The rule of divisibility for 6 says that if a number is divisible by 6, it is divisible by both 2 and 3. In another words, it must be an even number and the sum of all digits on each place of the number is divisible by 3.

For example, 432, all digits are added up to 9, which is divisible by 3 and 2. Hence, 432 is divisible by 6 without leaving remainder.

Then how to prove the rule in general form?

Suppose any given integer p divided by 6

Since 2 is a factor for 6, p must be an even number to be divisible by 6.

Let p = 2q, then

So if p is divisible by 6, its factor p must be divisible by 3. Let’s first evaluate 3-digit number that is denoted as \overline{abc}, in which a, b, c are three digits at hundreds, tens and ones places respectively. Since a number can be expressed as the sum of its place values, \overline{abc} is rewritten as

\overline{abc}=100a+10b+c

=99a+a+9b+b+c

=(99a+9b)+(a+b+c)

Divide both sides by 3

\dfrac{\overline{abc}}{3} =\dfrac{99a+9b}{3}+ \dfrac{a+b+c}{3}

=33a+3b+\dfrac{a+b+c}{3}

Now it’s clear that if left hand side is divisible by 3, the sum of a, b, c in the right hand side must be divisible by 3.

In a more general form, any integers can be expressed as

q=a_n10^n+a_{n-1}10^{n-1}+\dots+a_210^2+a_110+a_0

=a_n(10^n-1+1)+a_{n-1}(10^{n-1}-1+1)+\dots+a_1(10-1+1)+a_0

=a_n(10^n-1)+a_{n-1}(10^{n-1}-1)+\dots+9a_1+ ( a_n+a_{n-1}+\dots+a_1+a_0)

\because 10^n-1

=9\cdotp 10^{n-1}+9\cdotp 10^{n-2}+\dots+9

which is divisible by 3, that is

\dfrac{10^n-1}{3}

=3\cdotp 10^{n-1}+3\cdotp 10^{n-2}+\dots+3

\therefore \dfrac{q}{3}=3\cdotp 10^{n-1}+3\cdotp 10^{n-2}+\dots+3+\dfrac{a_n+a_{n-1}+\dots+a_1+a_0}{3}

Therefore, if q is divisible by 3 on the left hand side, the sum of all digits of the number must be divisible by 3 on the right hand side.

In summary, if the number p is divisible by 6, it must be an even number and the sum of all digits on each place of the number is divisible by 3.