Dhaka Secondary 2009/1

Problem for Secondary Group from Divisional Mathematical Olympiad will be solved here.
Forum rules
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
Posts: 134
Joined: Tue Jan 18, 2011 1:31 pm

Dhaka Secondary 2009/1

Unread post by BdMO » 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 $x+y$.

Hasib
Posts: 238
Joined: Fri Dec 10, 2010 11:29 am
Location: খুলনা, বাংলাদেশ
Contact:

Re: Dhaka Secondary 2009/1

Unread post by Hasib » 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.
A man is not finished when he's defeated, he's finished when he quits.

User avatar
Moon
Site Admin
Posts: 751
Joined: Tue Nov 02, 2010 7:52 pm
Location: Dhaka, Bangladesh
Contact:

Re: Dhaka Secondary 2009/1

Unread post by Moon » 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!
It is easier (and less time consuming) to solve this problem with checking.
First notice $[x][y]-[y][x]=[\frac{x-2}{2}][2(x+y)] \iff 2(x-y)=(x+y)$
Also we have $[2(x+y)]$ as a digit. So it is enough to check the primes from $11$ to $47$ (why?), also notice that $2|(x+y)$. So $13$ is the prime, and $x+y=4$. :)
"Inspiration is needed in geometry, just as much as in poetry." -- Aleksandr Pushkin

Please install LaTeX fonts in your PC for better looking equations,
learn how to write equations, and don't forget to read Forum Guide and Rules.

Shifat
Posts: 53
Joined: Sun Jul 31, 2011 12:21 pm
Location: Dhaka, Bangladesh

Re: Dhaka Secondary 2009/1

Unread post by Shifat » Thu Aug 25, 2011 4:12 am

answer maybe 2(x-y). :?:

User avatar
Tahmid Hasan
Posts: 665
Joined: Thu Dec 09, 2010 5:34 pm
Location: Khulna,Bangladesh.

Re: Dhaka Secondary 2009/1

Unread post by Tahmid Hasan » 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 :D
Last edited by Tahmid Hasan on Thu Aug 25, 2011 4:18 pm, edited 2 times in total.
বড় ভালবাসি তোমায়,মা

Shifat
Posts: 53
Joined: Sun Jul 31, 2011 12:21 pm
Location: Dhaka, Bangladesh

Re: Dhaka Secondary 2009/1

Unread post by Shifat » 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?? : :oops: :?:

Shifat
Posts: 53
Joined: Sun Jul 31, 2011 12:21 pm
Location: Dhaka, Bangladesh

Re: Dhaka Secondary 2009/1

Unread post by Shifat » Thu Aug 25, 2011 4:34 pm

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

User avatar
nafistiham
Posts: 829
Joined: Mon Oct 17, 2011 3:56 pm
Location: 24.758613,90.400161
Contact:

Re: Dhaka Secondary 2009/1

Unread post by nafistiham » 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 $13,31$
\[\sum_{k=0}^{n-1}e^{\frac{2 \pi i k}{n}}=0\]
Using $L^AT_EX$ and following the rules of the forum are very easy but really important, too.Please co-operate.
Introduction:
Nafis Tiham
CSE Dept. SUST -HSC 14'
http://www.facebook.com/nafistiham
nafistiham@gmail

Post Reply