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Difference between revisions of "2017 AMC 12B Problems/Problem 25"
m (→‎Solution 1.5)
Line 55: Line 55:
 
\implies& 84T = \binom{n}{5}
 
\implies& 84T = \binom{n}{5}
 
\end{align*}
 
\end{align*}
−
Therefore <math>84 \divides \binom{n}{5}</math> and proceed as shown in solution 1.
+
Therefore <math>84 \mid \binom{n}{5}</math> and proceed as shown in solution 1.
  
−
FYI, to find n such that <math>32 \divides n(n-1)(n-2)(n-3)(n-4)</math> without bashing everything:
+
FYI, to find n such that <math>32 \mid n(n-1)(n-2)(n-3)(n-4)</math> without bashing everything:
 
Clearly <math>n \equiv 0, 1, 2, 3, 4 \pmod{32}</math> works. Then, do cases.
 
Clearly <math>n \equiv 0, 1, 2, 3, 4 \pmod{32}</math> works. Then, do cases.
  
−
Case 1: <math>4 \divides n</math>.
+
Case 1: <math>4 \mid n</math>.
−
This will always work, as <math>4 \divides n</math>, <math>4 \divides n-4</math>, and <math>2 \divides n-2</math>, so <math>4 \cdot 4 \cdot 2 = 32 \divides n(n-1)(n-2)(n-3)(n-4)</math>. So, <math>n \conj 0, 4, 8, 12, 16, 20, 24, 28 \pmod{32}</math> are also solutions.
+
This will always work, as <math>4 \mid n</math>, <math>4 \mid n-4</math>, and <math>2 \mid n-2</math>, so <math>4 \cdot 4 \cdot 2 = 32 \mid n(n-1)(n-2)(n-3)(n-4)</math>. So, <math>n \equiv 0, 4, 8, 12, 16, 20, 24, 28 \pmod{32}</math> are also solutions.
  
−
Case 2: <math>n \conj 2 \pmod{4}</math>
+
Case 2: <math>n \equiv 2 \pmod{4}</math>
−
Then <math>n-4 \conj 2 \pmod{4}</math> as well. <math>n</math> and <math>n-4</math> can only contribute one 2 each, since <math>4 \not \divides n, n-4</math>. We need a factor of <math>2^3</math> in <math>n-2</math> then. So, <math>n \conj 2 \pmod{8}</math>. Then, we get <math>n \conj 2, 10, 18, 26 \pmod{32}</math> are solutions.
+
Then <math>n-4 \equiv 2 \pmod{4}</math> as well. <math>n</math> and <math>n-4</math> can only contribute one 2 each, since <math>4 \not \mid n, n-4</math>. We need a factor of <math>2^3</math> in <math>n-2</math> then. So, <math>n \equiv 2 \pmod{8}</math>. Then, we get <math>n \equiv 2, 10, 18, 26 \pmod{32}</math> are solutions.
  
−
Case 3: <math>4 \not \divides n</math>
+
Case 3: <math>4 \not \mid n</math>
−
We can only get twos from <math>n-1</math> and <math>n-3</math>. Note that one of them can only contribute a single factor of 2, because otherwise <math>4 \divides n-1</math> and <math>4 \divides n-3</math>, so <math>n</math> has a remainder of both 1 and 3 mod 4, which is impossible. So, one must have a factor of 16. We get <math>n \conj 1, 3 \pmod{16}</math> so <math>n \conj 1, 3, 17, 19 \pmod{32}</math>.
+
We can only get twos from <math>n-1</math> and <math>n-3</math>. Note that one of them can only contribute a single factor of 2, because otherwise <math>4 \mid n-1</math> and <math>4 \mid n-3</math>, so <math>n</math> has a remainder of both 1 and 3 mod 4, which is impossible. So, one must have a factor of 16. We get <math>n \conj 1, 3 \pmod{16}</math> so <math>n \conj 1, 3, 17, 19 \pmod{32}</math>.
  
 
In all, we get the solutions
 
In all, we get the solutions

Revision as of 17:21, 3 October 2024

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 $4 \not \mid n, n-4$. 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: $4 \not \mid n$ 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 $n \conj 1, 3 \pmod{16}$ (Error compiling LaTeX. Unknown error_msg) so $n \conj 1, 3, 17, 19 \pmod{32}$ (Error compiling LaTeX. Unknown error_msg).

In all, we get the solutions 

\[n \conj 0, 1, 2, 3, 4, 8, 10, 16, 17, 18, 19, 20, 24, 26, 28 \pmod{32}\] (Error compiling LaTeX. Unknown error_msg)

~CrazyVideoGamez

Video Solution by Dr. Nal

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

See Also

2017 AMC 12B (Problems • Answer Key • Resources)
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