## Brain-Teaser 492

**From The Sunday Times, 1st November 1970** [link]

Six football sides, A, B, C, D, E and F are all to play each other once, with 2 points for a win, 1 point for a draw.

After some of the matches have been played a battered piece of paper is discovered on which has obviously been written particulars of the results so far.

All that can be read is shown below:

What were the details (opponents and scores) of E’s matches? (Example: EvP = 4-3, EvQ = 0-2, etc. E’s score first).

[teaser492]

## Jim Randell 2:26 pm

on8 August 2019 Permalink |F has 7 points and has played 5 games, but they only have 2 goals for, so cannot have won more than 2 games. Their 7 points must be made by 2w + 3d, which must be 1-0, 1-0, 0-0, 0-0, 0-0, so their goals against is 0. Hence the total number of goals scored is 25, and between them C and E have scored 17 goals.

This Python program chooses values for C and E’s “goals for” values and then uses the [[

`Football()`

]] helper class from theenigma.pylibrary to determine the outcomes and scorelines of each match.It runs in 550ms.

Run:[ @repl.it ]Solution:The outcomes of E’s matches are: E vs A = not yet played; E vs B = 0-1; E vs C = 3-5; E vs D = not yet played; E vs F = 0-1.There is only one scenario that works, which is when C has 14 goals for, and E has 3.

The full scores are:

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## John Crabtree 6:24 pm

on8 August 2019 Permalink |This teaser has a logical solution which leads directly to the answer, with no other paths.

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## John Crabtree 2:36 pm

on13 August 2019 Permalink |F has 7 points, but only scores 2 goals, and so F’s line reads 5 2 0 3 2 0 7

C has 7 points, but loses a game, and so C’s line reads 5 3 1 1 ? 7 7

As A plays 2 games, E must play 3 games, not 5.

The total of the Played column = 22 = the total of the Points column.

A has a minimum of 2 points and so the minimum Points total = 22

And so A has 2 points and so A’s line reads 2 1 1 0 4 2 2

And so E has 0 points and so B’s line reads 3 0 3 0 ? 7 0

B can only draw one game not three, and so B’s line reads 4 1 2 1 1 4 3

D can only draw one game,not three, and so D’s line reads 3 1 1 1 1 5 3

By inspection A v F 0-1, B v F 0-0, C v F 0-0, D v F 0-0, E v F 0-1.

Then A beats C. C beats B, D and E. B loses to D and beats E.

From A’s GF and GA, A v C 4-1.

From D”s GF = 1, , B v D 0-1

From D’s GF and GA, C v D 5-0

From B’s GF = 1, B v E 1-0

From B’s GF and GA, C v B 3-0

From C’s GA and E’s GA, C v E 5 -3

And so E played 3 matches: E v B 0-1, E v C 3-5 and E v F 0-1

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