If $8A5146B$ is divisible by $88$, then what is the value of $A^B$?

For a number to be divisible by $88$, it must be divisible by both $8$ and $11$.1. Divisibility by 8:A number is divisible by 8 if its last three digits are divisible by 8. Here, the last three digits form the number $46B$.Dividing $46B$ by 8:
$$460÷8=57\text{ with a remainder of }4$$
For $46B$ to be perfectly divisible by 8, the number $4B$ must be a multiple of 8. The only single digit that satisfies this condition is $B=4$ ($44÷8$ leaves a remainder, but $48÷8$ is exact, wait: $464÷8=58$). Thus, $B=4$.2. Divisibility by 11:The number is now $8A51464$. A number is divisible by 11 if the difference between the sum of the digits at odd positions and even positions is either 0 or a multiple of 11.Sum of digits at odd positions (from right): $4+4+5+8=21$ Sum of digits at even positions (from right): $6+1+A=7+A$ $
$\text{Difference}=21-(7+A)=14-A$$For$ (14 - A)$tobeamultipleof11where$ A$isasingledigit($ 0 \le A \le 9$):$$14-A=11⟹A=3$$3.Valueof$ A^B$:$$A^B=3^4=81$ $