Difference between revisions of "2017 AMC 12B Problems/Problem 25"
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Solution by Pieater314159 | Solution by Pieater314159 | ||
+ | Minor edits by Zeric | ||
Let there be <math>T</math> teams. For each team, there are <math>{n-5\choose 4}</math> different subsets of <math>9</math> players including that full team, so the total number of team-(group of 9) pairs is | Let there be <math>T</math> teams. For each team, there are <math>{n-5\choose 4}</math> different subsets of <math>9</math> players including that full team, so the total number of team-(group of 9) pairs is | ||
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It is obvious that <math>5</math> divides the RHS, and that <math>7</math> does iff <math>n\equiv 0,1,2,3,4\mod 7</math>. Also, <math>3^2</math> divides it iff <math>n\not\equiv 5,8\mod 9</math>. One can also bash out that <math>2^5</math> divides it in <math>16</math> out of the <math>32</math> possible residues <math>\mod 32</math>. | It is obvious that <math>5</math> divides the RHS, and that <math>7</math> does iff <math>n\equiv 0,1,2,3,4\mod 7</math>. Also, <math>3^2</math> divides it iff <math>n\not\equiv 5,8\mod 9</math>. One can also bash out that <math>2^5</math> divides it in <math>16</math> out of the <math>32</math> possible residues <math>\mod 32</math>. | ||
− | + | Note that <math>2016 = 7*9*32</math> so by using all numbers from <math>2</math> to <math>2017</math>, inclusive, it is clear that each possible residue <math>\mod 7,9,32</math> is reached an equal number of times, so the total number of working <math>n</math> in that range is <math>5\cdot 7\cdot 16 = 560</math>. However, we must subtract the number of "working" <math>2\leq n\leq 8</math>, which is <math>3</math>. Thus, the answer is <math>\boxed{\textbf{(D) } 557}</math>. | |
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+ | Alternatively, it is enough to approximate by finding the floor of <math>2017 \cdot \frac57 \cdot \frac79 \cdot \frac12 - 3</math> to get <math>\boxed{\textbf{(D) } 557}</math>. | ||
Video Solution by Dr. Nal: | Video Solution by Dr. Nal: |
Revision as of 14:51, 7 November 2021
Problem
A set of people participate in an online video basketball tournament. Each person may be a member of any number of -player teams, but no two teams may have exactly the same members. The site statistics show a curious fact: The average, over all subsets of size of the set of participants, of the number of complete teams whose members are among those people is equal to the reciprocal of the average, over all subsets of size of the set of participants, of the number of complete teams whose members are among those people. How many values , , can be the number of participants?
Solution
Solution by Pieater314159 Minor edits by Zeric
Let there be teams. For each team, there are different subsets of players including that full team, so the total number of team-(group of 9) pairs is
Thus, the expected value of the number of full teams in a random set of players is
Similarly, the expected value of the number of full teams in a random set of players is
The condition is thus equivalent to the existence of a positive integer such that
Note that this is always less than , so as long as is integral, is a possibility. Thus, we have that this is equivalent to
It is obvious that divides the RHS, and that does iff . Also, divides it iff . One can also bash out that divides it in out of the possible residues .
Note that so by using all numbers from to , inclusive, it is clear that each possible residue is reached an equal number of times, so the total number of working in that range is . However, we must subtract the number of "working" , which is . Thus, the answer is .
Alternatively, it is enough to approximate by finding the floor of to get .
Video Solution by Dr. Nal:
https://www.youtube.com/watch?v=2p2qYRWbvV4&feature=emb_logo
See Also
2017 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 24 |
Followed by Last Problem |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
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