Enigma 1693 - Inseparable

[Chappy]

Enigma 1693 - Inseparable
New Scientist magazine, 14 April 2012.
By Richard England.

Albion, Borough, City, Rangers and United have
staged a tournament in which each team played
each of the others once. No two matches had
the same score, and no team scored more than
three goals in any match. Each played two home
and two away fixtures.

Albion drew against Borough; both won their
other three matches. Albion scored the same
number of goals at home as Borough, and conceded
the same number of goals at home as Borough.
Albion also scored the same number of goals
away as Borough scored away, and conceded the
same number of goals away as Borough.

If I told you how many goals Albion and Borough
each scored at home you could deduce with
certainty the score in the fixture between
Albion and Borough.

Tell me what the score was in that match.

Ciao,
Chappy.

[Ted Schuerzinger]

First, let's abbreviate the five teams A, B, C, R, and U.

All different results with nobody scoring more than 3 means there are 10
possibilities: 0:0, 1:0, 1:1, 2:0, 2:1, 2:2, 3:0, 3:1, 3:2, and 3:3.
Six of those are victories; four are draws. We therefore know that A
and B are responsible for all six victories, with C-R, C-U, and R-U
being draws.

We also know that there are 14 goals scored by winning teams in matches
which aren't draws, so one of A/B scored 3 once and 2 twice, while the
other scored 3 twice and 2 once. There are four goals given up by
winning teams in victories, and those are in the 2:1, 3:1, and 3:2
matches. Obviously each of A and B has to give up two of them, which
means one of them won 3:1, 2:1, and 2:0, while the other won 3:2, 3:0,
and 1:0.

Three of these results are home results, with the fourth home fixture
being a draw; one of the teams won twice at home. We have six
possilibites for those six home victories:

a) 3:1 and 2:1 Five home goals scored, two home goals conceded. This
would require the other team to win 3:0 at home and draw 2:2.
b) 3:1 and 2:0 Home goal difference 5:1. No possibility of 5:1 for the
othe rteam.
c) 2:1 and 2:0 Home goal difference 4:1. Other team has 3:0 and 1:1 at
home.
d) 3:2 and 3:0 Home goal difference 6:2; no possibility for 6:2 for the
other team
e) 3:2 and 1:0 Home goal difference 4:2; other team could have 2:0 and
2:2 or 3:1 and 1:1
f) 3:0 and 1:0 Home goal difference 4:0; impossible for other team to
achieve.

Therefore each of Albion and Borough scored five at home, and their
joint fixture must have ended 2:2.

And now on to the pools news.

--
Ted S.
fedya at hughes dot net
Now blogging at http://justacineast.blogspot.com