   SEARCH HOME Math Central Quandaries & Queries from Ray: There is a rule that a number is divisible by 3 if the sum of its digits are divisible by 3 (for example, 81=8+1=9 {divisible by 3} and 33=3+3=6 {again, divisible by 3}) I know this works but I don't know why! Please help. Hi Ray,

I'm going to illustrate with an example, $n = 4357.$

\begin{eqnarray*}
n = 4357 &=& 4000 + 300 + 50 + 7\\
&=& 4 \times 1000 + 3 \times 00 + 5 \times 10 + 7\\
&=& 4(999 + 1) + 3(99 + 1) + 5(9 + 1) + 7\\
&=& 4 \times 999 + 3 \times 99 + 5 \times 9 + 4 + 3 + 5 + 7\\
&=& a + s
\end{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,
Penny       * Registered trade mark of Imperial Oil Limited. Used under license. Math Central is supported by the University of Regina and the Imperial Oil Foundation.