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### Dhaka Secondary 2009/1

Posted: Tue Feb 01, 2011 11:51 pm
$x$ 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 $x+y$.

### Re: Dhaka Secondary 2009/1

Posted: Wed Feb 02, 2011 12:36 am
$[y][x]$ 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.

### Re: Dhaka Secondary 2009/1

Posted: Tue Feb 15, 2011 8:00 pm
Sorry, is should be $[\frac{x-y}{2} ]$ not $[\frac{x-2}{2} ]$

Now the problem is correct!

### Re: Dhaka Secondary 2009/1

Posted: Thu Aug 25, 2011 4:12 am

### Re: Dhaka Secondary 2009/1

Posted: Thu Aug 25, 2011 12:13 pm
$x,y$ 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

### Re: Dhaka Secondary 2009/1

Posted: Thu Aug 25, 2011 3:18 pm
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?? :

### Re: Dhaka Secondary 2009/1

Posted: Thu Aug 25, 2011 4:34 pm
Moon vai er solution eo deklam, but why??? please describe....

### Re: Dhaka Secondary 2009/1

Posted: Fri Jan 06, 2012 6:53 pm
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 $13,31$