How to solve this in function of the remainder

Asked Nov 28 '16 at 18:15

Active Nov 28 '16 at 18:50

Viewed 21 times

I am trying to solve an exercise but I don't know how to do it at all :

Given $m,n \in \mathbb N$ with $m>n>0$

and $r = m \bmod n$

Find ($2^m-1) \bmod (2^n-1)$ (as a function of $r$)

I don't know at all where to start or what to do...

How can the value of the remainder help me to find the required result ?

Thank you.

edited Jun 12 '20 at 10:38

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asked Nov 28 '16 at 18:15

Pop Flamingo

If $m=n q+r$ with $0\leq r<n$ then $2^{n q}=1$ so $2^m-1=2^r-1 $ mod $2^n-1 $ – Lozenges Nov 28 '16 at 18:38
See in particular this answer for the key idea in the remainder calculation. – Bill Dubuque Nov 28 '16 at 18:48

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