Question

Suppose a 3-digit number abc is divisible by 3. Prove that abc+bca+cab is divisible by 9.

Answer 1

abc is divisible by 3   =>  a+b+c = 3 k   for some integer k.

sum of three numbers abc , bca , cab
  = 100 a+ 10 b+ c + 100 b+ 10 c+ a+ 100 c+ 10 a+ b 
  = 100 (a+b+c) + 10 (b+c+a) + (c+a+b)
  = (a+b+c) * 111
   = 3 k * 3 * 37
   = 9 * 37 * k

Hence the answer..