27. How many squares are there ~ above a chessboard or chequerboard?? (the answer is no 64)Can you extend your technique to calculate the variety of rectangles top top a chessboard?

Another puzzle that was e-mailed to me through this website. Mine instinct was the the prize was just a lot, however I thought about it and the systems is actually relatively simple...

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Before reading the answer deserve to I interest you in a clue?The very first thing is why the price is not just 64...
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All the red squares in the above picture would count as valid squares, so we are asking how numerous squares of any type of dimension indigenous 1x1 to 8x8 there space on a chess board.The vital is to think how plenty of positions there space that each size of square can be located... A 2x2 square, for example, can, by virtue of it"s size, be situated in 7 places horizontally and 7 areas vertically. In other words in 49 various positions. A 7x7 square though can only to the right in 2 location vertically and 2 horizontally. Consider what"s below...
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sizehorizontal positionsvertical positionspositions204
1x18864
2x27749
3x36636
4x45525
5x54416
6x6339
7x7224
8x8111
total
In total there are 204 squares ~ above a chessboard. This is the amount of the variety of possible location for all the squares of size 1x1 come 8x8.

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Formula because that n x n Chessboard?

It"s clean from the analysis over that the solution in the instance 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 that is created as follows:
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The proof of the explicit systems is past the border of this site, however if you want to look it up a mathematician would describe it as "the sum of the squares that the first n organic numbers." The last answer is given byn3/3 + n2/2 + n/6

Can you prolong your technique to calculation the variety of rectangles ~ above a chessboard?

Below space some instances of possible rectangles...
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All that the above examples would certainly be vailid rectanges...There is much more than one method of solving this. However it renders sense to expand our method from the squares difficulty first. The crucial to this is come think of each rectangle individually and consider the variety of positions it have the right to be located. For example a 3x7 rectangle deserve to be located in 6 positions horizontally and 2 vertically. From this us can build a matrix of every the feasible rectangles and sum. 1296
Dimensions12345678
Positions87654321
18645648403224168
27564942352821147
36484236302418126
45403530252015105
5432282420161284
632421181512963
72161412108642
8187654321
In complete then there are 1296 feasible rectangles.

Elegant method to rectangles, think about the vertices and also diagonals.

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I"ve been sent creative solution come the difficulty of the number of rectangles on a chessboard through Kalpit Dixit. This solution tackles the concern from a various approach. Rather than spring at particular sizes that rectangles and also working out whereby they have the right to be located we start at the various other end and look at areas 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 define a rectangle. In bespeak to be a diagonal and not a upright or horizontal line we may start anywhere but the end point must not have actually the exact same vertical or horizontal coordinate. Because of this there are 64 (8 x 8) feasible end points.There are thus 81 x 64 = 5184 acceptable diagonals.However, whilst every diagonal explains a unique rectangle, every rectangle walk not explain a distinctive diagonal.
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We see trivially the each rectangle have the right to be represented by 4 diagonals.So our variety of rectangles is given 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,) problems can currently be calculated. The variety of vertices being offered by (n + 1)2 and (n + 1).(m + 1) respectively. Thus 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 can obviously it is in arranged right into something more complicated.

Rectangles in Maths Nomenclature

It"s constantly my intentionally to explain the problems without formal maths nomenclature, through reasoning and also common sense. But there is quite a neat solution here if you do know about combinations, as in permutations and combinations. Horizontally us are choosing 2 vertices native the 9 available. The order walk not matter so it"s combinations rather than permutations. And also the very same vertically. For this reason the answer come the rectangle difficulty can be answered by:9C2•9C2 = 362 = 1296
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