17 votes

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?

- $64$
- $65$
- $204$
- $144$
- $256$

26 votes

Best answer

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: $C$

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

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?

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?

17 votes

No. of squares on chessboard of $n\times n$ is equal to sum of squares of $n$ terms for $8\times 8$ chessboard,

$\begin{align} &=\frac{n\left(n+1\right)\left(2n+1\right)}{6} \\&=\frac{8\times 9\times 17}{6}\\&=204.\end{align}$

$\begin{align} &=\frac{n\left(n+1\right)\left(2n+1\right)}{6} \\&=\frac{8\times 9\times 17}{6}\\&=204.\end{align}$

3 votes

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 + 2^{2}

^{}

Similarly T(n,n) = T(n-1,n-1) + n^{2 } if n>2 with base conditions T(1,1) = 1 and T(2,2) = 5

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

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.

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

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

sum of squares of numbers.

1 vote

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

0 votes

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

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