2012 AIME I 第 1 题

Divisibility by $4$ depends only on the last two digits, so $4$ must divide both $10b + c$ and $10b + a$

Subtracting shows $a - c$ is divisible by $4$ with $a$ and $c$ even, so both lie in $\{2, 6\}$ or both in $\{4, 8\};$ then $c$ forces the parity of $b$

An integer is a multiple of $4$ exactly when its last two digits form a multiple of $4,$ so we need $10b + c$ and $10b + a$ to be divisible by $4.$ In particular $a$ and $c$ are even, and subtracting the two conditions shows $a - c$ is divisible by $4.$ The even nonzero digits split by remainder mod $4$ into $\{2, 6\}$ and $\{4, 8\},$ so $a$ and $c$ must both come from the same one of these sets: $4$ ordered pairs $(a, c)$ from each.

If $c \equiv 2 \pmod 4,$ then $10b + c \equiv 2b + 2 \pmod 4$ requires $b$ odd ($5$ choices), and the condition on $10b + a$ holds automatically since $a \equiv c \pmod 4.$ If $c \equiv 0 \pmod 4,$ then $b$ must be even ($5$ choices).

The count is $4 \cdot 5 + 4 \cdot 5 = 40.$