AMC12 2018 B

Answer (A): Let $abc$ be a 3-digit positive odd multiple of 3 that does not include the digit 3. There are 8 possible values for $a$, namely 1, 2, 4, 5, 6, 7, 8, and 9, and 4 possible values for $c$, namely 1, 5, 7, and 9. The possible values of $b$ can be put into three groups of the same size: {0, 6, 9}, {1, 4, 7}, and {2, 5, 8}. Recall that an integer is divisible by 3 if and only if the sum of its digits is divisible by 3. Thus for every possible pair of digits $(a, c)$, the choices for $b$ such that $abc$ is divisible by 3 constitute one of those groups. Hence the answer is $8\cdot4\cdot3=96$.