Let $m,n$ be positive integers. Prove that $(2^m-1)^2|(2^n-1)$ if and only if $m(2^m-1)|n$.
Source:Russia 1997
m and n
- Phlembac Adib Hasan
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Re: m and n
Hint:
- zadid xcalibured
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Re: m and n
$2^m-1|2^n-1 \longleftrightarrow m|n$.Let $n=km$. As $2^m-1|\frac{2^n-1}{2^m-1}$
and $2^{m(k-1)}+2^{m(k-2)}+.............+1 \equiv k \equiv 0 (mod 2^m-1)$
\[\Longleftrightarrow 2^m-1|k\]
\[\Longleftrightarrow m(2^m-1)|n\]
and $2^{m(k-1)}+2^{m(k-2)}+.............+1 \equiv k \equiv 0 (mod 2^m-1)$
\[\Longleftrightarrow 2^m-1|k\]
\[\Longleftrightarrow m(2^m-1)|n\]
Re: m and n
https://www.awesomemath.org/assets/PDFs ... _Lemma.pdf
A general version, and applicable in many criteria. Particularly, this problem becomes one liner
A general version, and applicable in many criteria. Particularly, this problem becomes one liner
One one thing is neutral in the universe, that is $0$.
- Fm Jakaria
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Re: m and n
Ithink problem 3 has a crack in the solution. How can we assume a and d relatively prime?Masum wrote:https://www.awesomemath.org/assets/PDFs ... _Lemma.pdf
A general version, and applicable in many criteria. Particularly, this problem becomes one liner
You cannot say if I fail to recite-
the umpteenth digit of PI,
Whether I'll live - or
whether I may, drown in tub and die.
the umpteenth digit of PI,
Whether I'll live - or
whether I may, drown in tub and die.