Difference between revisions of "2017 AMC 12B Problems/Problem 25"
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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>. | 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>. | ||
− | ==Solution 1. | + | ==Solution 1.0== |
Another way to find the average number of teams in a group of 9 or 8 people is as follows: | Another way to find the average number of teams in a group of 9 or 8 people is as follows: | ||
Revision as of 16:55, 16 October 2024
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
Let there be teams. For each team, there are
different subsets of
players that includes a given 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 if
. 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
.
Solution 1.0
Another way to find the average number of teams in a group of 9 or 8 people is as follows:
Let there be teams. There are
total possible teams. So, the probability of any 5 people being a team is
. There are
possible teams in a group of 9. So, on average, there will be
teams in any group of 9 people. Similarly, on average, there will be
teams in any group of 8 people. So,
and proceed as shown in solution 1.
FYI, to find n such that without bashing everything:
Clearly
works. Then, do cases.
Case 1: .
This will always work, as
,
, and
, so
. So,
are also solutions.
Case 2:
Then
as well.
and
can only contribute one 2 each, since
. We need a factor of
in
then. So,
. Then, we get
.
Case 3:
We can only get twos from
and
. Note that one of them can only contribute a single factor of 2, because otherwise
and
, so
has a remainder of both 1 and 3 mod 4, which is impossible. So, one must have a factor of 16. We get
so
In all, we get the solutions
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.