For n ? N, expressed in base 10, prove that n is divisible by 3 if and only if the sum of the digits of n is divisible by 3. Solution Let n be a four digit number n=10^3p+10^2q+10r+s=(999+1)p+(99+1)q+(9+1)r+s=(999p+99q+9r)+(p+q+r+s)=3(333p+33q+ 3r)+(p+q+r+s), so when you divide n by 3, you.