2005 Alabama ARML TST Problems/Problem 4

Problem

A number is divisible by $12$ if it is divisible by $3$ and $4$ . A number is divisible by $4$ if its last two digits are divisible by $4$ , so $4|\overline{B8} \Longrightarrow B = 0,2,4,6,8$ . A number is divisible by $3$ if the sum of its digits is $3$ , so

$3| s(2A08) = 10 + A \Longrightarrow A = 2,5,8$ - 3 solutions
$3| s(2A28) = 12 + A \Longrightarrow A = 0,3,6,9$ - 4 solutions
$3| s(2A48) = 14 + A \Longrightarrow A = 1,4,7$ - 3 solutions
$3| s(2A68) = 16 + A \Longrightarrow A = 2,5,8$ - 3 solutions
$3| s(2A88) = 18 + A \Longrightarrow A = 0,3,6,9$ - 4 solutions

These sum to $17$ .