Another puzzle the was e-mailed come me with this website. Mine instinct was the the prize was simply a lot, yet I thought around it and also the solution is actually fairly simple...

You are watching: How many squares in a checkerboard

Before analysis the answer can I interest you in a clue?The an initial thing is why the price is not just 64... All the red squares in the above photo would count as valid squares, therefore we are asking how countless squares of any type of dimension indigenous 1x1 come 8x8 there room on a chess board.The vital is come think how countless positions there are that each dimension of square can be located... A 2x2 square, for example, can, through virtue of it"s size, be located in 7 locations horizontally and also 7 areas 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 |

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## Formula for n x n Chessboard?

It"s clean from the analysis above that the equipment in the situation of n x n is the sum of the squares indigenous n2 come 12 the is to say n2 + (n-1)2 + (n-2)2 ... ... 22 + 12Mathematically the is composed as follows:The proof of the explicit systems is past the border of this site, however if you desire to look at it up a mathematician would describe it as "the amount of the squares the the very first n herbal numbers." The last answer is offered byn3/3 + n2/2 + n/6## Can you extend your technique to calculate the variety of rectangles on a chessboard?

Below room some examples of possible rectangles... All of the above examples would be vailid rectanges...There is much more than one means of resolving this. Yet it renders sense to extend our method from the squares difficulty first. The an essential to this is to think of every rectangle individually and also consider the variety of positions it deserve to be located. For instance a 3x7 rectangle can be located in 6 positions horizontally and also 2 vertically. From this us can build a procession 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 technique to rectangles, consider the vertices and diagonals.

I"ve been sent an innovative solution to the problem of the number of rectangles top top a chessboard through Kalpit Dixit. This equipment tackles the issue from a various approach. Quite than feather at particular sizes that rectangles and working out wherein they deserve to be located we start at the other end and also look at places first.The vertices room the intersections. Because that our chessboard there space 81 (9 x 9). A diagonal beginning at one vertex and ending at one more will uniquely describe a rectangle. In order to it is in a diagonal and not a upright or horizontal heat we may start anywhere yet the end point must not have the very same vertical or horizontal coordinate. Thus there room 64 (8 x 8) possible end points.There are because of this 81 x 64 = 5184 agree diagonals.However, whilst every diagonal explains a distinct rectangle, every rectangle walk not explain a unique diagonal. We watch trivially the each rectangle deserve to be represented by 4 diagonals.So our variety 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 variety of vertices being offered by (n + 1)2 and (n + 1).(m + 1) respectively. Therefore the last solutions are as follows.n x n: (n + 1)2 x n2 / 4n x m: (n + 1) x (m + 1) x (n x m) / 4Which have the right to obviously it is in arranged into something an ext complicated.## Rectangles in Maths Nomenclature

It"s constantly my intentionally to define the problems without officially maths nomenclature, with reasoning and common sense. However there is fairly a practiced solution below if you perform know about combinations, as in permutations and also combinations. Horizontally we are choosing 2 vertices indigenous the 9 available. The order walk not issue so it"s combinations fairly than permutations. And also the very same vertically. So the answer to the rectangle difficulty can be answered by:9C2•9C2 = 362 =**1296**

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