



 
Hi Ray, I'm going to illustrate with an example, $n = 4357.$ \begin{eqnarray*} where $a = 4 \times 999 + 3 \times 99 5 \times 9$ which is divisible by $3,$ and $s$ is the sum of the digits of $n.$ The above development shows that any integer $n$ can be written \[n = a + s\] where $a$ is divisible by $3$ and $s$ is the sum of the digits of $n.$ If $n$ is divisible by 3 then, since $a$ is divisible by 3, $s$ is also divisible by 3. If $s$ is divisible by 3 then, since $a$ is divisible by 3, $n$ is also divisible by 3. I hope this helps, 



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