I am taking this problem from the book The Great Book of Mind Teasers & Mind Puzzlers
by George J. Summers. In the book, it is called the Dart
Game. The solution process that will be given differs from that
given in the book and has, I hope you will agree, a certain degree of
elegance.
Three people play a game of darts, each throwing a dart nine
times. The possible scores for a given throw are 1, 5, 10, 25, 50
and 100. Each of the six dart scores is thrown by one person, i.e, one person
throws all the 1's, one person throws all the 5's, etc.The total score for the nine dart throws
is the same for each player. As a start, show that
there must be five 1's, which is a crucial part of the solution in the
book and in the solution given here.
Summers required that each of the six scores was achieved. As
another problem, suppose we lift this requirement. Show that the
problem cannot be solved for fewer than the six scores.
Answer.
Let us define MaxScore to be the maximum score of a given player.
Whoever scored 100 has a MaxScore of 100. The next highest
MaxScore is at most 50, and the third highest MaxScore, which will be
called the MinMaxScore, is at most 25, which willl be the case when
100, 50 and 25 are each thrown by different players.
What does this tell us? Someone has scores no greater than 25, so that
person's score total is at most 9 x 25 = 225, and since all the score
totals are the same, 225 is an upper bound for the common total score.
Can we improve on this? If 100, 50 and 25 are scored by two
players, the MinMaxScore is at most 10, and the associated score total
is at most 9 x 10 = 90. This obviously can't be the case, since
the person who scored 100's scored over 100. We conclude that
each of 100, 50 and 25 was scored by a different person and that 225 is
an upper bound for the score total.
In an analogous fashion we can define MinScore to be the minimum score
of a given player, and MaxMinScore to be the highest MinScore, which
must be at least 10, and which occurs when 1, 5 and 10 are thrown by
different players. This tells us that the total score is at least
9 x 10 = 90, which is not particularly useful since we already know
that the total score is at least 100. Nevertheless, the
requirement that 1, 5 and 10 are thrown by different
players makes this case relatively easy to analyze.
As pointed out be Summers, there must be 5 throws of 1, since all
the other scores are divisible by 5. In the case being considered, the
100's thrower cannot be the person who threw the 1's since the total
score of 5 + 4 x 100 = 405, greater than the upper bound of
225. The 1's are thrown by either the 25 scorer or the 50 scorer,
giving possible total scores of 105 and 205. The 100's player
must have either one 100 or two of them. If two were thrown, the total
score would be at least 2 x 100 + 7 x 5 = 235, which goes over the 225
limit. With one 100 the possible scores are 100 + 8x5 = 140 and 100 + 8
x 10 = 180, neither of which match the possible scores of 105 and 205.
It must be the case that the scores of 1, 5 and 10 are thrown by two of
the players and that the MaxMinScore is at least 25. The
MaxMinScore cannot be greater than the MinMaxScore, since that would
give one player with maximum score less than the minimum score of
another player, meaniing the total score of one player would be less
than that of the other.
We conclude that the MaxMinScore = 25, the same as the
MinMaxScore. The person who scored 25 has no greater and lesser
score.The person threw nine 25's for a total of 9 x 25= 225, which must
be the total score of the others. We need to distribute the
remaining scores of 1,5 snf 10 between the 100's scorer and the 50's
scorer.
If the 50's scorer scored the 1's then the total score is at most 5 + 4
x 50 = 205, which is less than the requried 225 total. Therefore
the 100's scorer must have also scored the 1's. If the 100's
scorer only scored one 100, the total is at most 100 + 5 + 3 x 10 = 135
< 225. With two 100's, the total for seven throws is 200 + 5 =
205. Adding two 10's gives the required total of 225.
The 50's thrower must have thrown the remaining score of 5. We could
determine algebraically how many 5's and 50's were thrown from x+y=9
and 50x + 5y = 225. This should not be necessary. It is not
too hard to determine that there must be four 50s and five 5's to get
the desired total of 4 x 50 + 5 = 225.
The three sets of scores are:
Nine 25's
Four 50's and five 5's
Two 100's, two 10's and five 1's
Consider the cases where fewer than six scores are used. It is
obvious that the number of scores is greater than three. Four
scores can also be dismissed, since the MinMaxScore is at or below
the second lowest score and the MaxMinScore is at or above the
third lowest score, creating a case where the
MinMaxScore must be less than the MaxMinScore.
If six scores are not all used then five of them must be. The
MinMaxScore and the MaxMinScore must both be the middle score, which
well be either 10 or 25, depending on which score is omittted..
One player will only have only the middle score, with an associated
total of 90 or 225, and the other players will have one each of
the high scores and one each of the low scores.
Consider first the cases where the omitted score is other than 1.
The score total is either 90 or 225. Someone has five 1's and
four of another score. Let s stand for this other score.
Then 5 + 4s = 90 or 5 + 4s = 225 Then either s is not a whole number or
equals 55, which does not match any of the scores.
Therefore no solution is possible in these cases..
If the 1's are omitted then the middle value is 25 and the total must
be 225. 50's and 100's combine with 5's and 10's. The 10's
must be used with either the 50's or the 100's, requiring the total
score to be divisible by 10, which is not true for 225, so there is no
solution in this case either.