## 2014-national

For students of class 9-10 (age 14-16)
Mahfuz Sobhan
Posts: 25
Joined: Sat Feb 07, 2015 5:40 pm

### 2014-national

Closing his eyes Towsif begins to place knights on a Chess board of ${19} \times {21}$. After placing how many knights
Towsif will be sure that on the next move at least one knight will attack another one. (In one move knight goes
straight for ${2}$ steps and the 3rd step should be at right angle to the previous path.)

tanmoy
Posts: 301
Joined: Fri Oct 18, 2013 11:56 pm

### Re: 2014-national

It is a nice problem.Have you solved this?If not,try to solve it.I haved enjoyed it.I think you will also enjoy.
"Questions we can't answer are far better than answers we can't question"

Mahfuz Sobhan
Posts: 25
Joined: Sat Feb 07, 2015 5:40 pm

### Re: 2014-national

no bhai. i did not solve it. i need help

seemanta001
Posts: 13
Joined: Sat Jun 06, 2015 9:31 am
Location: Chittagong

### Re: 2014-national

You can place the knights in this order.This in an $8 \times 7$ chessboard.Hope this helps you.
Attachments
8 X 7 Chessboard
Chessboard.png (14.76 KiB) Viewed 4392 times
Last edited by seemanta001 on Fri Jul 10, 2015 8:10 pm, edited 1 time in total.
"Sometimes it's the very people who no one imagines anything of who do the things no one can imagine"

seemanta001
Posts: 13
Joined: Sat Jun 06, 2015 9:31 am
Location: Chittagong

### Re: 2014-national

The solution is:
For an $m \times n$ chessboard, the number of the knights is $n \times \lceil\dfrac{m}{3}\rceil$.
"Sometimes it's the very people who no one imagines anything of who do the things no one can imagine"

asif e elahi
Posts: 183
Joined: Mon Aug 05, 2013 12:36 pm

### Re: 2014-national

seemanta001 wrote:The solution is:
For an $m \times n$ chessboard, the number of the knights is $n \times \lceil\dfrac{m}{3}\rceil$.

RJHridi
Posts: 3
Joined: Tue Jun 30, 2015 1:22 pm

### Re: 2014-national

Imagine such a chessboard. A knight placed at a black square will go to a white square on its next move and vice versa. So the maximum number of knights placed in the board will be the number of black or white squares, assuring that none attacks anyone. Assuming that the board starts with a black square, the number of black squares will be= (ceiling of (19.21)/2) = 250. So after placing at least 251 knights, it can be made sure that on the next move at least one knight will attack another.

Kazi_Zareer
Posts: 86
Joined: Thu Aug 20, 2015 7:11 pm
Location: Malibagh,Dhaka-1217

### Re: 2014-national

seemanta001 wrote:The solution is:
For an $m \times n$ chessboard, the number of the knights is $n \times \lceil\dfrac{m}{3}\rceil$.
How? Can you please tell it?
We cannot solve our problems with the same thinking we used when we create them.

abmonim
Posts: 1
Joined: Sun Feb 14, 2016 1:07 pm

### Re: 2014-national

19.21/2=250 ??? Confused!!!

asif e elahi
Posts: 183
Joined: Mon Aug 05, 2013 12:36 pm
This isn't obvious. You have to prove it. (This isn't even true for a $2\times 2$ chessboard.)