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.