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Difference between revisions of "2021 Fall AMC 12A Problems/Problem 20"

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==Solution 1==
 
==Solution 1==
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First, we can test values that would make <math>f(x)=12</math> true. For this to happen <math>x</math> must have <math>6</math> divisors, which means its prime factorization is in the form <math>pq^2</math> or <math>p^5</math>, where <math>p</math> and <math>q</math> are prime numbers. Listing out values less than <math>50</math> which have these prime factorizations, we find <math>12,20,28,44,18,45,50</math> for <math>pq^2</math>, and just <math>32</math> for <math>p^5</math>. Here <math>12</math> especially catches our eyes, as this means if one of <math>f_i(n)=12</math>, each of <math>f_{i+1}(n), f_{i+2}(n), ...</math> will all be <math>12</math>. This is because <math>f_{i+1}(n)=f(f_i(n))</math> (as given in the problem statement), so were <math>f_i(n)=12</math>, plugging this in we get <math>f_{i+1}(n)=f(12)=12</math>, and thus the pattern repeats. Hence, as long as for a <math>i</math>, such that <math>i\leq 50</math> and <math>f_{i}(n)=12</math>, <math>f_{50}(n)=12</math> must be true, which also immediately makes all our previously listed numbers, where <math>f(x)=12</math>, possible values of <math>n</math>.
+
First, we can test values that would make <math>f(x)=12</math> true. For this to happen <math>x</math> must have <math>6</math> divisors, which means its prime factorization is in the form <math>pq^2</math> or <math>p^5</math>, where <math>p</math> and <math>q</math> are prime numbers. Listing out values less than <math>50</math> which have these prime factorizations, we find <math>12,18,20,28,44,45,50</math> for <math>pq^2</math>, and just <math>32</math> for <math>p^5</math>. Here <math>12</math> especially catches our eyes, as this means if one of <math>f_i(n)=12</math>, each of <math>f_{i+1}(n), f_{i+2}(n), ...</math> will all be <math>12</math>. This is because <math>f_{i+1}(n)=f(f_i(n))</math> (as given in the problem statement), so were <math>f_i(n)=12</math>, plugging this in we get <math>f_{i+1}(n)=f(12)=12</math>, and thus the pattern repeats. Hence, as long as for a <math>i</math>, such that <math>i\leq 50</math> and <math>f_{i}(n)=12</math>, <math>f_{50}(n)=12</math> must be true, which also immediately makes all our previously listed numbers, where <math>f(x)=12</math>, possible values of <math>n</math>.
  
−
We also know that if <math>f(x)</math> were to be any of these numbers, <math>x</math> would satisfy <math>f_{50}(n)</math> as well. Looking through each of the possibilities aside from <math>12</math>, we see that <math>f(x)</math> could only possibly be equal to <math>20</math> and <math>18</math>, and still have <math>x</math> less than or equal to <math>50</math>. This would mean <math>x</math> must have <math>10</math>, or <math>9</math> divisors, and testing out, we see that <math>x</math> will then be of the form <math>p^4q</math>, or <math>p^2q^2</math>. The only two values less than or equal to <math>50</math> would be <math>48</math> and <math>36</math> respectively. From here there are no more possible values, so tallying our possibilities we count <math>\boxed{\textbf{(D) }10}</math> values (Namely <math>12,20,28,44,18,45,50,32,36,48</math>).
+
We also know that if <math>f(x)</math> were to be any of these numbers, <math>x</math> would satisfy <math>f_{50}(n)</math> as well. Looking through each of the possibilities aside from <math>12</math>, we see that <math>f(x)</math> could only possibly be equal to <math>20</math> and <math>18</math>, and still have <math>x</math> less than or equal to <math>50</math>. This would mean <math>x</math> must have <math>10</math>, or <math>9</math> divisors, and testing out, we see that <math>x</math> will then be of the form <math>p^4q</math>, or <math>p^2q^2</math>. The only two values less than or equal to <math>50</math> would be <math>48</math> and <math>36</math> respectively. From here there are no more possible values, so tallying our possibilities we count <math>\boxed{\textbf{(D) }10}</math> values (Namely <math>12,18,20,28,32,36,44,45,48,50</math>).
  
