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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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