BdMO National 2013: Junior 4

Discussion on Bangladesh Mathematical Olympiad (BdMO) National
BdMO
Posts: 134
Joined: Tue Jan 18, 2011 1:31 pm

BdMO National 2013: Junior 4

Unread post by BdMO » Fri Jan 10, 2014 1:24 am

Let $a$ be an integer divisible by $2$ but not divisible by $4$. What is the largest positive integer $n$ such that $2^n$ divides $a^{2012} + a^{2013} + \cdots + a^{3012}$?

Tahmid
Posts: 110
Joined: Wed Mar 20, 2013 10:50 pm

Re: BdMO National 2013: Junior 4

Unread post by Tahmid » Sun Jan 12, 2014 1:54 am

2 divides a, but 4 doesn't .
so a is form as $a=2(2k+1)$
so,
$a^{2012}+a^{2013}+a^{2014}+.........+a^{3012}$

$=\left \{ 2(2k+1) \right \}^{2012}+\left \{ 2(2k+1) \right \}^{2013}+\left \{ 2(2k+1) \right \}^{2014}+.........+\left \{ 2(2k+1) \right \}^{3012} $

$=2^{2012}(2k+1)^{2012}+2^{2013}(2k+a)^{2013}+2^{2014}(2k+1)^{2014}+.........+2^{3012}(2k+1)^{3012}$

$=2^{2012}\left \{ (2k+1)^{2012}+2(2k+1)^{2013}+2^{2}(2k+1)^{2014}+.........+2^{1000}(2k+1)^{3012} \right \}$

now let $\left \{ (2k+1)^{2012}+2(2k+1)^{2013}+2^{2}(2k+1)^{2014}+.........+2^{1000}(2k+1)^{3012} \right \}=x$
and here x is a odd number.

so,$2^{n}$ does not divide x.
that imply $2^{n}$ must divide $2^{2012}$.
so,the maxium value of n is 2012 for which $2^{n}$ divides $a^{2012}+a^{2013}+a^{2014}+.........+a^{3012}$
n=2012 ...(ans:)... :)

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