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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?

1. $64$
2. $65$
3. $204$
4. $144$
5. $256$

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0
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. Now lets see how many $7 \times 7$ squares are possible  These two patterns can shift to right as well as follows:  So, $7\times 7$ squares possible is $4$

Now lets see how many $6 \times 6$ squares are possible So, $6\times 6$ squares possible is $9$

Now lets see how many $5 \times 5$ squares are possible: $4$ vertical moves $\times 4$ horizontal moves $=4^2$ possibilities.

Proceeding like this,

• $8\times 8$  squares possible $: 1\times 1=1$
• $7\times 7$  squares possible $:2\times 2=4$
• $6\times 6$  squares possible $:3\times 3=9$
• $5\times 5$  squares possible $:4\times4=16$
• $4\times 4$  squares possible $: 5\times5=25$
• $3\times 3$  squares possible $: 6\times6 =36$
• $2\times 2$  squares possible $: 7\times7 =49$
• $1\times 1$  squares possible $:8\times8=64$

Total squares $\quad: 204$

Now we can generalize like with $n \times n$ chess board total squares $=1^2+2^2+3^2+\ldots +n^2 = \frac{n(n+1)(2n+1)}{6}$

Correct Answer: $C$

by Veteran (119k points)
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I could not understand your method. Can u please explain a bit more
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Can u tell me how $2\times 2$ chess board 5 square possible?
+2
This is trivial.. I can count it by looking on it.. 4 squares of size 1*1. One big square of size 4*4
0
yes

but it is not $4\times 4$

it is $2\times 2$

right?
+1
Yes there are 4 squares in total in a grid of 2*2
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Similarly I have done

In $8\times 8$ chess board $1\times 1$ square possible 64, i.e. $8\times 8$

$2\times 2$    square possible  $7\times 7$

Say if  in a chess board we give number of each point like this

1              2                3               4................................

2

3

4

.

.

then $2\times 2$ square go like this

1 to 3 is one square, 2 to 4 next square like this.

like this u go and get $7\times 7$ squares for $2\times 2$ matrix.

got it?
0
Ok.. Got it... Thanks for detailed explanation..

I have few more queries in combinatorics... Can I ask u by sending message
No. of squares on chessboard of $n\times n$ is equal to sum of squares of $n$ terms for $8\times 8$ chessboard,

\begin{align} &=\frac{n\left(n+1\right)\left(2n+1\right)}{6} \\&=\frac{8\times 9\times 17}{6}\\&=204.\end{align}
by Boss (31.4k points)
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Can u please explain why it worked this way

Let T(n,n) be the number of squares in a n*n chess board.

T(1,1) = 1

T(2,2) (i.e) number of squares in a 2*2 chessboard = number of squares in a 1*1 chessboard + 22 Similarly T(n,n) = T(n-1,n-1) + n2  if n>2 with base conditions T(1,1) = 1 and T(2,2) = 5

Solving, we will get T(8,8) = 204.

by Loyal (8k points)
edited
0
After solving the relation -:

T(n,n) = $n*(n+1)*(2n+1)/6$
+1 vote
By simple observation it will be-

1^2 + 2^2+ 3^2+ ...................+ n^2.

where n is the size given, in the question its 8.

sum of squares of numbers.
by Boss (17.7k points)
+2

Yeah!.

# squres in 8 ⨉ 8 chessboard = 82 (1 ⨉1 squres) +72 (2⨉2 squares)+62 (3⨉3 squares)+52 (4⨉4 squares)

+ 42 (5⨉5 squares)+32 (6 ⨉6 squares)+22 (7⨉7 squares)+12 (8 ⨉8 squares) = 204

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 = $\frac{(8)(8+1)((2 \times 8)+1)}{6} = 204$

by Junior (839 points)
edited