 
~Ericsz
 
~Ericsz
  
−
== Solution 2 ==
+
== Solution 2 (''Rigorous'' reasoning on why there cannot be any other solutions) ==
 +
First, take note that the maximum possible value of <math>f_1(n)</math> for <math>1 \le n \le k</math> increases as <math>k</math> increases (it is a step function), i.e. it is increasing. Likewise, as <math>k</math> decreases, the maximum possible value of <math>f_1(n)</math> decreases as well. Also, let <math>f_1(n) = 2d(n)</math> where <math>d(n)</math> is the number of divisors of n.
 +
 
 +
Since <math>n \le 50</math>, <math>f_1(n) <= 20</math>. This maximum occurs when <math>d(n) = 10 \implies n = 2^4 \cdot 3 = 48</math>. Next, since <math>f_1(n) <=20</math>, <math>f_1(f_1(n)) \le 12 \implies f_2(n) \le 12</math>. This maximum occurs when <math>d(f_1(n)) = 6 \implies n = 2 \cdot 3^2 = 18, n = 2^2 \cdot 3 = 12</math>. Since <math>f_2(n) \le 12</math>, <math>f_1(f_2(n)) \le 12 \implies f_3(n) \le 12</math>, once again. This maximum again occurs when <math>d(f_2(n)) = 6 \implies f_2(n) = 2^2 \cdot 3 = 12</math>. Now, suppose for the sake of contradiction that <math>f_2(n) < 12</math>. Then, <math>f_3(n) < 12</math> (since <math>f_2(n) = 12</math> was the only number that would maximize <math>f_3(n))</math> for <math>f_2(n) \le 12</math>). As a result, since <math>f_1(n)</math> is increasing, and because <math>12</math> is where <math>f_1</math> steps down from a maximum of <math>6 \cdot 2 = 12</math>, we must have that <math>f_1(f_3(n)) < f_1(12) = 12 \implies f_4(n) < 12</math>. We continue applying <math>f_1</math> on both sides (which is possible since <math>f_1</math> is increasing) until we reach <math>f_{50}</math>, giving us that <math>f_{50}(n) < 12</math>. However, <math>f_{50}(n) = 12</math>, which is a contradiction. Thus, <math>f_2(n) = 12</math>.
 +
 
 +
Now, let us finally solve for the solutions. <math>f_2(n) = 12 \implies f_1(f_1(n)) = 12 \implies d(f_1(n)) = 6</math>. <math>d(f_1(n)) = 6</math>. This gives us two cases. First, we have the case where <math>f_1(n) = p^2 \cdot q</math> where <math>p</math> and <math>q</math> are primes. Second, we have the case where <math>f_1(n)=p^5</math> where <math>p</math> is a prime. For both cases, since <math>f_1(n) \le 20</math>, <math>f_1(n)</math> can only be <math>12</math>, <math>18</math>, or <math>20</math>. If <math>f_1(n) = 12</math>, then <math>d(n) = 6 \implies n = p^5, p^2 \cdot q \implies n \in \{ 12, 18, 20, 28, 32, 44, 45, 50 \}</math>, resulting in 8 solutions. If <math>f_1(n) = 18</math>, then <math>d(n) = 9 \implies n = p^8, p^2 \cdot q^2 \implies n = 36</math>, giving us one more solution. Finally, <math>f_1(n) = 20 \implies d(n) = 10 \implies n = p^9, p^4 \cdot q \implies n = 48</math>. Thus, in total, we have <math>\boxed{\textbf{(D)} 10}</math> solutions.
 +
 
 +
~[https://artofproblemsolving.com/wiki/index.php/User:CrazyVideoGamez CrazyVideoGamez]
 +
 
 +
== Solution 3 ==  
 
<math>\textbf{Observation 1}</math>: <math>f_1 \left( 12 \right) = 12</math>.
 
<math>\textbf{Observation 1}</math>: <math>f_1 \left( 12 \right) = 12</math>.
  
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We have <math>f_1 \left( n \right) = 4</math>, <math>f_2 \left( n \right) = f_1 \left( 4 \right) = 6</math>, <math>f_3 \left( n \right) = f_1 \left( 6 \right) = 8</math>. Hence, Observation 2 implies <math>f_{50} \left( n \right) = 8</math>.
 
