BDMO NATIONAL Junior 2013/04

Discussion on Bangladesh Mathematical Olympiad (BdMO) National
Math Mad Muggle
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BDMO NATIONAL Junior 2013/04

Unread post by Math Mad Muggle » Mon Jan 30, 2017 9:43 pm

please , give me the reply quickly
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ahmedittihad
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Re: BDMO NATIONAL Junior 2013/04

Unread post by ahmedittihad » Mon Jan 30, 2017 10:02 pm

$a^{2012} + a^{2013}+.....+a^{3012}=a^{2012}(1+a+a^2+.....+a^{1000})$.
The number in the bracket is odd. So, the highest power of two dividing the whole number is the highest power of two dividing $a^{2012}$. Which is clearly $2012$. So, $n=2012$.
Frankly, my dear, I don't give a damn.

Math Mad Muggle
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Re: BDMO NATIONAL Junior 2013/04

Unread post by Math Mad Muggle » Mon Jan 30, 2017 10:47 pm

tnx,bro....bt one person said me that the ans will be the sum of powers....my ans is same with u

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ahmedittihad
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Re: BDMO NATIONAL Junior 2013/04

Unread post by ahmedittihad » Mon Jan 30, 2017 10:50 pm

That person is incorrect.
Frankly, my dear, I don't give a damn.

andaleeb
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Re: BDMO NATIONAL Junior 2013/04

Unread post by andaleeb » Wed Feb 08, 2017 5:09 am

I would like to add some corrections from "ahmedittihad" 's answer.

According to question, $a = 2^k$ Where, $k$ is any odd natural number

Now, $ a^{2012} + a^{2013} + ...... + a^{3012} $

= $ a^{2012} ( 1 + a + a^{2} +.........+ a^{1000} )$
= $ 2^{2012 k} ( 1 + 2^{k} + 2^{2 k} +.........+ 2^{1000 k} )$

Since, of the two factors only $ 2^{2012 k} $ is even, so maximum power of 2 dividing $a^{2012} = 2^{2012k} $ is the required answer.

Therefore $ n = 2012 k $ Where, $k$ is any odd natural number

Where, $ n = 2012 $ is only one of the possible answers.

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