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# [#42] PS Challenge : Math Contest

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CEO
Joined: 15 Aug 2003
Posts: 3454
[#42] PS Challenge : Math Contest [#permalink]

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09 May 2004, 19:02
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Code: md ( its for my reference, dont worry about it)
Level : Advanced (if you do it under time limits )

1. Time yourself
2. Solve as fast as you can

Three problems were given to participants of a math contest. Each participant got 0,1,2, or 3 points for each problem. After the papers were graded, it turned out that no pair of participants received matching scores for more than one problem. What is the largest possible number of participants?

a. 8
b. 9
c. 12
d. 16
e. 24
Senior Manager
Joined: 06 Dec 2003
Posts: 366
Location: India
Re: [#42] PS Challenge : Math Contest [#permalink]

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10 May 2004, 00:29
Praetorian wrote:
Code: md ( its for my reference, dont worry about it)
Level : Advanced (if you do it under time limits )

1. Time yourself
2. Solve as fast as you can

Three problems were given to participants of a math contest. Each participant got 0,1,2, or 3 points for each problem. After the papers were graded, it turned out that no pair of participants received matching scores for more than one problem. What is the largest possible number of participants?

a. 8
b. 9
c. 12
d. 16
e. 24

Answer should be 24:

Solution:

total results possible 4 * 4 * 4 = 64
(3 questions and for each question 4 grades possible)

Now apply limitation that no pair of participants received matching scores for more than one problem,
__
000 |
001 | ---> Select only one pair from these, and remove other 3
002 |
003__|

__
010 |
020 | ---> Select only one from these, and remove other 2
030__|

__
100 |
200 | ---> Select only one from these, and remove other 2
300__|

Total remove = 3 + 2 + 2 = 7
If we do for each grade points, total removal = 7 * 4 = 28
So we are left with 64 - 28=36
out of 36, remove:

_
012 |--------> Remove any 1
013_|

_
021 |--------> Remove any 1
023_|

_
031 |--------> Remove any 1
032_|

Total removal will become 3 * 4=12 ( Pairs could be - 01, 02, 03, 10, 12, 13, 20, 21, 23, 30, 31, 32)

Remaining would be 36 - 12 = 24

If there is any flaw in solution, please let me know.
Dharmin
SVP
Joined: 30 Oct 2003
Posts: 1790
Location: NewJersey USA

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10 May 2004, 08:17
This problem is not clear to me. I see only following combinations

Student 1 = 000
Student 2 = 111 ( cannot have 00 )
Student 3 = 222 ( cannot have 00, 11 )
Student 4 = 333 ( cannot have 00, 11, 22 )
Student 5 = 012 ( cannot have 00, 11, 22, 33 )
Student 6 = 023 ( cannot have 00, 11, 22, 33, 01 )
what is meant by no pair had same score in more than one problem ?
Does it mean a pair can have one score in common and other two scores should differ ? ( This is what I think )

I believe anyone can be paired with anyone else.
Intern
Joined: 06 May 2004
Posts: 15
Location: Toronto

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10 May 2004, 09:18
16 is my answer

I am assuming that : when comparing scrores , its w.r.t a particular question . or in another words if you compare scores for any two students, only one of the (corresponding question)score would match.

possible combinations
000
012
021
033
103
122
110
131
201
213
220
232
302
311
323
330

--------
none of the two students have more than one score matching for corresponsing questions.
_________________

lets do it together....

SVP
Joined: 30 Oct 2003
Posts: 1790
Location: NewJersey USA

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10 May 2004, 11:16
I think I got the problem

if a student gets 012 then no one else can get
01x, x12 and 1x2
I get following combinations

001
100
010
112
211
121
023
302
230
203
320
032
222
333

14 combinations
CEO
Joined: 15 Aug 2003
Posts: 3454

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10 May 2004, 14:51
Problem courtesy: University of Maryland High School Math Competition

First, it is possible to have 16 participants (their scores could be 000,011,022,033,101,112,123,130,202,213,220,231,303,310,321,332).

The general pattern is that the scores are of the form (a,b,c) where a and b are anything (i.e., 16 possibilities) and c=a+b (mod 4). So, for any two participants, if their scores are the same on one of the first two problems but are different on the other, then their score on the third problem will also be different. To see that more than 16 participants is not possible, suppose that there were at least 17 participants. Since there are only 4 possible scores on the first question, there would be a group of at least 5 participants that got the same score on the first question. But then, since there are only 4 answers for the second question, at least two of these 5 would have the same score on the second question. Thus, this pair would have the same score on at least two of the questions, contradiction.

Thus, the answer is 16 (thats D)
10 May 2004, 14:51
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# [#42] PS Challenge : Math Contest

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