In cross-country races with teams of six runners, the team scores are calculated by adding together the finishing positions of the first four runners in each team. The lowest scoring team is the winner. Individuals never tie for any position.
The fifth and sixth runners to finish in each team do not contribute to the team score, but if they finish ahead of scoring runners in other teams they make the positions of those scoring runners and their teams' scores that much worse.
In a race involving seven teams of six runners, the scores of the first four teams formed a geometric progression, as did the scores of the last four teams. what were the scores of the first and last teams?
Let the seven scores be:
S1 S2 S3 S4 S5 S6 S 7
Let the ratio of the geometric progression of the first four teams be
w (w for winners),
and the ratio of the last four teams be
l (l for losers),
\ S2 = w S1, S3 = w S2 = w2 S1, S4 = w S3 = w3 S1
and S5 = l S4, S6 = l S5 = l2 S4, S 7 = l S6 = l3 S4 = l3 w3 S1
Taking account of the highest and lowest possible scores, we have:
S 7 ≤ 154 and S1 ≥ 10
\ l3 w3 = S 7 / S1 ≤ 15.4
\ l w ≤ 3√ 15.4 < 5/2
Neither w or l is necessarily integer, but both are necessarily rational and greater than 1.
Given that all the scores are integers, especially S4,
the denominator of w must occur as a factor in S1
at least 3 times.
Similarly, the denominator of l must occur as a factor in S4
at least 3 times.
Let us define:
| | w | = | Nw Dw | and | l | = | Nl Dl |
S1 has to be a multiple of Dw3 in order that S4 is integer.
S4 has to be a multiple of Dl3 in order that S 7 is integer.
This can either occur because S1 is a multiple of Dl3 Dw3,
or because Nw is a multiple of Dl.
If either denominator is 1, its corresponding numerator must be 2
as the ratio must exceed 1, and the other ratio must then be less than 5/4.
So the denominator must be at least 5.
If Dl = 1 and Dw = 5, S1 has to be a multiple of 125, which is far too a big a score for the
winning team.
If Dw = 1 and Dl = 5, S1 still has to be a multiple of 125 because Nw is not a multiple of Dl.
It seems very likely that
Nw = Dl = 3 ,
Dw = 2 ,
and the winning score is 16.
l < 5/3, \ Nl = 4.
The scores are then:
16, 24, 36, 54, 72, 96, 128
I've not worked out how you make up the scores, but my hunch looks right.
It meets all the constraints that are listed above.
If you swap the ratios round, the losing total becomes 216, i.e. too big.
w = 4/3 and l = 5/4 works with a winning score of 27:
27, 36, 48, 64, 80, 100, 125
If the denominators get any bigger, it looks like the losing score gets too large.
Also the winning score tends to look a bit large.
To a first approximation, the largest winning score (and also the smallest losing score) will occur when the scores are shared approximately evenly.
Label the teams A, B, C, D, E and F, and
consider a situation when the runners in the first batch of 7 are all from different teams, e.g.:
ABCDEFG ABCDEFG ABCDEFG ABCDEFG
All the runners that count are in the first four batches.
The teams finish in the order A, B, C, D, E, F with scores of:
team A: S1 = 1 + 8 + 15 + 22 = 46
team B: S2 = S1 + 4 = 50
team C: S3 = S2 + 4 = 54
team D: S4 = S3 + 4 = 58
team E: S5 = S4 + 4 = 62
team F: S6 = S5 + 4 = 66
team G: S 7 = S6 + 4 = 70
But we only need a winning margin of 1, so we can push the last runner
from team A back 2 places, so that teams B and C each move up 1:
ABCDEFG ABCDEFG ABCDEFG BCADEFG
team A: S1 = 1 + 8 + 15 + 24 = 48
team B: S2 = 2 + 9 + 16 + 22 = 49
team C: S3 = 3 + 10 + 17 + 23 = 53
team D: S4 = 4 + 11 + 18 + 25 = 58
If we allow teams to tie (and there seems no reason not to), although
individual runners cannot, the reversal of the sequence in two of the batches will give:
GFEDCBA ABCDEFG GFEDCBA ABCDEFG
team A: S1 = 7 + 8 + 21 + 22 = 58
team B: S2 = 6 + 9 + 20 + 23 = 58
team C: S3 = 5 + 10 + 19 + 24 = 58
team D: S4 = 4 + 11 + 18 + 25 = 58
team E: S5 = 3 + 12 + 17 + 26 = 58
team F: S6 = 2 + 13 + 16 + 27 = 58
team G: S 7 = 1 + 14 + 15 + 28 = 58
We can actually have separation between the teams by using the
non-counting runners.
If we put the two non-counting A team runners ahead of the last finisher from team B,
we get:
GFEDCBA ABCDEFG GFEDCBA AAABCDEFG
team A: S1 = 7 + 8 + 21 + 22 = 58
team B: S2 = 6 + 9 + 20 + 25 = 60
team C: S3 = 5 + 10 + 19 + 26 = 60
team D: S4 = 4 + 11 + 18 + 27 = 60
team E: S5 = 3 + 12 + 17 + 28 = 60
team F: S6 = 2 + 13 + 16 + 29 = 60
team G: S 7 = 1 + 14 + 15 + 30 = 60
Similarly we can put the two non-counting B team runners ahead of the last finisher from team C,
and so on, we get:
GFEDCBA ABCDEFG GFEDCBA AAABBBCCCDDDEEEFFFG
team A: S1 = 7 + 8 + 21 + 22 = 58
team B: S2 = 6 + 9 + 20 + 25 = 60
team C: S3 = 5 + 10 + 19 + 28 = 62
team D: S4 = 4 + 11 + 18 + 31 = 64
team E: S5 = 3 + 12 + 17 + 34 = 66
team F: S6 = 2 + 13 + 16 + 37 = 68
team G: S 7 = 1 + 14 + 15 + 40 = 70
Team A now has a clear lead of 2 over team B, so that if we swap their finishers
in the second batch, they will each be equal on 59.
In fact, we can reverse the order of the second batch and get:
GFEDCBA GFEDCBA GFEDCBA AAABBBCCCDDDEEEFFFG
team A: S1 = 7 + 14 + 21 + 22 = 64
team B: S2 = 6 + 13 + 20 + 25 = 64
team C: S3 = 5 + 12 + 19 + 28 = 64
team D: S4 = 4 + 11 + 18 + 31 = 64
team E: S5 = 3 + 10 + 17 + 34 = 64
team F: S6 = 2 + 9 + 16 + 37 = 64
team G: S 7 = 1 + 8 + 15 + 40 = 64