BdMO National Secondary 2015/2

Discussion on Bangladesh Mathematical Olympiad (BdMO) National
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samiul_samin
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BdMO National Secondary 2015/2

Unread post by samiul_samin » Sun Feb 18, 2018 12:26 pm

How many pairs of integers ($m,n$) satisfy the equation $m+n=mn$ ?

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samiul_samin
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Re: BdMO National Secondary 2015/2

Unread post by samiul_samin » Sun Feb 18, 2018 12:46 pm

Hint
1.Multiplication is nothing but long addition
2.Two conseqative numbers are always Co-prime
Answer
$\fbox 0$

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samiul_samin
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Re: BdMO National Secondary 2015/2

Unread post by samiul_samin » Sun Feb 18, 2018 12:55 pm

Solution
$m+n=mn$
$\Rightarrow m+n=m+m+m+m+m+m+...+m $ (n times)
$\Rightarrow n=m+m+m+m+m....+m+m$ {(n-1)times}
$\Rightarrow n=m(n-1)\Rightarrow \dfrac {n}{n-1}=m$
But $n$ and $n-1$ are co-prime.
So, $m$ cannot be an integer.
A Contrudiction
So,there are no such $(m,n) $that satisfies the equation $m+n=mn$

mac0220
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Re: BdMO National Secondary 2015/2

Unread post by mac0220 » Mon Mar 05, 2018 9:38 am

The solution is ( m,n ) = (2,2)
n and n-1 are co prime unless n-1=1
which makes n=m=2

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samiul_samin
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Re: BdMO National Secondary 2015/2

Unread post by samiul_samin » Mon Mar 05, 2018 12:46 pm

mac0220 wrote:
Mon Mar 05, 2018 9:38 am
The solution is ( m,n ) = (2,2)
n and n-1 are co prime unless n-1=1
which makes n=m=2
Yes,you are right.It was a very foolosh mistake of me.Sorry. :oops: :oops: :oops:

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