividing both sides by $10$, we get:
$$A^2 + B^2 \equiv -\frac{1}{10} \pmod{101}$$
Since $10^{-1} \equiv 91 \pmod{101}$ (because $10 \cdot 91 \equiv 1 \pmod{101}$), we have:
$$A^2 + B^2 \equiv -91 \pmod{101}$$
$$A^2 + B^2 \equiv 10 \pmod{101}$$
9. We need to find digits $A$ and $B$ (between $0$ and $9$) such that:
$$A^2 + B^2 = 10$$
10. Checking possible values, we find:
- $A = 1$ and $B = 3$ (since $1^2 + 3^2 = 1 + 9 = 10$)
- $A = 3$ and $B = 1$ (since $3^2 + 1^2 = 9 + 1 = 10$)

11. Therefore, the possible values for $A$ and $B$ are $(1, 3)$ and $(3, 1)$.

12. Calculating $k$ for these values:
- For $A = 1$ and $B = 3$:
$$k = (10 \cdot 1 + 3)(10 \cdot 3 + 1) = 13 \cdot 31 = 403$$
- For $A = 3$ and $B = 1$:
$$k = (10 \cdot 3 + 1)(10 \cdot 1 + 3) = 31 \cdot 13 = 403$$

Thus, in both cases, we get the same value for $k$.

The final answer is $\boxed{403}$