25 golfers have a tournament. They split up into 5 groups each with
5 players. After 6 playing rounds, every person has played against all
of the others. How was the 25-5-6 tournament arranged?
I don't see any clever ways of solving it.
1: abcde, fghij, klmno, pqrst, uvwxy
2: afkpu, bglqv, chmrw, dinsx, ejoty
3: agmsy, bhntu, ciopv, djkqw, eflrx
4: ahoqx, bikry, cjlsu, dfmtv, egnpw
5: ailtw, bjmpx, cfnqy, dgoru, ehksv
6: ajnrv, bfosw, cgktx, dhlpy, eimqu
--Ed Pegg Jr, www.mathpuzzle.com
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Dear Ed,
There is a simple mathematical way of constructing the structure you want.
Let the 25 peope be represented by the points (x,y) where each of x and y
is an integer from 0 to 4 inclusive.
We use arithmetic modulo 5. There are six families of parallel lines:
(1) Slope 0: y=0, y=1, y=2, y=3, y=4.
(2) Slope 1: y=x, y=x+1, y=x+2, y=x+3, y=x+4.
(3) Slope 2: y=2x, y=2x+1, y=2x+2, y=2x+3, y=2x+4.
(4) Slope 3: y=3x, y=3x+1, y=3x+2, y=3x+3, y=3x+4.
(5) Slope 4: y=4x, y=4x+1, y=4x+2, y=4x+3, y=4x+4.
(6) Slope infinite: x=0, x=1, x=2, x=3, x=4.
Such a structure is called a finite affine plane. You can of course go
to higher dimensions.
Best regards, Andy Liu.
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label the puzzlers with pairs of integers (i, j) with 0 <= i,j < 5.
on round k , group the golfers according to the residue of
f_k(i,j) modulo 5 , where the linear functions are
f_k(i, j) = i + k * j (for k = 1, 2, 3, 4, 5)
and
f_6(i, j) = j .
note: this kind of construction generalizes very easily. the sets are
basically lines in an affine plane over a finite field.
mike reid