TIFR CSE 2015 | Part A | Question: 7

Question

A $1 \times 1$ chessboard has one square, a $2 \times 2$ chessboard has five squares. Continuing along this fashion, what is the number of squares on the regular $8 \times 8$ chessboard?

• A. $64$
• B. $65$
• C. $204$
• D. $144$
• E. $256$

Comments

I can solve this que with  the help of present of mind but why this que is the part of Permutations and combinations, where we are arranging and selecting.

Answer 1

An $n \times n$ chessboard is made up of $n$ adjacent units on either side. Here, unit means side of each small square.

Now, $n$ adjacent units on horizontal side and $n$ adjacent units on vertical side make a square of $n \times n$ dimension.

Note: The horizontal and vertical sides must have common originating point.

 

Therefore, for a $n x n$ chessboard, following observations can be made-

Dimension i	No. of Horizontal Sides of i Adjacent Units	No. of Vertical Sides of i Adjacent Units	Total No. of Squares
$n \times n$	$n - n + 1 = 1$	$n - n + 1 = 1$	$1^{2}$
$(n-1) \times (n-1)$	$n - (n-1) + 1 = 2$	$n - (n-1) + 1 = 2$	$2^{2}$
$(n-2) \times (n-2)$	$n - (n-2) + 1 = 3$	$n - (n-2) + 1 = 3$	$3^{2}$
...	...	...	...
$3 \times 3$	$n - 3+ 1 = (n-2)$	$n - 3+ 1 = (n-2)$	$(n-2)^{2}$
$2 \times 2$	$n - 2 + 1 = (n-1)$	$n - 2 + 1 = (n-1)$	$(n-1)^{2}$
$1 \times 1$	$n - 1 + 1 = n$	$n - 1 + 1 = n$	$n^{2}$

Therefore, total no. of squares = $1^{2} + 2^{2} + 3^{2} + ... + (n-2)^{2} + (n-1)^{2} + n^{2} = \frac{(n)(n+1)(2n+1)}{6}$  

Therefore,  total no. of squares in 8 x 8 chessboard = $\frac{(8)(8+1)((2 \times 8)+1)}{6} = 204$ 

Answer 2

It can be solved by
1^2 + 2^2 + ... + 8^2 = (8 * 9 * 17) / 6 = 204

Answer 3

Ans : C

This can be solved by recognizing the pattern.

Let f(n) = Number of squares in n x n chessboard.

By drawing the chessboards and calculating number of squares manually for n=1,2,3,4 we get:-

f(1) =  1

f(2) = 4 + 1

f(3) = 9 + 4 + 1

f(4) = 16 + 9 + 4 + 1

These four observations are enough to conclude that f(n) = 1^2 + 2^2 + 3^2 +....+ n^2 = n(n+1)(2n+1)/6

Hence, f(8) = 204

Answer 4

Consider $n\times n$ where

$n$ = 1 $\Rightarrow$ 1

$n$ = 2 $\Rightarrow$ $2^{2}$ + 1 = 5

$n$ = 3 $\Rightarrow$ $3^{2}$ + 5 = 14

$n$ = 4 $\Rightarrow$ $4^{2}$ + 14 = 30...and so on.

So basically, I counted this manually until $n=4$ and understood the pattern of adding up squares, and we can do that by using this,

$\sum_{1}^{n} n^{{2}} = \tfrac{n(n+1)(2n+1)}{6}$

Working out for $n=8$,

$= \tfrac{8\times 9\times17}{6}$

$= 204$

We get the value 204.

Hence, Option (C).