Let $A$ and $B$ be digits between $0$ and $9$, and suppose that the product of the two-digit numbers $\overline{AB}$ and $\overline{BA}$ is equal to $k$. Given that $k+1$ is a multiple of $101$, find $k$.

[i]Proposed by Andrew Wu[/i]
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 $\boxed{403}$