Prove the divisibility test for p 3 Specifically If 3 divide

Prove the divisibility test for p= 3.

Specifically: If 3 divides the sum of the digits of n, then n is divisible by 3

Suppose that you have a four-digit number n that is written abcd. Then

n = 10^3a+10^2b+10c+d

= (999+1)a+(99+1)b+(9+1)c+d

=(999a+99b+9c)+(a+b+c+d)

=3(333a+33b+3c)+(a+b+c+d)

divide n by 3, we’ll get

333a+33b+3c+(a+b+c+d)/3

The remainder is (a+b+c+d)/3, since 333a+33b+3c is an integer.

Thus proved