A number theory problem by Nikkil V

For $abc$ to be divisible by $4$ we require only that $bc$ be divisible by $4$ . Allowing for $b = 0$ there are $25$ such two-digit "constructs", $5$ of which have $c = 0$ , (namely $00, 20, 40, 60$ and $80$ ), leaving us with $20$ allowable constructs $bc$ .

Then for $cba$ to also be divisible by $4$ we require that $ba$ be divisible by $4$ . For each of the previous $20$ constructs $bc$ there will be $2$ corresponding constructs $ba$ that will be divisible by $4$ such that $a \ne 0$ , (i.e., $2$ for each value of $b$ from $0$ through $9$ ), giving us a total of $2 \times 20 = \boxed{40}$ desired integers.