Another puzzle the was e-mailed to me v this website. My instinct was the the prize was just a lot, yet I thought about it and the equipment is actually reasonably simple...

You are watching: How many squares are there on a checker board

Before analysis the answer can I interest you in a clue?The first thing is why the answer is not simply 64... All the red squares in the above snapshot would count as valid squares, for this reason we are asking how numerous squares of any dimension native 1x1 come 8x8 there are on a chess board.The an essential is come think how plenty of positions there are that each dimension of square can be located... A 2x2 square, for example, can, through virtue that it"s size, be located in 7 places horizontally and 7 places vertically. In other words in 49 different positions. A 7x7 square though deserve to only right in 2 location vertically and also 2 horizontally. Take into consideration what"s below...

sizehorizontal positionsvertical positionspositions204

1x1 | 8 | 8 | 64 |

2x2 | 7 | 7 | 49 |

3x3 | 6 | 6 | 36 |

4x4 | 5 | 5 | 25 |

5x5 | 4 | 4 | 16 |

6x6 | 3 | 3 | 9 |

7x7 | 2 | 2 | 4 |

8x8 | 1 | 1 | 1 |

total |

See more: 1 Liter Is How Many Cups - How Many Cups Of Water Make A Liter

## Formula for n x n Chessboard?

It"s clear from the analysis over that the solution in the instance of n x n is the sum of the squares from n2 come 12 the is come say n2 + (n-1)2 + (n-2)2 ... ... 22 + 12Mathematically that is created as follows:The proof of the explicit systems is past the limit of this site, but if you desire to watch it up a mathematician would refer to it as "the sum of the squares that the an initial n natural numbers." The last answer is given byn3/3 + n2/2 + n/6## Can you extend your method to calculation the number of rectangles top top a chessboard?

Below are some instances of possible rectangles... All the the over examples would certainly be vailid rectanges...There is an ext than one way of fixing this. Yet it provides sense to prolong our technique from the squares difficulty first. The crucial to this is come think of every rectangle individually and also consider the variety of positions it deserve to be located. For example a 3x7 rectangle have the right to be situated in 6 location horizontally and 2 vertically. From this us can develop a matrix of all the possible rectangles and also sum. 1296Dimensions | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | ||

Positions | 8 | 7 | 6 | 5 | 4 | 3 | 2 | 1 | ||

1 | 8 | 64 | 56 | 48 | 40 | 32 | 24 | 16 | 8 | |

2 | 7 | 56 | 49 | 42 | 35 | 28 | 21 | 14 | 7 | |

3 | 6 | 48 | 42 | 36 | 30 | 24 | 18 | 12 | 6 | |

4 | 5 | 40 | 35 | 30 | 25 | 20 | 15 | 10 | 5 | |

5 | 4 | 32 | 28 | 24 | 20 | 16 | 12 | 8 | 4 | |

6 | 3 | 24 | 21 | 18 | 15 | 12 | 9 | 6 | 3 | |

7 | 2 | 16 | 14 | 12 | 10 | 8 | 6 | 4 | 2 | |

8 | 1 | 8 | 7 | 6 | 5 | 4 | 3 | 2 | 1 | |

## Elegant method to rectangles, take into consideration the vertices and also diagonals.

I"ve been sent an innovative solution to the trouble of the variety of rectangles on a chessboard by Kalpit Dixit. This equipment tackles the problem from a various approach. Fairly than spring at specific sizes of rectangles and working out whereby they can be situated we start at the other end and also look at places first.The vertices room the intersections. For our chessboard there space 81 (9 x 9). A diagonal beginning at one vertex and ending at one more will uniquely define a rectangle. In order to it is in a diagonal and not a vertical or horizontal line we might start anywhere however the end allude must not have the exact same vertical or horizontal coordinate. As such there are 64 (8 x 8) possible end points.There are as such 81 x 64 = 5184 agree diagonals.However, whilst each diagonal describes a unique rectangle, each rectangle go not explain a distinct diagonal. We watch trivially the each rectangle deserve to be stood for by 4 diagonals.So our number of rectangles is provided by 81 x 64 /4 =**1296**

## n x n or n x m?

The n x n (eg. 9x9,) or n x m (eg 10x15,) troubles can currently be calculated. The number of vertices being given by (n + 1)2 and (n + 1).(m + 1) respectively. Thus the last solutions space as follows.n x n: (n + 1)2 x n2 / 4n x m: (n + 1) x (m + 1) x (n x m) / 4Which deserve to obviously be arranged right into something an ext complicated.## Rectangles in Maths Nomenclature

It"s constantly my intentionally to define the troubles without official maths nomenclature, with reasoning and also common sense. But there is rather a practiced solution below if you do know around combinations, together in permutations and also combinations. Horizontally we are selecting 2 vertices native the 9 available. The order go not issue so it"s combinations rather than permutations. And also the same vertically. So the answer to the rectangle difficulty can be answered by:9C2•9C2 = 362 =**1296**

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