Tricky Track

Werg22

Three High Schools—Washington, Lincoln and Roosevelt— competed in a track meet. Each school entered one man, and one only, in each event. Susan, a student at Lincoln High, sat in the bleachers to cheer her boyfriend, the school's shot-put champion. When Susan returned home later in the day, her father asked how her school had done.

"We won the shot-put all right," she said, "but Washington High won the track meet. They had a final score of 22. We finished with 9. So did Roosevelt High."
"How were the events scored?" her father asked.
"I don't remember exactly," Susan replied, "but there was a certain number of points for the winner of each event, a smaller number for second place and a still smaller number for third place. The numbers were the same for all events." (By "number" Susan of course meant a positive integer.)
"How many events were there altogether?"
"Gosh, I don't know, Dad. All I watched was the shot-put."
"Was there a high jump?" asked Susan's brother.

Susan nodded.

"Who won it?"

Susan didn't now. As incredible as it may seem, this last question can be answered with only the information given. Which school won the high jump?

upendrasonu

Hi Werg22,

Washington High may have won the high jump event.

If we consider this situation wherein there are 5 events such as the winner takes 5 points and the schools at 2nd and 3rd places take 2 and 1 points each respectively.

Schools
Events Washington Lincoln Roosevelt
Shotput 2 5 1
HighJump 5 1 2
etc 5 1 2
etc 5 1 2
etc 5 1 2

Total 22 9 9

neutrino

>>>>>
I assume that no school was disqualified in any of the events.

• Lincoln High did not win
  In addition to the short-put, if LH had won the high-jump, then the restrictions on the scores would imply that they scored 4, 4 and 1. (1 in the track meet.) This also means that the team placed third in any event scored just 1 point, and more importantly, there were only three events. Given this, it is impossible for RH to have scored 9.
  
  
• Neither did Roosevelt High
  If RH had won it, then both RH and LH have scored 9 points with at least one win.
  
  In this case, the maximum points that could have been awarded is from the range 3-8, but in any case, the scores for WH and/or RH wouldn't add up to what they are. There are just too many cases to type here, but I just know they don't add up (due to a combination of the restrictions on the possible scoring system and the number of events .)

<<<<<

Werg22

What makes you assume there are only three events?

neutrino

Do you mean in the first case? >>>>4+4+1 = 9 Three events. In the first case, I assumed that LH had won the high-jump (therefore, LH won at least 2 events). If max. points = 3, then there were either 3 or 4 events. The case with 3 events is not possible, since that would mean LH won all events, which is not true. In the case with 4, the scores of WH cannot possibly add up to 22.

Werg22

All you are doing here is eliminating the distribution you have (4 - 4 - 1). Did you consider 3 - 3 - 1 - 1 - 1? Or 3 - 3 - 2 - 1?

neutrino

Yes, of course. As I said, it will contradict the fact that WH has 22 points.

LH 3,3,1,1,1

Even in the case where WH has won the rest of the three events and came second in other two. 3+3+3+2+2 = 13 != 22. Similarly for the case 3,3,2,1.

Werg22

Ok, but eliminating the possibility of RH having won one event is not very thorough. In the end, you'd have to list all possible cases, which makes it the least simple solution.

neutrino

That's true. I'll have to mull over that one.

K.J.Healey

Maximum value for 1st place is 8, (8,1=9)
Minimum number of games needed for each maximum value from (3,8) as follows (to equal 22):
8 8 6
7 7 7 1
6 6 6 4
5 5 5 5 2
4 4 4 4 4 2
3 3 3 3 3 3 3 1

Now the maximum usable for each to hit 9:
8 1
7 1 1
6 1 1 1
5 1 1 1 1
4 1 1 1 1 1
3 1 1 1 1 1 1


work our way up the numbers. (3 wont work, not enough to sum to.)
3 1 1 1 1 1 1 (7 places * 3 maximum < 22)


4 1 1 1 1 1 (9)
2 4 4 4 4 4 (22) this is the only way to get 22 with 4 as 1st place.
1 2 2 2 2 2 (11) close but no cigar

4 1 1 1 1 1 (9)
3 3 4 4 4 4 (22)
1 4 3 3 3 3 (nope) just getting larger

5 1 1 1 1 (9) Lincoln
2 5 5 5 5 (22) Washington
1 2 2 2 2 (9) Roosevelt

Looks like a solution to me. Washington won everything but the shot-put.

If you continue this way:
6 1 1 1 (9)
4 6 6 6 (22)
1 4 4 4 (>9)

And it continues to get messed larger. Theres only a certain number of limited possibilities for each sum to 9 and sum to 22 with remaining less than the 1st place level.

Werg22

Everyone here failed to make a simple realization which makes the very straightforward. Here is how I personally proceeded in solving it:

In each event, the total number of point given stays the same. For example, with the distribution 3 - 2 - 1, the number of points given would always be 6. This said, there was a total of 40 points given (22 + 9 + 9), which means that sum of the points attributed for the first, second and third place must be a factor of 40. A little thinking let's us realize that this sum can't be 2, 4 nor 5. Next comes 8, with two possible distributions: 5 - 2 -1 and 4 - 3 - 1. It can obviously not be the latter; as the maximum number of points a team could earn would be 4*5 = 20 (there are obviously five games in this case, as 8*5 = 40), which is under 22. We examine the other possibility: the only possible result for the winning team is ranking first in four events and second in other. This other event would obviously have to be the shot-put, in which one of the losing team won first place. This leaves the other losing team with 1 point on that event, meaning they have earned 8 points in the other 4 events; they obviously ended second in which one of them. This leaves us with the shot-put winning team having earned third place in the four other events; it adds up: 5 + 4*1 = 9. The puzzle is solved. Of course, this seems a bit lengthy, but it really is so simple it can be done without the use of any paper, just reflection.