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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.
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