Suppose that $A$ and $B$ are digits with

| $A A A$ |
| ---: |
| $A A B$ |
| $A B B$ |
| $+\quad B B B$ |
| 1503 |

What is the value of $A^{3}+B^{2}$ ?
The integer with digits $A A A$ is equal to $100 A+10 A+A=111 A$.

Similarly, the integer with digits $A A B$ is equal to $110 A+B$, the integer with digits $A B B$ is equal to $100 A+11 B$, and the integer with digits $B B B$ is equal to $111 B$.

From the given addition, we obtain $111 A+(110 A+B)+(100 A+11 B)+111 B=1503$ or $321 A+123 B=1503$.

Dividing both sides by 3 , we obtain $107 A+41 B=501$.

Rearranging gives $41 B=501-107 A$.

Since $A$ and $B$ are positive integers, then $107 A<501$.

Since $107 \cdot 5=535$, then the possible values of $A$ are $1,2,3,4$.

If $A=1$, then $501-107 A=394$ which is not a multiple of 41 .

If $A=2$, then $501-107 A=287$ which equals $7 \times 41$.

(Neither $A=3$ nor $A=4$ makes $501-107 A$ a multiple of 41.)

Therefore, $A=2$ and $B=7$.

(Checking, we see that $222+227+277+777=1503$.)

Thus, $A^{3}+B^{2}=2^{3}+7^{2}=8+49=57$.

ANSWER: 57