# The Proof of the Divisibility by 3

In the previous post, we have discussed about divisibility by 2. In this post, we discuss about divisibility by 3.

Rule: A number is divisible by 3 if the sum of the digits is divisible by 3.

The number 321 is divisible by 3 because 3 + 2 + 1 = 6 is divisible by 3. On the other hand, the number 185 is not divisible by 3 because 1 + 8 + 5 = 14 is not divisible by 3. Now, why does this rule work?

Notice how the numbers are represented in expanded notation: $534 = 5(100) + 3(10) + 4$ $185 = 1(100) +8(10) + 5$

This means that number in hundreds can be represented as $100h + 10t + u$

where h, t, u are the hundreds, tens, and units digits. Now, we can represent $100h + 10t + u$ as $(99h + h) + (9t + t) + u$ and regroup the terms as $(99h + 9t) + (h + t + u)$.

Of course, $99h + 9t$ is divisible by 3, so it only remains to show that $h + t + u$ is divisible by 3. But, $h + t + u$ is the sum of the digits of a 3-digit number. This proves (for three digit numbers) that the rule above is true.

Although the proof above works only for 3-digit numbers,  it can be done to any number of digits.