If the sum of the digits is divisible by nine, so is the number.

Proof:

Every integer n can be expressed as n = a1*(10^k) + a2*(10^k-1)+ .....+ a_k+1 where a1, a2, a3, ...a_k+1 are integers between 0 and 9. Note that 10 is congruent to 1 (mod 9). Thus 10^k is congruent to 1 (mod 9) for every k >= 0.Thus n is congruent to (a1+a2+a3+....+a_k+1) mod(9). Hence (a1+a2+...+a_k+1) is divisible by 9 iff n is divisible by 9.

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