Number Theory - Proof of divisibility by $3$ Prove that for every positive integer $x$ of exactly four digits, if the sum of digits is divisible by $3$, then $x$ itself is divisible by 3 (i.e., consider $x = 6132$, the sum of digits of $x$ is $6+1+3+2 = 12$, which is divisible by 3, so $x$ is also divisible by $3$.)
How could I approach this proof? I'm not sure where I would even begin. 

Actually, this is true for an integral number with any digits. The proof is quite easy. Let's denote the integral number by $\overline{a_n a_{n-1} \ldots a_1}$. If the sum of its digits $\sum_{i=1}^n{a_i}$ is divisible by 3, then $\sum_{i=1}^n{(1+\overline{9...9}_{i-1})*a_i}$ is too. Here $\overline{9...9}_{i-1}$ denotes the integer with $i-1$ 9's. But this second sum is just the original number $\overline{a_n a_{n-1} \ldots a_1}$ expanded.
