sum digits

Question

an integer n is divisible by 3 if and only if the sum of digits of n is divisible by 3. In other words, if n= a₀+ 10 a₁ +10² a₂ +…+ 10^k a_k, then n is divisible by 3 if and only if a₀+a₁+a₂+…+a_k is divisible by 3. Example: 12765312 is divisible by 3 because 1+2+7+6+5+3+1+2=27 is divisible by 3; indeed, 12765312 / 3 = 4255104.

 

Answer 1

10=9+1; 100=99+1, 1000=999+1, and so on.

3 goes into (10^k)-1, where k is a positive integer.

n=a0+9a1+a1+99a2+a2+...

So, if we divide by 3 we can ignore (10^k)-1 because we know they are all divisible by 3.

That leaves us with a0+a1+a2+... which is simply the sum of the digits. If this sum is divisible by 3 then so is n.