Rangpur Secondary 2011/6

Problem for Secondary Group from Divisional Mathematical Olympiad will be solved here.
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Moon
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Rangpur Secondary 2011/6

Unread post by Moon » Wed Feb 02, 2011 6:49 pm

Problem 6:
When $N$ is divided by $7$, the quotient is twice the remainder. What is the number greater than $7$ that must divide $N$?
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Tahmid Hasan
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Re: Rangpur Secondary 2011/6

Unread post by Tahmid Hasan » Wed Feb 02, 2011 7:51 pm

is the wanted number a prime?
if it isn't there is no soln.
let's express $N=7.2a+a$
let's input the possible values of $a=1,2,3,4,5,6$
then the numbers are not divisible by any prime greater than 7 :(
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Mehfuj Zahir
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Re: Rangpur Secondary 2011/6

Unread post by Mehfuj Zahir » Thu Feb 03, 2011 4:46 am

N=15a then must be divided by 15

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leonardo shawon
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Re: Rangpur Secondary 2011/6

Unread post by leonardo shawon » Thu Feb 03, 2011 11:36 am

no..actually it is if $N=15x$ [$x=1,2] and divisible by 7 then quotient is twice the remainder.
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Re: Rangpur Secondary 2011/6

Unread post by Mehfuj Zahir » Thu Feb 03, 2011 11:58 am

For any natural number it is true.U should read the question carefully.

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Re: Rangpur Secondary 2011/6

Unread post by leonardo shawon » Thu Feb 03, 2011 6:41 pm

sorry. Mistake
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BRAC University.

long way to go .....

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Re: Rangpur Secondary 2011/6

Unread post by Moon » Thu Feb 03, 2011 8:07 pm

No problem. We are here so that we may discuss and learn from our mistakes. :)
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Re: Rangpur Secondary 2011/6

Unread post by *Mahi* » Thu Feb 03, 2011 9:07 pm

Here,$N$ must be equal to $15x$ where $x\neq7y$.Then $15$ is the answer.
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