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True or false: If b & m are positive integers, then b|m iff the last b-adic digit $d_{0}$ of $m$ is $0$.

Here $m = d_{k}b^k + d_{k-1}b^{k-1} + ... + d_{0}$, & the numbers $d_{k}, ..., d_{0}$ are called b-adic digits of m.

Proof: I know I need to show proof of this for each way, (-->) & (<--). I have a Lemma which states that if $a$,$b$, & d are integers then $d|a + b$ iff $d | b$, which I think I could make use of. I don't understand why $d_{0}$ should be $0$ though, does this somehow offset potential division by integers?

Sorry if my formatting is off. I'm attempting to learn it properly!

Siminore
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BigD4J
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2 Answers2

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Hint: Recall that the decomposition of $m$ in $b$-ary means that $m=\sum_{k=0}^\infty d_k b^k$, where each $d_k\in\{0,\dots,b-1\}$ (and only finitely many of them are non-zero). What happens when you take the RHS modulo $b$?

(remember that $b\mid m$ is equivalent to $m = 0 \mod b$)

Clement C.
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  • We haven't dealt with "modulo" in my Algebra 1 class yet, is there another approach you might recommend? Thanks for your helpful edits! – BigD4J Feb 04 '15 at 14:25
  • Instead of the modulos, you can use the lemma you mention, it is the same up to the syntax: $m = d_0 + d_1 b + d_2 b^2 +\dots + d_K b^K$ (for some $K$). Clearly, $b\mid d_1 b$, $b\mid d_2 b^2$, ..., , $b\mid d_K b^K$.

    For the first direction, $if b\mid m$ then $b\mid d_0 + d_1 b + d_2 b^2 +\dots + d_K b^K$, applying the lemma shows that $b\mid d_0$.

    For the converse, do the same but on $d_0 = m - d_1 b + d_2 b^2 +\dots + d_K b^K$: if $b\mid d_0$, then $b\mid m - d_1 b + d_2 b^2 +\dots + d_K b^K$ and by the lemma $b\mid m$.

     – Clement C. Feb 04 '15 at 14:33
  • Great, thank you. What I still don't understand though is why d_0 must be 0. – BigD4J Feb 04 '15 at 14:43
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    If $b\mid d_0$ and $0 \leq d_0 \leq b-1$, then $d_0=0$: can you see why? (how many natural integers strictly smaller than $b$ can be divided by $b$?) – Clement C. Feb 04 '15 at 15:28
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Hint $\,\ \color{#c00}b\mid \color{#c00}b\, n+d_0\iff b\mid d_0.\ $ More conceptually we can exploit the polynomial form:

Remark $\ $ More generally $\ b\mid f(b)\iff b\mid f(0)\ $ for any polynomial $\,f(x)\,$ with integer coefficients, because, by the Polynomial Congruence Rule, $\ {\rm mod}\ b\!:\,\ b\equiv 0\,\Rightarrow\,f(b)\equiv f(0),\,$ therefore $\,f(b)\equiv 0\iff f(0)\equiv 0.\,$ This is simply a number theoretical form of the well known Polynomial Factor Theorem: $\ x\mid f(x)\iff x\mid f(0) \iff f(0) = 0.\,$

The above applies because radix notation is an integer-coefficient polynomial function of the radix.

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Bill Dubuque
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