The first thing to note is that, for a given league of N teams, how many matches will be
played in that league? Well, a league of two teams has 1 match, of three teams has 3
matches, of four teams has 6 matches, and so on. This creates an arithmetic series which is
the sum of all natural numbers from 1 to n, where n = N-1, the sum of which is n/2*(n+1).
[The proof for this, briefly, being that the series is [1 + 1+1 + ... + 1+(n-2) + 1+(n-1)] From
here we see that taking terms from the beginning and end and pairing them up gives us n+1
consistently, meaning if we had two of this sequence and paired up each corresponding
term, we would have n lots of n+1, and therefore half of that for one sequence, namely
n/2*(n+1) ]
Now that that is established, we can apply this to the situation. Taking X as the n value for
one league, and Y for the other, we know that:
[X/2(X+1)] - [Y/2(Y+1)] = 171
[(X^2+X)/2] - [(Y^2+Y)/2] = 171
(X^2-Y^2+X-Y)/2 = 171
X^2 - Y^2 + X - Y = 342
We also know that (X+1)+(Y+1) = 39, as X & Y represent one less than the number of teams
in each league, and therefore X+Y = 37
From here, we can substitute Y = 37-X into the equation, as follows:
X^2 - (37-X)^2 + X - (37-X) = 342
X^2 - (1369+X^2-74X) + X - 37 + X = 342
X^2 - X^2 + 2X + 74X -1369 -37 = 342
76X - 1406 = 342
76X = 1748
X = 23
We also know from this that Y = 14, as X+Y = 37. This means that one league has 15 teams
in it, and the other has 24.
From the Formula established at the beginning, we can also know that the smaller league
has a total of 105 matches, and the larger has 276 matches, which matches the difference of
171, but is otherwise not actually important for the last stage of the question.
We know that Chelchestpool played either 14 or 23 matches, depending on which league
they are in. 3 of these matches earned them 0 points, and the rest of the matches earned
them either 1 or 3 points, in unknown proportions. Let us first assume that they played in the
smaller league:
3*0 + 1*d + 3*w = 32
AND
d+w+3 = 14
Quickly substituting in w = 11-d shows us that the team playing in the smaller league is
actually impossible, as:
d + 3(11-d) = 32
d + 33 -3d = 32
-2d = -1
d = 0.5
As it is impossible to draw half of a game, we know that Chelchestpool must be playing in
the larger league. From here, it is an easy solution, as:
3*0 + 1*d + 3*w = 32
AND
d+w+3 = 23
This time, substituting in w = 20-d
d + 3(20-d) = 32
d + 60 -3d =32
-2d = -28
d = 14
Which is a logical answer for a number of draws, unlike 0.5.
Due to this we know that Chelchestpool played in the larger of the two leagues (One
consisting of 24 teams, 276 total matches, and each team playing in 23 matches), and it:
Lost only 3 of its games.
Won a respectable 6 of them.
Drew a whopping 14 of them.