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Let $a=(a_na_{n-1}…a_o)_b$ and $s=\sum_{i=0}^na_i$ Show that if $b-1 \mid a \iff b-1\mid s $.

I’ve found this problem in the number theory book that i read.

I think if the Base was $10$, it would be more simple, however i don’t know how to prove this.

PNT
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  • Write down a formula for the value of $a$ in terms of $b$ and the $a_k$. The proof is the same in base $10$ and base $b$, so do it first in base $10$ if that's easier for you. – saulspatz Feb 21 '21 at 13:04
  • This is simply the radix $b$ analog of casting out nines. The proof is exactly the same as the well-known proof in radix $10$ - see the linked dupe. – Bill Dubuque Feb 21 '21 at 18:41

1 Answers1

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You have

$$a=a_nb^n + a_{n-1}b^{n-1} + ... + a_1b + a_0$$

and because $b \equiv 1$ modulo $b-1$, you get $$a \equiv a_n + a_{n-1} + ... + a_1 + a_0\quad [b-1]$$ i.e. $$a \equiv s \quad [b-1]$$

The result follows.

TheSilverDoe
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