We have <math>f_1 \left( n \right) = 4</math>, <math>f_2 \left( n \right) = f_1 \left( 4 \right) = 6</math>, <math>f_3 \left( n \right) = f_1 \left( 6 \right) = 8</math>. Hence, Observation 2 implies <math>f_{50} \left( n \right) = 8</math>.
  
−
<math>\textbf{Case 3}</math>: the prime factorization of <math>n</math> takes the form <math>p_1^2</math>.
+
<math>\textbf{Case 3}</math>: The prime factorization of <math>n</math> takes the form <math>p_1^2</math>.
  
 
We have <math>f_1 \left( n \right) = 6</math>, <math>f_2 \left( n \right) = f_1 \left( 6 \right) = 8</math>. Hence, Observation 2 implies <math>f_{50} \left( n \right) = 8</math>.
 
We have <math>f_1 \left( n \right) = 6</math>, <math>f_2 \left( n \right) = f_1 \left( 6 \right) = 8</math>. Hence, Observation 2 implies <math>f_{50} \left( n \right) = 8</math>.
  
−
<math>\textbf{Case 4}</math>: the prime factorization of <math>n</math> takes the form <math>p_1^3</math>.
+
<math>\textbf{Case 4}</math>: The prime factorization of <math>n</math> takes the form <math>p_1^3</math>.
  
 
We have <math>f_1 \left( n \right) = 8</math>. Hence, Observation 2 implies <math>f_{50} \left( n \right) = 8</math>.
 
We have <math>f_1 \left( n \right) = 8</math>. Hence, Observation 2 implies <math>f_{50} \left( n \right) = 8</math>.
  
−
<math>\textbf{Case 5}</math>: the prime factorization of <math>n</math> takes the form <math>p_1^4</math>.
+
<math>\textbf{Case 5}</math>: The prime factorization of <math>n</math> takes the form <math>p_1^4</math>.
  
 
We have <math>f_1 \left( n \right) = 10</math>, <math>f_2 \left( n \right) = f_1 \left( 10 \right) = 8</math>. Hence, Observation 2 implies <math>f_{50} \left( n \right) = 8</math>.
 
We have <math>f_1 \left( n \right) = 10</math>, <math>f_2 \left( n \right) = f_1 \left( 10 \right) = 8</math>. Hence, Observation 2 implies <math>f_{50} \left( n \right) = 8</math>.
  
−
<math>\textbf{Case 6}</math>: the prime factorization of <math>n</math> takes the form <math>p_1^5</math>.
+
<math>\textbf{Case 6}</math>: The prime factorization of <math>n</math> takes the form <math>p_1^5</math>.
  
 
We have <math>f_1 \left( n \right) = 12</math>. Hence, Observation 1 implies <math>f_{50} \left( n \right) = 12</math>.
 
We have <math>f_1 \left( n \right) = 12</math>. Hence, Observation 1 implies <math>f_{50} \left( n \right) = 12</math>.
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In this case the only <math>n</math> is <math>2^5 = 32</math>.
 
In this case the only <math>n</math> is <math>2^5 = 32</math>.
  
−
<math>\textbf{Case 7}</math>: the prime factorization of <math>n</math> takes the form <math>p_1 p_2</math>.
+
<math>\textbf{Case 7}</math>: The prime factorization of <math>n</math> takes the form <math>p_1 p_2</math>.
  
 
We have <math>f_1 \left( n \right) = 8</math>. Hence, Observation 2 implies <math>f_{50} \left( n \right) = 8</math>.
 
We have <math>f_1 \left( n \right) = 8</math>. Hence, Observation 2 implies <math>f_{50} \left( n \right) = 8</math>.
  
−
<math>\textbf{Case 8}</math>: the prime factorization of <math>n</math> takes the form <math>p_1 p_2^2</math>.
+
<math>\textbf{Case 8}</math>: The prime factorization of <math>n</math> takes the form <math>p_1 p_2^2</math>.
  
 
We have <math>f_1 \left( n \right) = 12</math>. Hence, Observation 1 implies <math>f_{50} \left( n \right) = 12</math>.
 
We have <math>f_1 \left( n \right) = 12</math>. Hence, Observation 1 implies <math>f_{50} \left( n \right) = 12</math>.
  
