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Topic: chess puzzler
Replies: 11   Last Post: Aug 26, 2004 10:01 PM

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Alan Lipp

Posts: 41
Registered: 12/6/04
Re: chessboard rectangles
Posted: Oct 11, 1998 2:13 PM
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I agree with your total (1296). With a little rearranging your method
will produce the general formula for the number of rectangles on an n by n

Cont the rectangles by rows.

In row 1 there are n 1 by 1s , n-1 1 by 2s, ... 2 1 by n-1s and 1 1 by n:

Row Total = n + (n-1) + (n-2) + ... + 3 + 2 + 1 = n(n+1)/2 rectangles.

But of course there are n rows giving n ( row sum )

Now count all rectangels of height 2 . Start in the bottom row.

There are n 2 by 1s and n-1 2 by 2s and ... and 1 2 by n for a

Row Total = n + (n-1) + (n-2) + ... + 3 + 2 + 1 = n (n+1)/2

The row total is the same, as it will be for all the rectangles of height
3, 4, ... n because they all share the same bases at the bottom of the
board. However, there are only n-1 rows of height 2 and n-2 rows of
height 3 etc. Thus,

# 1 by any totals: n [n(n+1)/2]
# 2 by any totals: (n-1)[n(n+1)/2]
# 3 by any totals: (n-2) [n(n+1)/2]


#n by any totals: 1 [n(n+1)/2]

So the total number of rectangles is [n(n+1)/2](n + n-1) + ... + 3+ 2+1)

or Total = [n(n+1)/2]^2

which, as you mentioned, is the sum of the cubes from 1 to n.

Alan Lipp
On Sun, 11 Oct 1998, Sarah Seastone wrote:
> Someone asked the number of rectangles on a chessboard. The answer is
> "Many, in fact 1296."


> Have I missed any? Or added wrong? What if we reduce or augment the
> size of the chessboard? Can we find a general formula? What is it? How
> do we use induction to show that it is indeed general?

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