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dimension of the chess board

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by jaxis » Tue Jan 18, 2011 2:20 am
the answer is 10 by 10

Soln:
Let the Chess board be of dimension N by N.


No of 1 by 1 squares = N^2.
No of 2 by 2 squares = (N-1)^2
No of 3 by 3 squares= (N-2)^2
.
.
.
No of (N-1) by (N-1) squares = 2^2
No of N by N squares = 1^2

So,


total possiblesquares in a N BY N chess board = Sum of thesquares of first N ntural numbers.

=> N*(N+1)*(2N+1)/6 = 385
N=10.

Thanks,
Jaxis.
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by prachich1987 » Tue Jan 18, 2011 10:00 pm
jaxis wrote:the answer is 10 by 10

Soln:
Let the Chess board be of dimension N by N.


No of 1 by 1 squares = N^2.
No of 2 by 2 squares = (N-1)^2
No of 3 by 3 squares= (N-2)^2
.
.
.
No of (N-1) by (N-1) squares = 2^2
No of N by N squares = 1^2

So,


total possiblesquares in a N BY N chess board = Sum of thesquares of first N ntural numbers.

=> N*(N+1)*(2N+1)/6 = 385
N=10.

Thanks,
Jaxis.
brilliant !!!
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