# A Martin Gardner problem about constructing a scoreboard given minimal information

The following is a paraphrasing of a problem from Martin Gardner called "Tricky Track."

Three schools $-$ $A,B,C$ $-$ participated in a track meet with several events. Each school was awarded some amount of points for each event (a positive integer). Each event was scored in the same way, and of course the first place winner of an event would receive more points than the second place one, and so on.

Susan, a student of school $B$, won the shot-put for her school. However, they were still beaten handily by school $A$, with the final result of the track meet being $22-9-9$, respectively.

How many events took place at the track meet? Can you reconstruct the entire scoreboard?

--

There is no solution provided, just an answer.

• The total score for an individual event must be a divisor of $22+9+9=40$. The points for first place has to be at most 9.
– Thomas Andrews
Aug 5, 2014 at 0:08
• from Thomas' comment you get that there were 5 games played with 8 total points in each one. After this you also have two different possible ways to distribute the points: 5-2-1 or 4-3-1 (do the nth place for n>3 also gets points?) Aug 5, 2014 at 0:39

1. There are $22 + 9 + 9 = 40$ total points.
2. As each school receives a positive integer number of points for each event, and placing strictly better gives strictly more points, there must be at least $6 = 1 + 2 + 3$ points per event.
3. The divisors of $40$ that are greater than $6$ are $8,10,20,$ and $40$, and so there can be $5,4,2,$ or $1$ events.
4. As schools $B$ and $C$ are tied at the end, there must be more than one event.
5. As school $B$ won an event and there is more than one event, first place must give strictly fewer than $9$ points.
6. If there are exactly two events, then first place must give exactly $8$ points, as $8,7,5$ is the only possible distribution of points that gives first place fewer than $9$ points, sums to $20$ points per event, and gives strictly more points to schools placing strictly better.
7. Note, however, that this makes it impossible for school $B$ to have $9$ points.
8. Similarly, if there are exactly four events, then the only possible distribution of points is $1,3,6$.
9. Then, giving school $B$ 9 points can happen in only one way — $1,1,1,6$, and now it is impossible for school $C$ to have only $9$ points.
10. Thus, there are five events. There are two possible distributions of points, now — $1,2,5$ and $1,3,4$. The latter does not let school $B$ get exactly $9$ points, and so the answer is that there are $5$ events, with $5$ points for first place, $2$ points for second place, and $1$ point for third place.
This works, with school $A$ getting $5,5,5,5,2$, school $B$ getting $1,1,1,1,5$, and school $C$ getting $2,2,2,2,1$.

$40$ total points were scored. As the bottom two schools get at least $3$ points per event and $B$ won one, there were at most $5$ events. So far, we could have $5$ events with a total score of $8$ per event or $4$ events with a total score of $10$ per event.

The first thing I would try is five events with scores $5,2,1$. Then $B$ can score $5,1,1,1,1, C$ can score $1,2,2,2,2$ and $A$ can score $2,5,5,5,5$ Assuming there is only one solution (and you can trust Gardner for that) here we are.