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Divisibility

Posted: Fri Dec 15, 2017 9:36 pm
by mathlover007
Find n such that 2n31024 − 1.

Re: Divisibility

Posted: Thu Mar 01, 2018 7:38 pm
by samiul_samin
mathlover007 wrote:
Fri Dec 15, 2017 9:36 pm
Find n such that 2n31024 − 1.
The correct question is,
Find the highest value of $n$ such that $2^n|3^{1024}-1$.
Hint
$2^{10}=1024$
$3^{2k}+1$ is divided by $2$.
Use the formula $x^2-y^2=(x+y)(x-y)$
Answer
$9+2+1=12$