Divisibility rule:

If a number is evenly divisible by three, the sum of its digits (and the (recursive) sum of its digits, known as the number's digital root) will be evenly divisible by three.

Example: 6927 is divisible by 3 because 6+9+2+7 = 24, 2+4 = 6, and both 24 and 6 are divisible by 3.

Why does this work?

It works because in a base-10 number system, the place values correspond to powers of 10:

The ones place consists of a single digit (0-9) multiplied by 10 to the zero power, the tens place consists of a single digit (0-9) multiplied by 10 to the first power, the hundreds place consists of a single digit (0-9) multiplied by 10 to the second power, the thousands place consists of a single digit (0-9) multiplied by 10 to the third power, and so on...

So, 6927 = 6(10^3) + 9(10^2) + 2(10^1) + 7(10^0) = 6(1000) + 9(100) + 2(10) +7(1) = 6000 + 900 + 20 + 7

If we generalize this pattern for all base-10 numbers, we get:

a(10^n)...v(10^4) + w(10^3) + x(10^2) + y(10^1) + z(10^0)

For the sake of simplicity, I will continue my explanation using the generic four digit number "wxyz" which expands to w(10^3) + x(10^2) + y(10^1) + z(10^0).

Now, it should be clear that...

w(10^3) + x(10^2) + y(10^1) + z(10^0) = w(1000) + x(100) + y(10) + z(1)

Using the distributive law of real numbers, we can transform w(1000) + x(100) + y(10) + z(1) into its equivalent w(999 + 1) + x(99 + 1) + y(9+1) + z(1) and further split this out into w(999) + w(1) + x(99) + x(1) + y(9) + y(1) + z(1). Because we know that any number evenly divisible by 9 is evenly divisible by 3, we can disregard w(999), x(99), and y(9) when trying to determine if the original number "wxyz" is divisible by 3.

When we toss out w(999), x(99), and y(9), we are left with w(1) + x(1) + y(1) + z(1) = w+ x + y + z. Thus, the sum of the digits will determine whether or not the number itself is evenly divisible by three.

And if w+x+y+z > 9, we can optionally reapply the digital summation process in order to find the single-digit digital root.