Find all pairs of digits $(a, b)$ such that the integer whose four digits are ab32 is divisible by 99.
Let $n$ be the integer formed by the two digits $a b$. We seek $a$ and $b$ such that $100 n + 32$ is divisible by 99.

Now, $100 n + 32 = n + 32 + 99 n$, so the condition is equivalent to $n + 32$ being divisible by 99.

Now, $0 \leqslant n \leqslant 99$, so $32 \leqslant n + 32 \leqslant 131$. The only integer between 32 and 131 that is divisible by 99 is 99, so the only solution is $n = 99 - 32 = 67$, that is, $a = 6$ and $b = 7$.