
Let a+b+c equal the number of 1st place
points, a+b equal the number of 2nd place points, and
a equal the number of 3rd place points. The total number
of points per event is 3a+2b+c and the
total number of points for the meet is 22+9+9 = 40 =
n(3a+2b+c).
Because there is one competitor per school per event, each school must
have a 1st, 2nd, or 3rd place finish in each event.
Possible values for n(point sum) are 10(4), 8(5), 5(8), 4(10)
Possible value for sums: >= 6 (3 2 1) hence n=5
or n=4.
Try n=5, sum=8, possible sums: 1+2+5 1+3+4

1+2+5:

Can we get 9 points with one win?
 1 win, 2 2nd place  no because we need 5 events
 1 win, 4 3rd place  OK, number of awards is 5
 4 2nd place, 1 3rd place  OK, number of awards is 5
Results for Washington: Lincoln must have a), Roosevelt b)
Washington: 4 wins, 1 2nd, 0 3rd  OK, number of awards is 5
sum = 20+2 = 22



1+3+4:

Can we get 9 points with two wins?
2 wins 1, 3rd  no since number of awards must be 5
Can we get 9 points with one win?
1 win, 1 2nd, 2 3rd  no, number of awards must be 5


Try n=4, sum = 10, possible sums = 1+2+7, 1+3+6, 1+4+5, 2+3+5

1+2+7:

Can we get 9 points with 1 win?
 1 win, 1 2nd  no, number of awards must be 4
 1 win, 2 3rd  no, number of awards must be 4



1+3+6:

Can we get 9 points with 1 win?
 1 win, 1 2nd  no, number of awards must be 4
 1 win, 3 3rd  OK
Can we get 9 points with 0 wins?
 3 2nd  no, number of awards must be 4
 2 2nd, 3 3rd  no, number of awards must be 4
 Cannot have L & R with 9 since #3rd is then 6



1+4+5:

Can we get 9 points with 1 win?
 1 win, 1 2nd  no, number of awards must be 4
 1 win, 4 3rd  no, number of awards must be 4



2+3+5:

Can we get 9 points with 1 win?
 1 win, 2 3rd  no, number of awards must be 4


The only results possible are n=5, points=1,2,5
Washington: 4 wins, 1 2nd (22)
Lincoln: 1 win, 4 3rd (9)
Roosevelt: 4 2nd, 1 3rd (9)
Since Lincoln wins shotput and nothing else, and Roosevelt wins
nothing, only Washington could have won the longjump. 