Mathletics
Thomas Koshy from Ramingham State College in Massachusetts, USA, has sent us a nice problem. It was proposed by Kenton B. Creuser of Atlanta, Georgia, and appeared in a syndicated column written by Marilyn vos Savant.
Three schools meet for an athletics match, and each enters one competitor for each event. The number of events is unknown, and so is the scoring system, except that the winner scores a certain number of points, second place scores fewer and third place fewer still (but not zero). Georgia won the meet with 22 points and Alabama and Florida each scored 9. Florida won the High Jump; who won the mile?
Let n and p be the number of events and the total points for each event respectively. Then np=40. Also, since the smallest value of p would be 6 (for a 3-2-1) system) we have p 
6 and thus n < 7. Hence, since n divides 40, n = 2, 4 or 5. Supposed n = 2, then p = 20. But since Florida scored 9, including a win, the maximum total score for an event would be 8 + 7 + 1 < 20. Thus n 
2. So n = 4, p = 10 or n = 5, p = 8. Consider these in turn.
If n = 4, p = 10, the possible scoring systems are 7–2–1, 6-3-1, 5-4-1, and 5-3-2.
With the 7-2-1 system the worst Florida could get (a win and three thirds) is 7 + 3 × 1 = 10 > 9, so reject this.
The best Georgia could get (3 wins and a second) would then be 3 × 6 + 3, 3 × 5 + 4 and 3 × 5 + 3. Since all these are less than 22 we must reject them as well, so we now know that n = 5, p = 8, with the possible systems 5-2-1 and 4-3-1.
Looking again at Georgia’s score, the best possible, for 5 events, is four wins and a second, which in the two possible systems would be 4 × 5 + 2 or 4 × 4 + 3. The first of these gives the required total of 22. So Georgia had four wins and a second, and since Florida won the High Jump, Georgia won the mile. [We deduce also that florida had 1 win and 4 thirds = 5 + 4 × 1 = 9; Alabama had 4 seconds and a third = 4 × 2 + 1 = 9.]
