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Topic: School match (Read 693 times) |
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TJMann
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School match
« on: Apr 19th, 2008, 9:21pm » |
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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?
Sorry, I meant to put this in another category.
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| « Last Edit: Apr 19th, 2008, 9:25pm by TJMann » |
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Grimbal
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Re: School match
« Reply #1 on: Apr 20th, 2008, 7:41am » |
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To get so many points, Georgia must have won some events. If so, you can assign the mile to be one of these events. If the solution is unique, it can only be if Alabama can not win with the points given. So the solution is: Georgia.
Well here is one possible score
G A F
22 9 9 Total
2 1 5 High Jump
5 2 1 Mile
5 2 1 Holding breath
5 2 1 Snail rodeo
5 2 1 Riddling
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Michael Dagg
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Re: School match
« Reply #2 on: Apr 20th, 2008, 2:44pm » |
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Nice solution Grimbal!
It is the case of that the priori probability
dominates! (i.e. the games are over). There is
another way to solve it if you work in the ring
mod 40 but what you did was much better.
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Regards,
Michael Dagg
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TJMann
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Re: School match
« Reply #3 on: Apr 20th, 2008, 5:15pm » |
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Grimbals answer is the correct one.
But when you say "priori probability" you have to mean as you say that the "games are over" and thieir is no other observations because "priori" refers as previous, but I don;t discount that we can expect the games "not to end" at any one moment and that there can be further results, or at least further results if you do think about it the last way.
Where do you get mod 40 from? Is is from the total number of events times the total number of points per event?
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TJMann
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Re: School match
« Reply #4 on: Apr 20th, 2008, 8:19pm » |
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All right I know know what you say by mod 40=4*10. The scores 5-3-2,7–2–1,6-3-1,5-4-1 work for these. There is also another pattern 5-2-1 and 4-3-1 but this one is 40=8*5. I am slightly confused but I think I am near the exact answer.
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| « Last Edit: Apr 20th, 2008, 8:19pm by TJMann » |
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Grimbal
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Re: School match
« Reply #5 on: Apr 21st, 2008, 1:22am » |
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Here is a more formal proof, if you don't know beforehand the answer is unique.
The total number of points is 40 so the points for each event must divide 40.
That is 1,2,4,5,8,10,20,40.
But the number of points per event is at least 1+2+3 = 6, so the only possibilities are
5 events with 8 points each,
4 events with 10 points each,
2 events with 20 points each,
or 1 event with 40 points.
1 event is not possible since F won once and G has the most points.
2 events is not possible because F won once and earned only 9 points. So the points for a 1st place must be <= 9 and G could not earn 22 points in 2 events.
With 4 events:
G made 22 points in 4 events, so there are at least 6 points for 1st place. (If 1st place won 5 points or less, G could not have more than 20 points).
F won once and made 9 points. He got at least 6 for his 1st place, so only 1 point remain for the 3 other events. F was 1st 1 time and 3rd 3 times.
If A also won once, he also would have been 3rd 3 times conflicting with F.
That means F won once and G won all other games.
(In fact 4 events is not possible but it is longer to prove).
With 5 events:
5 events means 8 points per event.
8 points can only be split as (5, 2, 1) or (4, 3, 1) for (1st, 2nd, 3rd).
Combining 4, 3, 1 you can make 9 points only with 3+1+1+1, so F could not earn 9 points and get 4 points for a win.
So the earnings for a game must be (5, 2, 1) for (1st, 2nd, 3rd).
Combining 1,2,5, you can make 22 points only as 5+5+5+5+2. Again, F won once and G won all other games.
Since F won the high jump, G won the mile.
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| « Last Edit: Apr 21st, 2008, 1:25am by Grimbal » |
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TJMann
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Re: School match
« Reply #6 on: Apr 21st, 2008, 9:27am » |
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Thank you Sir for the great detail! I couldn't exactly put my finger on it like you could. I guess I need more practice!
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Hippo
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Re: School match
« Reply #7 on: Apr 21st, 2008, 10:41am » |
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[spam]
Let (1), (2) resp (3) denotes the points for corresponding place. Let k be number of matches.
As Grimbal shown the divisibility simplifies things so we know k<=5.
As (1)>=22/k we get (1)>=5. And as (2)>=2 and (3)>=1 we get (1)=5,(2)=2 and (3)=1 is the unique solution for 40=5*8 division.
For k=4 we get (1)>=22/4 so (1)>=6. From Florida's we know (1)<=9-(k-1)(3)<=9-3*1=6 therefore (1)=6 and (3)=1. Georgia gives 3*6+(2)=22 so (2)=4 and Alabama scores 12 ... contradiction.
If we only need to show Georgia wins the mile, it's sufficient to show the following does not have a solution:
(k-2)(1)+2*(2)>=22 and 2*(1)+(k-2)*(2)+k*(3)<=18. ... k<=2+(18-2((1)+(3)))/((2)+(3)).
Knowing
(1)>(2)>(3) are positive integers, so (1)>=3, (2)>=2 and (3)>=1 we get k<=2+10/3 ... k<=5. Therefore (1)>=22/5 (1)>=5 so k<=2+6/3=4 and (1)>=22/4 (1)>=6 so k<=2+4/3 k<=3 and (1)>=22/3 (1)>=8 so k<=2+0/3=2. So from the first one 2*(2)>=22 and (2)>=11 so (1)>=12 and Florida scores more than 9 ... contradiction.
Actually we have solved
(k-1)(1)+(2)>=22 and (1)+(k-1)*(2)+k*(3)<=18. ... k<=1+(18-((1)+(3)))/((2)+(3)).
Start with assumption (2)+(3)>3 ... (1)>(2)>=3.
k<=1+13/4 k<=4 so (1)>22/4 (1)>=6 k<=1+11/4 k<=3 so from divisibility also k<=2 and (1)>=11 gives contradiction for Florida. So (2)=2, (3)=1.
From Florida we know (1)+(k-1)(3)<=9 so k<=9-(1)+1. Therefore 22<=(k-1)(1)+(2)<=(9-(1))(1)+2=22.25-((1)-4.5)^2 so |(1)-4.5|=0.5 and (1)=4 or (1)=5. Former is wrong (4+2+1 does not divide 40  ) so (5,5,5,5,2 2,2,2,2,1 1,1,1,1,5 is the only case)
[/spam]
on Apr 24th, 2008, 8:22pm, Michael Dagg wrote:What is the idea behind that spam tag or is it just
something playful? |
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 I just mark the post not having much value
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| « Last Edit: Apr 25th, 2008, 3:55am by Hippo » |
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Michael Dagg
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Re: School match
« Reply #8 on: Apr 24th, 2008, 8:22pm » |
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What is the idea behind that spam tag or is it just
something playful?
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Regards,
Michael Dagg
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TJMann
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Posts: 18
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Re: School match
« Reply #9 on: Apr 29th, 2008, 10:02pm » |
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That last one is good and I understand it, ... as well as Grimbal's solution which is quite nice! Gee you guys are solution factories!
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