27. How countless squares space there top top a chessboard or chequerboard?? (the prize is not 64)Can you prolong your technique to calculate the number of rectangles ~ above a chessboard?

Another puzzle that was e-mailed come me with this website. Mine instinct was that the prize was just a lot, however I thought around it and also the systems is actually reasonably simple...

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Before analysis the answer can I attention you in a clue?The very first thing is why the prize is not just 64...
All the red squares in the above snapshot would count as valid squares, therefore we are asking how many squares of any type of dimension from 1x1 come 8x8 there room on a chess board.The crucial is to think how many positions there room that each size of square can be located... A 2x2 square, for example, can, by virtue that it"s size, be situated in 7 places horizontally and 7 places vertically. Ie in 49 various positions. A 7x7 square though deserve to only fit in 2 location vertically and 2 horizontally. Take into consideration what"s below...

sizehorizontal positionsvertical positionspositions204
In complete there room 204 squares top top a chessboard. This is the amount of the variety of possible location for every the squares of size 1x1 come 8x8.

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Formula because that n x n Chessboard?

It"s clear from the analysis above that the systems in the situation 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 equipment is past the border of this site, however if you desire to look it increase a mathematician would refer to it together "the amount of the squares the the an initial n organic numbers." The final answer is offered byn3/3 + n2/2 + n/6

Can you prolong your method to calculate the variety of rectangles on a chessboard?

Below space some examples of possible rectangles...
All that the above examples would be vailid rectanges...There is more than one means of solving this. However it makes sense to extend our technique from the squares difficulty first. The vital to this is to think of every rectangle individually and also consider the number of positions it can be located. For instance a 3x7 rectangle can be situated in 6 location horizontally and 2 vertically. From this us can build a matrix of all the feasible rectangles and also sum. 1296
In total then there room 1296 possible rectangles.

Elegant strategy to rectangles, consider the vertices and diagonals.

I"ve to be sent creative solution come the trouble of the number of rectangles ~ above a chessboard by Kalpit Dixit. This solution tackles the worry from a different approach. Quite than spring at particular sizes of rectangles and also working out where they have the right to be located we start at the various other end and also look at places first.The vertices are the intersections. For our chessboard there room 81 (9 x 9). A diagonal beginning at one vertex and ending at one more will uniquely define a rectangle. In stimulate to be a diagonal and also not a vertical or horizontal heat we may start anywhere however the end point 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 acceptable diagonals.However, whilst every diagonal defines a distinctive rectangle, each rectangle does not explain a unique diagonal.
We watch trivially the each rectangle have the right to be represented by 4 diagonals.So our variety of rectangles is offered 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,) difficulties can currently be calculated. The number of vertices being offered by (n + 1)2 and also (n + 1).(m + 1) respectively. Hence the last solutions room 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 into something more complicated.

Rectangles in Maths Nomenclature

It"s always my intention to describe the troubles without official maths nomenclature, through reasoning and also common sense. Yet there is rather a practiced solution below if you do know about combinations, as in permutations and also combinations. Horizontally we are choosing 2 vertices from the 9 available. The order does not matter so it"s combinations rather than permutations. And also the very same vertically. For this reason the answer to the rectangle problem can it is in answered by:9C2•9C2 = 362 = 1296
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