### 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 $x+y$.

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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 $x+y$.

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.

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.

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!

Now the problem is correct!

Posted: **Thu Aug 25, 2011 4:12 am**

answer maybe 2(x-y).

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

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

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?? :

I mean x=3y but what about the logic behind 13 and 31 s?? :

Posted: **Thu Aug 25, 2011 4:34 pm**

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

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 $13,31$

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 $13,31$