(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

Given that [itex]b>1,b\in\mathbb{Z},c_{0},c_{1},...,c_{m}\in\{0,...,b-1\}[/itex], [itex]0\leq c_{m+1}\leq b-1[/itex], and

[itex]c_{m+1}b^{m+1}=(\sum\limits _{k=0}^{m+1}c_{k}b^{k})\text{mod }b^{m+2}-c_{0}-c_{1}b-c_{2}b^{2}-...-c_{m}b^{m}[/itex], show that [itex]c_{m+1}\in\mathbb{Z}[/itex].

Also,

[itex]\sum\limits _{k=0}^{n}c_{k}p^{k}=(\sum\limits _{k=0}^{n}(a_{k}+b_{k})p^{k})\text{mod }p^{n+1}[/itex] where [itex]\{a_{k}\},\{b_{k}\}\subseteq\{0,...,b-1\}[/itex] for n=0,1,2,...,m+1

2. Relevant equations

This problem was taken from adding numbers in a base other than 10. Some of the above in the case of what I am trying to prove holds n=0,1,2,... in terms of the upper limit of the sum. But, this is the core of what I am trying to show right now in order to complete a proof.

3. The attempt at a solution

Any help would be appreciated. I understand that given that this is true that b_(m+1) must divide the right hand side.

Been thinking about how to show this in general and even with specific examples, adding numbers in different bases as it relates to this, I have not yet been able to figure out exactly how this works out.

Came up with that I know that [itex](\sum\limits _{k=0}^{m+1}c_{k}b^{k})\text{mod }b^{m+2}\geq c_{0}+c_{1}b+...+c_{m}b^{m}[/itex] so I am thinking I could at least show this and maybe in doing so it may help me to come up with a way to show that c_(m+1) is in Z.

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# Homework Help: Adding in different bases proof

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