−
In this case, all <math>n</math> are <math>18, 50, 12, 20, 45, 28, 44</math>.
+
In this case, all <math>n</math> are <math>12, 18, 20, 28, 44, 45,</math> and <math>50</math>.
  
−
<math>\textbf{Case 9}</math>: the prime factorization of <math>n</math> takes the form <math>p_1 p_2^3</math>.
+
<math>\textbf{Case 9}</math>: The prime factorization of <math>n</math> takes the form <math>p_1 p_2^3</math>.
  
 
We have <math>f_1 \left( n \right) = 16</math>, <math>f_2 \left( n \right) = f_1 \left( 16 \right) = 10</math>, <math>f_3 \left( n \right) = f_1 \left( 10 \right) = 8</math>. Hence, Observation 2 implies <math>f_{50} \left( n \right) = 8</math>.
 
We have <math>f_1 \left( n \right) = 16</math>, <math>f_2 \left( n \right) = f_1 \left( 16 \right) = 10</math>, <math>f_3 \left( n \right) = f_1 \left( 10 \right) = 8</math>. Hence, Observation 2 implies <math>f_{50} \left( n \right) = 8</math>.
  
−
<math>\textbf{Case 10}</math>: the prime factorization of <math>n</math> takes the form <math>p_1 p_2^4</math>.
+
<math>\textbf{Case 10}</math>: The prime factorization of <math>n</math> takes the form <math>p_1 p_2^4</math>.
  
 
We have <math>f_1 \left( n \right) = 20</math>, <math>f_2 \left( n \right) = f_1 \left( 20 \right) = 12</math>. Hence, Observation 1 implies <math>f_{50} \left( n \right) = 12</math>.
 
We have <math>f_1 \left( n \right) = 20</math>, <math>f_2 \left( n \right) = f_1 \left( 20 \right) = 12</math>. Hence, Observation 1 implies <math>f_{50} \left( n \right) = 12</math>.
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In this case, the only <math>n</math> is <math>48</math>.
 
In this case, the only <math>n</math> is <math>48</math>.
  
−
<math>\textbf{Case 11}</math>: the prime factorization of <math>n</math> takes the form <math>p_1^2 p_2^2</math>.
+
<math>\textbf{Case 11}</math>: The prime factorization of <math>n</math> takes the form <math>p_1^2 p_2^2</math>.
  
 
We have <math>f_1 \left( n \right) = 18</math>, <math>f_2 \left( n \right) = f_1 \left( 18 \right) = 12</math>. Hence, Observation 1 implies <math>f_{50} \left( n \right) = 12</math>.
 
We have <math>f_1 \left( n \right) = 18</math>, <math>f_2 \left( n \right) = f_1 \left( 18 \right) = 12</math>. Hence, Observation 1 implies <math>f_{50} \left( n \right) = 12</math>.
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In this case, the only <math>n</math> is <math>36</math>.
 
In this case, the only <math>n</math> is <math>36</math>.
  
−
<math>\textbf{Case 12}</math>: the prime factorization of <math>n</math> takes the form <math>p_1 p_2 p_3</math>.
+
<math>\textbf{Case 12}</math>: The prime factorization of <math>n</math> takes the form <math>p_1 p_2 p_3</math>.
  
 
We have <math>f_1 \left( n \right) = 16</math>, <math>f_2 \left( n \right) = f_1 \left( 16 \right) = 10</math>, <math>f_3 \left( n \right) = f_2 \left( 10 \right) = 8</math>. Hence, Observation 2 implies <math>f_{50} \left( n \right) = 8</math>.
 
We have <math>f_1 \left( n \right) = 16</math>, <math>f_2 \left( n \right) = f_1 \left( 16 \right) = 10</math>, <math>f_3 \left( n \right) = f_2 \left( 10 \right) = 8</math>. Hence, Observation 2 implies <math>f_{50} \left( n \right) = 8</math>.
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==Video Solution by Mathematical Dexterity==
 
==Video Solution by Mathematical Dexterity==
 
https://www.youtube.com/watch?v=WQQVjCdoqWI
 
https://www.youtube.com/watch?v=WQQVjCdoqWI
 +
 +
==Video Solution by TheBeautyofMath==
 +
https://youtu.be/o2MAmtgBbKc
 +
 +
~IceMatrix
  
 
==See Also==
 
==See Also==

Latest revision as of 15:56, 5 June 2024

The following problem is from both the 2021 Fall AMC 10A #23 and 2021 Fall AMC 12A #20, so both problems redirect to this page.

