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24 Oct 2022, 01:30
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A
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Difficulty:
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32% (02:28) correct
68%
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There are n grandmasters in a certain chess tournament and each grandmaster should play each of the other grandmasters exactly once. Two of the grandmasters, after playing 5 games each, left the tournament but the remaining grandmasters still continued playing. If at the end of the tournament, the total number of games played was 45, what is the value of n ?
A. 11
B. 12
C. 13
D. 14
E. It cannot be determined from the information given.
A. 11
B. 12
C. 13
D. 14
E. It cannot be determined from the information given.
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24 Oct 2022, 09:07
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Two cases need to be considered -
1) The grandmasters who left the match didn't not play a tournament between them.
2) The grandmasters who left the match played a tournament between them.
Let's solve for each case -
Case 1 : The grandmasters who left the match didn't not play a tournament between them.
Number of grandmasters who played all the matches = n-2
Number of matches played by these grandmasters =\( n-2C_{2}\)
Number of extra matches (played by the two grandmasters who left the match) = 5 * 2 = 10
\(n-2 C_{2}\) + 10 = 45
\(\frac{(n-2)(n-3)}{2}\) + 10 = 45
(n-2)(n-3) = 70
'n' represents the number of grandmasters, hence it should be an integer. However for no integer value of n does the equation satisfy.
P.S: We can also substitute the option choices at this stage to verify if for any value of 'n', (n-2)(n-3) = 70.
As no value exists, we have to discard this case.
Case 2 : The grandmasters who left the match played a tournament between them.
Number of grandmasters who played all the matches = n-2
Number of matches played by these grandmasters =\( n-2C_{2}\)
Number of extra matches (played by the two grandmasters who left the match) = (5 * 2) - 1 = 9
Note: We have to subtract 1, as one match is counted for both the grandmasters.
\(n-2 C_{2}\) + 9 = 45
\(\frac{(n-2)(n-3)}{2}\) + 9 = 45
(n-2)(n-3) = 72
Let's look at the options available -
if n = 11
9 * 8 = 72
Therefore n = 11
IMO A
1) The grandmasters who left the match didn't not play a tournament between them.
2) The grandmasters who left the match played a tournament between them.
Let's solve for each case -
Case 1 : The grandmasters who left the match didn't not play a tournament between them.
Number of grandmasters who played all the matches = n-2
Number of matches played by these grandmasters =\( n-2C_{2}\)
Number of extra matches (played by the two grandmasters who left the match) = 5 * 2 = 10
\(n-2 C_{2}\) + 10 = 45
\(\frac{(n-2)(n-3)}{2}\) + 10 = 45
(n-2)(n-3) = 70
'n' represents the number of grandmasters, hence it should be an integer. However for no integer value of n does the equation satisfy.
P.S: We can also substitute the option choices at this stage to verify if for any value of 'n', (n-2)(n-3) = 70.
As no value exists, we have to discard this case.
Case 2 : The grandmasters who left the match played a tournament between them.
Number of grandmasters who played all the matches = n-2
Number of matches played by these grandmasters =\( n-2C_{2}\)
Number of extra matches (played by the two grandmasters who left the match) = (5 * 2) - 1 = 9
Note: We have to subtract 1, as one match is counted for both the grandmasters.
\(n-2 C_{2}\) + 9 = 45
\(\frac{(n-2)(n-3)}{2}\) + 9 = 45
(n-2)(n-3) = 72
Let's look at the options available -
if n = 11
9 * 8 = 72
Therefore n = 11
IMO A
28 Oct 2022, 12:55
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The number of matches played among N players is the well-known
N(N-1)/2
If the two players didn't play each other then each played 5 other players, for a total of 10 matches.
That would mean then 45-10 = 35 matches were played solely among the unknown number of remaining players, X. So
35=X(X-1)/2. Since 7*6=42 and 6*5=30, it is clear that this isn't possible.
So the two must have played each other, 1 match.
Then each also played 4 other players for a total of 8 matches. So the number of matches played solely among the unknown number of remaining players, X, must be
45-1-8= 36. So
36=X(X-1)/2 and therefore
X(X-1)=72. The only pair that works is 9 and 8. So the total number of players must have been
9+2=11
Posted from my mobile device
N(N-1)/2
If the two players didn't play each other then each played 5 other players, for a total of 10 matches.
That would mean then 45-10 = 35 matches were played solely among the unknown number of remaining players, X. So
35=X(X-1)/2. Since 7*6=42 and 6*5=30, it is clear that this isn't possible.
So the two must have played each other, 1 match.
Then each also played 4 other players for a total of 8 matches. So the number of matches played solely among the unknown number of remaining players, X, must be
45-1-8= 36. So
36=X(X-1)/2 and therefore
X(X-1)=72. The only pair that works is 9 and 8. So the total number of players must have been
9+2=11
Posted from my mobile device
