Question

Find the number of positive integers with three not necessarily distinct digits, $abc,$ with $a \ne 0$ and $c \ne 0$ such that both $abc$ and $cba$ are multiples of $4.$

Answer

Solution:

An integer is a multiple of $4$ exactly when its last two digits form a multiple of $4,$ so we need $4 \mid 10b + c$ and $4 \mid 10b + a.$ In particular $a$ and $c$ are even, and subtracting the two conditions shows $4 \mid a - c.$ 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 $4 \cdot 5 + 4 \cdot 5 = 40.$

2012 AIME I Problem 1

Solution:

Problem 1 in Other Years