Problem 947

Divide by 3. We now want to count the number of six-digit multiples of 9 that only have 1, 2, or 3 as their digits. Due to the divisibility rule for 9, we only need to consider when the digit sum is a multiple of 9. Note that $3 \cdot 6=18$ is the maximum digit sum. If the sum is 18, the only case is 333333. Otherwise, the digit sum is 9. The possibilities here, up to ordering of the digits, are 111222 and 111123. The first has $\binom{6}{3}=20$ cases, while the second has $6 \cdot 5=30$. Thus the final answer is $1+20+30=51$.