2017 AMC 12B Problems/Problem 25

Problem

A set of $n$ people participate in an online video basketball tournament. Each person may be a member of any number of $5$ -player teams, but no two teams may have exactly the same $5$ members. The site statistics show a curious fact: The average, over all subsets of size $9$ of the set of $n$ participants, of the number of complete teams whose members are among those $9$ people is equal to the reciprocal of the average, over all subsets of size $8$ of the set of $n$ participants, of the number of complete teams whose members are among those $8$ people. How many values $n$ , $9\leq n\leq 2017$ , can be the number of participants?

$\textbf{(A) } 477 \qquad \textbf{(B) } 482 \qquad \textbf{(C) } 487 \qquad \textbf{(D) } 557 \qquad \textbf{(E) } 562$

Solution

Let there be $T$ teams. For each team, there are ${n-5\choose 4}$ different subsets of $9$ players that includes a given full team, so the total number of team-(group of 9) pairs is

$T{n-5\choose 4}.$

Thus, the expected value of the number of full teams in a random set of $9$ players is

$\frac{T{n-5\choose 4}}{{n\choose 9}}.$

Similarly, the expected value of the number of full teams in a random set of $8$ players is

$\frac{T{n-5\choose 3}}{{n\choose 8}}.$

The condition is thus equivalent to the existence of a positive integer $T$ such that

$\frac{T{n-5\choose 4}}{{n\choose 9}}\frac{T{n-5\choose 3}}{{n\choose 8}} = 1.$

$T^2\frac{(n-5)!(n-5)!8!9!(n-8)!(n-9)!}{n!n!(n-8)!(n-9)!3!4!} = 1$

$T^2 = \big((n)(n-1)(n-2)(n-3)(n-4)\big)^2 \frac{3!4!}{8!9!}$

$T^2 = \big((n)(n-1)(n-2)(n-3)(n-4)\big)^2 \frac{144}{7!7!8\cdot8\cdot9}$

$T^2 = \big((n)(n-1)(n-2)(n-3)(n-4)\big)^2 \frac{1}{4\cdot7!7!}$

$T = \frac{(n)(n-1)(n-2)(n-3)(n-4)}{2^5\cdot3^2\cdot5\cdot7}$

Note that this is always less than ${n\choose 5}$ , so as long as $T$ is integral, $n$ is a possibility. Thus, we have that this is equivalent to

$2^5\cdot3^2\cdot5\cdot7\big|(n)(n-1)(n-2)(n-3)(n-4).$

It is obvious that $5$ divides the RHS, and that $7$ does if $n\equiv 0,1,2,3,4\mod 7$ . Also, $3^2$ divides it iff $n\not\equiv 5,8\mod 9$ . One can also bash out that $2^5$ divides it in $16$ out of the $32$ possible residues $\mod 32$ .

Note that $2016 = 7*9*32$ so by using all numbers from $2$ to $2017$ , inclusive, it is clear that each possible residue $\mod 7,9,32$ is reached an equal number of times, so the total number of working $n$ in that range is $5\cdot 7\cdot 16 = 560$ . However, we must subtract the number of "working" $2\leq n\leq 8$ , which is $3$ . Thus, the answer is $\boxed{\textbf{(D) } 557}$ .

Alternatively, it is enough to approximate by finding the floor of $2017 \cdot \frac57 \cdot \frac79 \cdot \frac12 - 3$ to get $\boxed{\textbf{(D) } 557}$ .

Solution 1.5

Another way to find the average number of teams in a group of 9 or 8 people is as follows:

Let there be $T$ teams. There are $\binom{n}{5}$ total possible teams. So, the probability of any 5 people being a team is $\frac{T}{\binom{n}{5}}$ . There are $\binom{9}{5}$ possible teams in a group of 9. So, on average, there will be $\frac{T}{\binom{n}{5}} \binom{9}{5}$ teams in any group of 9 people. Similarly, on average, there will be $\frac{T}{\binom{n}{5}} \binom{8}{5}$ teams in any group of 8 people. So, \begin{align*} \frac{T}{\binom{n}{5}} \binom{9}{5} = \frac{1}{\frac{T}{\binom{n}{5}} \binom{8}{5}} \implies& \binom{9}{5}\binom{8}{5} T^2 = \binom{n}{5}^2 \\ \implies& 84T = \binom{n}{5} \end{align*} Therefore $84 \mid \binom{n}{5}$ and proceed as shown in solution 1.

FYI, to find n such that $32 \mid n(n-1)(n-2)(n-3)(n-4)$ without bashing everything: Clearly $n \equiv 0, 1, 2, 3, 4 \pmod{32}$ works. Then, do cases.

Case 1: $4 \mid n$ . This will always work, as $4 \mid n$ , $4 \mid n-4$ , and $2 \mid n-2$ , so $4 \cdot 4 \cdot 2 = 32 \mid n(n-1)(n-2)(n-3)(n-4)$ . So, $n \equiv 0, 4, 8, 12, 16, 20, 24, 28 \pmod{32}$ are also solutions.

Case 2: $n \equiv 2 \pmod{4}$ Then $n-4 \equiv 2 \pmod{4}$ as well. $n$ and $n-4$ can only contribute one 2 each, since $4 \nmid n, n-4$ . We need a factor of $2^3$ in $n-2$ then. So, $n \equiv 2 \pmod{8}$ . Then, we get $n \equiv 2, 10, 18, 26 \pmod{32}$ are solutions.

Case 3: $4 \nmid n$ We can only get twos from $n-1$ and $n-3$ . Note that one of them can only contribute a single factor of 2, because otherwise $4 \mid n-1$ and $4 \mid n-3$ , so $n$ has a remainder of both 1 and 3 mod 4, which is impossible. So, one must have a factor of 16. We get $n \equiv 1, 3 \pmod{16}$ so $n \equiv 1, 3, 17, 19 \pmod{32}$ .

In all, we get the solutions $n \equiv 0, 1, 2, 3, 4, 8, 10, 16, 17, 18, 19, 20, 24, 26, 28 \pmod{32}$

~CrazyVideoGamez

Video Solution by Dr. Nal

https://www.youtube.com/watch?v=2p2qYRWbvV4&feature=emb_logo

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2017 AMC 12B Problems/Problem 25

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Solution

Solution 1.5

Video Solution by Dr. Nal

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