A five digit number is divisible by 3 if the sum of its digits is divisible by 3 proof

Let 10000a+1000b+100c+10d+e be a 5-digit number, where a, b, c, d, e are integers between 0 to 9 inclusively.

Suppose a+b+c+d+e is divisible by 3.

Then there exists an integer M such that a+b+c+d+e=3M.

Then 10000a+1000b+100c+10d+e

=9999a+999b+99c+9d+3M

=3 (3333a+333b+33c+3d+M)

Since (3333a+333b+33c+3d+M) is an integer, 10000a+1000b+100c+10d+e is divisible by 3.