The rule of the divisibility for 3 says that if a number is divisible by 3, the sum of all digits in each place of the number is divisible by 3.

For example, 123, all digits are added up to 6, which’s divisible by 3. Hence, 123 is divisible by 3 without leaving remainder.

123 \div 3=41

Then how to prove the rule in general form?

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

a=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

Let’s denote the quotient as q_n

\therefore \dfrac{a}{3}=a_nq_n+a_{n-1}q_{n-1}+\dots+3a_1+\dfrac{a_n+a_{n-1}+\dots+a_1+a_0}{3}

Therefore, if a 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.