Problem

For each positive integer $n$, let $f_1(n)$ be twice the number of positive integer divisors of $n$, and for $j \ge 2$, let $f_j(n) = f_1(f_{j-1}(n))$. For how many values of $n \le 50$ is $f_{50}(n) = 12?$

$\textbf{(A) }7\qquad\textbf{(B) }8\qquad\textbf{(C) }9\qquad\textbf{(D) }10\qquad\textbf{(E) }11$

Solution 1

First, we can test values that would make $f(x)=12$ true. For this to happen $x$ must have $6$ divisors, which means its prime factorization is in the form $pq^2$ or $p^5$, where $p$ and $q$ are prime numbers. Listing out values less than $50$ which have these prime factorizations, we find $12,18,20,28,44,45,50$ for $pq^2$, and just $32$ for $p^5$. Here $12$ especially catches our eyes, as this means if one of $f_i(n)=12$, each of $f_{i+1}(n), f_{i+2}(n), ...$ will all be $12$. This is because $f_{i+1}(n)=f(f_i(n))$ (as given in the problem statement), so were $f_i(n)=12$, plugging this in we get $f_{i+1}(n)=f(12)=12$, and thus the pattern repeats. Hence, as long as for a $i$, such that $i\leq 50$ and $f_{i}(n)=12$, $f_{50}(n)=12$ must be true, which also immediately makes all our previously listed numbers, where $f(x)=12$, possible values of $n$.

We also know that if $f(x)$ were to be any of these numbers, $x$ would satisfy $f_{50}(n)$ as well. Looking through each of the possibilities aside from $12$, we see that $f(x)$ could only possibly be equal to $20$ and $18$, and still have $x$ less than or equal to $50$. This would mean $x$ must have $10$, or $9$ divisors, and testing out, we see that $x$ will then be of the form $p^4q$, or $p^2q^2$. The only two values less than or equal to $50$ would be $48$ and $36$ respectively. From here there are no more possible values, so tallying our possibilities we count $\boxed{\textbf{(D) }10}$ values (Namely $12,18,20,28,32,36,44,45,48,50$).

~Ericsz

Solution 2 (Rigorous reasoning on why there cannot be any other solutions)

First, take note that the maximum possible value of $f_1(n)$ for $1 \le n \le k$ increases as $k$ increases (it is a step function), i.e. it is increasing. Likewise, as $k$ decreases, the maximum possible value of $f_1(n)$ decreases as well. Also, let $f_1(n) = 2d(n)$ where $d(n)$ is the number of divisors of n.

Since $n \le 50$, $f_1(n) <= 20$. This maximum occurs when $d(n) = 10 \implies n = 2^4 \cdot 3 = 48$. Next, since $f_1(n) <=20$, $f_1(f_1(n)) \le 12 \implies f_2(n) \le 12$. This maximum occurs when $d(f_1(n)) = 6 \implies n = 2 \cdot 3^2 = 18, n = 2^2 \cdot 3 = 12$. Since $f_2(n) \le 12$, $f_1(f_2(n)) \le 12 \implies f_3(n) \le 12$, once again. This maximum again occurs when $d(f_2(n)) = 6 \implies f_2(n) = 2^2 \cdot 3 = 12$. Now, suppose for the sake of contradiction that $f_2(n) < 12$. Then, $f_3(n) < 12$ (since $f_2(n) = 12$ was the only number that would maximize $f_3(n))$ for $f_2(n) \le 12$). As a result, since $f_1(n)$ is increasing, and because $12$ is where $f_1$ steps down from a maximum of $6 \cdot 2 = 12$, we must have that $f_1(f_3(n)) < f_1(12) = 12 \implies f_4(n) < 12$. We continue applying $f_1$ on both sides (which is possible since $f_1$ is increasing) until we reach $f_{50}$, giving us that $f_{50}(n) < 12$. However, $f_{50}(n) = 12$, which is a contradiction. Thus, $f_2(n) = 12$.

