Numbers Divisible By Three

Why is it that if a numbers is divisible by three, then the digits are also divisible by three?

Consider every integer $x$ can be written as

$x = \sum_{i \in Z} a_i 10^i$

where $Z$ is the set of all integers, and \(a_i \in {0, 1, …, 9}\) for all $i \in Z$. This can be rewritten as

$x = \sum_{i \in Z} a_i (10^i-1+1)$

It follows that if $10^i -1$ is divisible by three for all $i \in Z^+$ then it follows that $x$ is divisible by three if the sum of the digits are also divisible by three since:

$\sum_{i \in Z} a_i (10^i-1+1) \mod 3 = \sum_{i \in Z} a_i \mod 3$

Lets begin by proving that $10^i -1$ is divisible by three for all $i \in Z$.

If $i = 0$, then we know that $0$ is clearly divisible by three. We can also easily verify this is true for $i = 1$ since $10-1 = 9$ is clearly divisible by three.

Now assuming that case $i=k$ holds, we need to demonstrate that case $k+1$ also hold.

So assuming that

$10^k -1 \mod 3 = 0$

then

$10^{k+1} - 1 \mod 3 = (10^{k} - 1 + 1) \times 10 - 1 \mod 3$

Since we have assumed that $10^k -1 \mod 3 = 0$, the expression above simplifies to:

$10-1 \mod 3 = 9 \mod 3 = 0$

as required.

Therefore since

$x = \sum_{i \in Z} a_i 10^i = \sum_{i \in Z} a_i (10^i-1+1)$

Then

$x = \sum_{i \in Z} a_i (10^i-1+1) \mod 3 = \sum_{i \in Z} a_i \mod 3$

as required.