Using the fact that (b^n)-1 is divisible by (b-1) for any positive integer n and any integer b1 (the proof was given before), show that if the sum of the digits of a number in base b is divisible by (b-1), then the number itself is divisible by (b-1).

In the particular case of base 10, this becomes the well-known fact that if the sum of the digits is divisible by 9, then the number itself is divisible by 9.

Question

Using the fact that (b^n)-1 is divisible by (b-1) for any positive integer n and any integer b>1 (the proof was given before), show that if the sum of the digits of a number in base b is divisible by (b-1), then the number itself is divisible by (b-1).

In the particular case of base 10, this becomes the well-known fact that if the sum of the digits is divisible by 9, then the number itself is divisible by 9.