Suppose a 3 digit number abc is divisible by 3. Prove that abc + bca + cab is divisible by 9.
We have a 3 digit number abc that is divisible by 3.
We know by divisibility rule of 3, if we take sum of the digits and the result is divisible by 3, then the original number is divisible by 3.
Therefore, we can say that:
(a+b+c) is also divisible by 3 ......... (1)
As given abc+bca+cab, so we can write it as:
100a+10b+c+100b+10c+a+100c+10a+b⇒111a+111b+111c⇒111(a+b+c)
Now we conclude that, 111 is divisible by 3 and
(a+b+c) is also divisible by 3 [ From (1) ]
Therefore, 111(a+b+c) is divisible by 9.
Hence, abc+bca+cab is divisible by 9.