No one taught me this in school, but if you add up the digits (5+7=12) and the result is divisible by 3, then so is the original number. It works recursively.
Induction seems like an odd way to prove this. Do you need to preform induction over all the numbers that are not multiples of three?
The converse claim (every multiple of 3 has a digit sum that is a multiple of 3) is a more natural one for induction, though that's not the most standard proof there either.
If induction is the only tool you have then it’s ok. What you really want is modular arithmetic. If you don’t have modular arithmetic then pricing it directly is hard:
n = Sum(d_k 10^k)
Let S = sum of digits = Sum(d_k)
n - S = Sum(d_k (10^k - 1))
But (10^k - 1) = ((9 + 1)^k - 1) = ((1 + k*9 + (2 choose k)*9^2 + ... + 9^k) - 1)
by binomial expansion so is divisible by 9.
Therefore n - S is divisible by 9, so n is divisible by 9 iff S is.
The same is true when you replace divisibility by 9 with divisibility by 3.
I remember being asked this in an interview for a place at university and I solved it by induction (which was all I had) and was then shown this direct proof.
Odd if you're looking for the simplest way to prove that. Less odd, perhaps, if you're looking for something that can be proved simply using induction, in order to make students do their first proof by induction.
I don't remember precisely what it was we were asked to prove. It may have been the converse.
As dan-robertson mentioned in another branch, you don't even need induction - you can sidestep it by writing X = sum_i x_i 10^i and noticing that all the 10^i are 1 modulo 9 (and therefore modulo 3).
"What is Mathematics" by Courant and Robbins teaches some more divisibility tricks, might be useful sometimes. btw, we learned divisibility by 3 in 5th or 6th grade (not the proof, of course)
