27. How countless squares room there ~ above a chessboard or chequerboard?? (the prize is no 64)Can you extend your technique to calculation the variety of rectangles on a chessboard?

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

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Before analysis the answer deserve to I interest you in a clue?The an initial thing is why the price is not simply 64... All the red squares in the above snapshot would count together valid squares, therefore we space asking how plenty of squares of any dimension native 1x1 come 8x8 there space on a chess board.The vital is to think how countless 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 locations horizontally and 7 areas vertically. Ie in 49 different positions. A 7x7 square though have the right to only fit in 2 positions vertically and also 2 horizontally. Consider what"s below... sizehorizontal positionsvertical positionspositions204
1x18864
2x27749
3x36636
4x45525
5x54416
6x6339
7x7224
8x8111
total
In full there are 204 squares on a chessboard. This is the sum of the variety of possible positions for every the squares of size 1x1 come 8x8.

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## 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 amount of the squares indigenous n2 come 12 the is to say n2 + (n-1)2 + (n-2)2 ... ... 22 + 12Mathematically the is written as follows: The proof of the explicit equipment is beyond the scope of this site, yet if you want to look it up a mathematician would refer to it as "the amount of the squares that the very first n herbal numbers." The final answer is given byn3/3 + n2/2 + n/6

## Can you extend your an approach to calculation the number of rectangles on a chessboard?

Below room some instances of feasible rectangles... All the the over examples would certainly be vailid rectanges...There is much more than one means of resolving this. But it provides sense to expand our method from the squares difficulty first. The key 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 deserve to be located in 6 positions horizontally and 2 vertically. Indigenous this us can construct a procession of every the possible rectangles and sum. 1296
 Dimensions 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
In complete then there space 1296 feasible rectangles.

## Elegant strategy to rectangles, consider the vertices and diagonals. I"ve been sent an innovative solution come 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 certain sizes of rectangles and working out whereby they can be located we start at the other end and also look at locations first.The vertices are the intersections. For our chessboard there are 81 (9 x 9). A diagonal starting at one vertex and ending at one more will uniquely explain a rectangle. In stimulate to it is in a diagonal and not a vertical or horizontal heat we might start anywhere however the end point must not have the same vertical or horizontal coordinate. Because of this there space 64 (8 x 8) feasible end points.There are thus 81 x 64 = 5184 agree diagonals.However, whilst every diagonal defines a unique rectangle, each rectangle does not describe a unique diagonal. We see trivially that each rectangle have the right to be stood for by 4 diagonals.So our number 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,) troubles can now be calculated. The variety of vertices being given by (n + 1)2 and also (n + 1).(m + 1) respectively. Thus the final 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 be arranged right into something much more complicated.

## Rectangles in Maths Nomenclature

It"s constantly my intentionally to explain the difficulties without formal maths nomenclature, v reasoning and also common sense. Yet there is quite a succinct solution here if you carry out know about combinations, as in permutations and combinations. Horizontally we are choosing 2 vertices from the 9 available. The order does not matter so it"s combinations fairly than permutations. And also the exact same vertically. Therefore the answer come the rectangle trouble can it is in answered by:9C2•9C2 = 362 = 1296
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