Re: Enigma 1380 - Points for a win

Stephen Merriman wrote:
Mark J. Tilford wrote:

On 19 Apr 2006 00:57:46 -0700, Chappy <petergregorychapman@xxxxxxxxxxx> wrote:


Enigma 1380 - Points for a win
New Scientist magazine, 25 February 2006.
by Susan Denham.

Our local football team consists of five teams and in
the course of the season each team plays each of the
others once. One point is awarded to each team for a
draw and a higher whole number of points (I forget how
many) are awarded for a win. Below is part of the league
table sometime during the season before all the matches
had been completed.

(Fixed width font required)

Won Lost Drawn Goals Goals Points Position
against for in league
Authors N
Bankers P O I N T S
Chemists F O R
Doctors A
Editors W I N

The teams are shown in alphabetical order rather than
order of merit. No teams have the same number of points.
In the table, digits have been replaced consistently by
letters, with different letters used for different digits.

How many points for a win? Which team or teams (if any)
have the Authors beaten so far?

Ciao,
Chappy.








As Richard pointed out, there was one possibility you missed. Continuing from where you left off:
A must have a win (4) and a draw, or a win (3) and two draws.
For B, we have one of the following:
1 win (4) 2 draws = 6 points
1 win (4) 3 draws = 7 points
2 wins (4) 1 draw = 9 points
1 win (3) 3 draws = 6 points
2 wins (3) 1 draw = 7 points

Lets suppose it is 3 points for a win. Then we have the following:

Won Lost Drawn Goals Goals Points Position
against for in league
Authors 1 2 5
Bankers P 0 I 5 T S
Chemists . 0 F 0 R
Doctors 0 0 4 4
Editors 0 . I W I 5

P = 1, 2
I = 1, 2, 3
F = 2, 3
T,R,W,S = 6,7,8,9

The total number of draws must be even, so F = 2, so P = 1, so I = 3, so S = 6.
So A has 2 draws, B has 3, C has 2, D has 4, E has 3.

D must have drawn with everybody. That leaves 1 draw for A, 2 for B, 1 for C, 2 for E. A cannot have drawn with C (B and E can't draw with each other twice), and we know A beat E, so A and B drew. That forces the other two draws to be B and E, and C and E. But that means that B drew with A, B drew with D, B drew with E. That only leaves C for B to beat, which is impossible as C lost no games.

So its 4 points for a win.

You can continue with the same sort of logic for the 3 point case, ending up somewhere like this:

I meant 4-point case, of course.

Stephen
.

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