## Dhaka Secondary 2009/1

Problem for Secondary Group from Divisional Mathematical Olympiad will be solved here.
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Please don't post problems (by starting a topic) in the "Secondary: Solved" forum. This forum is only for showcasing the problems for the convenience of the users. You can post the problems in the main Divisional Math Olympiad forum. Later we shall move that topic with proper formatting, and post in the resource section.
BdMO
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### Dhaka Secondary 2009/1

$x$ and $y$ are two digits and $[x][y]$ represents the number $10x+y$. If $[x][y]$ and $[y][x]$ are both primes and $[x][y]-[y][x]=[\frac{x-y}{2}][2(x+y)]$ find $x+y$.

Hasib
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### Re: Dhaka Secondary 2009/1

$[y][x]$ is a prime. So $x$ is odd number. So, $x-2$ is also odd number.
So how its possible that $[\frac{x-2}{2}]$ be a digit.

I assume that $[x][y]$ represent $\overline{xy}$. Is my assume correct? Or, i will try again.
A man is not finished when he's defeated, he's finished when he quits.

Moon
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### Re: Dhaka Secondary 2009/1

Sorry, is should be $[\frac{x-y}{2} ]$ not $[\frac{x-2}{2} ]$

Now the problem is correct!
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Shifat
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### Re: Dhaka Secondary 2009/1

answer maybe 2(x-y). Tahmid Hasan
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### Re: Dhaka Secondary 2009/1

$x,y$ are both odd.so $x-y$ is even.
simplifying the equation we get
$x=3y$
so the primes are $31y,13y$.it is possible iff $y=1$.
so $x=3$
hence $x+y=4$
Shifat,thank you for finding my folly Last edited by Tahmid Hasan on Thu Aug 25, 2011 4:18 pm, edited 2 times in total.
বড় ভালবাসি তোমায়,মা

Shifat
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### Re: Dhaka Secondary 2009/1

But akhon to baki part ta ami buztasi na, x+y=2(x-y) er por ki korlen??
I mean x=3y but what about the logic behind 13 and 31 s?? :  Shifat
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### Re: Dhaka Secondary 2009/1

Moon vai er solution eo deklam, but why??? please describe....

nafistiham
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### Re: Dhaka Secondary 2009/1

I did it just by checking.
as both $x,y$ are odd, and $[x][y],[y][x]$ both are primes, $x,y$ can be only $1,3,7,9$
so, we get $11,13,17,19,31,37,53,59,71,93,97$
now, let us take the pairs $13,31;71,17;37,73;97,79$
by just looking at these we can see,the only pair can be $13,31$
$\sum_{k=0}^{n-1}e^{\frac{2 \pi i k}{n}}=0$
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