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. 

It's due to radix representation being polynomial form in the radix, e.g. $\rm\ n = 4321 = p(10)\ $ for $\rm\ p(x) = 4\: x^3 + 3\: x^2 + 2\: x + 1\:.\:$ Thus mod $\rm\:3\::\ 10\equiv 1\ \Rightarrow\ p(10)\equiv p(1)\ =\ \sigma(n) :=$ sum of digits. Aternatively one may simply put $\rm\ x = 10\ $ in the $\:$ Factor Theorem $\rm\ \ x-1\ |\ p(x)-p(1)\:,\: $ hence $\rm\ 3\ |\ 9\ |\ p(10)-p(1) = n - \sigma(n)\:.\:$ This is a special case of casting out nines.
