Problem 832

Solution:
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 $1 + 20 + 30 = 51$.