Let $r$ be the remainder of $a$ when dividing by $b$. Prove: $2^r -1$ is the remainder of $2^a -1$ when dividing by $2^b -1$

Question

I am having trouble proving this question. Can anyone help me out?

Possible duplicate of How to prove $\gcd(a^m-b^m,a^n-b^n) = a^{\gcd(m,n)} - b^{\gcd(m,n)} $? — lab bhattacharjee, Feb 17 '19 at 12:25

score 1 · Answer 1 · edited Feb 17 '19 at 17:54

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Let $b\equiv r\bmod a$. This implies $a \mid b-r$ and $r<a$. So $2^a-1\mid 2^{b-r}-1$ and hence $2^a-1\mid(2^{b}-1)-(2^r-1)$. Now observe $2^r-1<2^a-1$ and hence the remainder is $2^r-1$.

edited Feb 17 '19 at 17:54

user26857

52,094

answered Feb 17 '19 at 12:31

user6

4,092

How do you get from $2^(b-r) -$1 to ($2^b - $1) - ($2^r - $1). These are not the same right? – Kath Feb 17 '19 at 13:14
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a=r mod b because we are dividing by b and not a – Feb 17 '19 at 13:22
@Kath $(2^{b}-1)-(2^r-1)=2^r(2^{b-r}-1)$ – user26857 Feb 17 '19 at 17:56

score 0 · Answer 2 · 2019-02-17T13:02:59.230

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here is my solution sorry I don't know how to write in latex I am new I hope that my solution is helpful :

edited Feb 17 '19 at 13:02

answered Feb 17 '19 at 12:52

Let $r$ be the remainder of $a$ when dividing by $b$. Prove: $2^r -1$ is the remainder of $2^a -1$ when dividing by $2^b -1$

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