Now, let us finally solve for the solutions. $f_2(n) = 12 \implies f_1(f_1(n)) = 12 \implies d(f_1(n)) = 6$. $d(f_1(n)) = 6$. This gives us two cases. First, we have the case where $f_1(n) = p^2 \cdot q$ where $p$ and $q$ are primes. Second, we have the case where $f_1(n)=p^5$ where $p$ is a prime. For both cases, since $f_1(n) \le 20$, $f_1(n)$ can only be $12$, $18$, or $20$. If $f_1(n) = 12$, then $d(n) = 6 \implies n = p^5, p^2 \cdot q \implies n \in \{ 12, 18, 20, 28, 32, 44, 45, 50 \}$, resulting in 8 solutions. If $f_1(n) = 18$, then $d(n) = 9 \implies n = p^8, p^2 \cdot q^2 \implies n = 36$, giving us one more solution. Finally, $f_1(n) = 20 \implies d(n) = 10 \implies n = p^9, p^4 \cdot q \implies n = 48$. Thus, in total, we have $\boxed{\textbf{(D)} 10}$ solutions.

~CrazyVideoGamez

Solution 3

$\textbf{Observation 1}$: $f_1 \left( 12 \right) = 12$.

Hence, if $n$ has the property that $f_j \left( n \right) = 12$ for some $j$, then $f_k \left( n \right) = 12$ for all $k > j$.

$\textbf{Observation 2}$: $f_1 \left( 8 \right) = 8$.

Hence, if $n$ has the property that $f_j \left( n \right) = 8$ for some $j$, then $f_k \left( n \right) = 8$ for all $k > j$.

$\textbf{Case 1}$: $n = 1$.

We have $f_1 \left( n \right) = 2$, $f_2 \left( n \right) = f_1 \left( 2 \right) = 4$, $f_3 \left( n \right) = f_1 \left( 4 \right) = 6$, $f_4 \left( n \right) = f_1 \left( 6 \right) = 8$. Hence, Observation 2 implies $f_{50} \left( n \right) = 8$.

$\textbf{Case 2}$: $n$ is prime.

We have $f_1 \left( n \right) = 4$, $f_2 \left( n \right) = f_1 \left( 4 \right) = 6$, $f_3 \left( n \right) = f_1 \left( 6 \right) = 8$. Hence, Observation 2 implies $f_{50} \left( n \right) = 8$.

$\textbf{Case 3}$: The prime factorization of $n$ takes the form $p_1^2$.

We have $f_1 \left( n \right) = 6$, $f_2 \left( n \right) = f_1 \left( 6 \right) = 8$. Hence, Observation 2 implies $f_{50} \left( n \right) = 8$.

$\textbf{Case 4}$: The prime factorization of $n$ takes the form $p_1^3$.

We have $f_1 \left( n \right) = 8$. Hence, Observation 2 implies $f_{50} \left( n \right) = 8$.

$\textbf{Case 5}$: The prime factorization of $n$ takes the form $p_1^4$.

We have $f_1 \left( n \right) = 10$, $f_2 \left( n \right) = f_1 \left( 10 \right) = 8$. Hence, Observation 2 implies $f_{50} \left( n \right) = 8$.

$\textbf{Case 6}$: The prime factorization of $n$ takes the form $p_1^5$.

We have $f_1 \left( n \right) = 12$. Hence, Observation 1 implies $f_{50} \left( n \right) = 12$.

In this case the only $n$ is $2^5 = 32$.

$\textbf{Case 7}$: The prime factorization of $n$ takes the form $p_1 p_2$.

We have $f_1 \left( n \right) = 8$. Hence, Observation 2 implies $f_{50} \left( n \right) = 8$.

$\textbf{Case 8}$: The prime factorization of $n$ takes the form $p_1 p_2^2$.

We have $f_1 \left( n \right) = 12$. Hence, Observation 1 implies $f_{50} \left( n \right) = 12$.

In this case, all $n$ are $12, 18, 20, 28, 44, 45,$ and $50$.

$\textbf{Case 9}$: The prime factorization of $n$ takes the form $p_1 p_2^3$.

We have $f_1 \left( n \right) = 16$, $f_2 \left( n \right) = f_1 \left( 16 \right) = 10$, $f_3 \left( n \right) = f_1 \left( 10 \right) = 8$. Hence, Observation 2 implies $f_{50} \left( n \right) = 8$.

$\textbf{Case 10}$: The prime factorization of $n$ takes the form $p_1 p_2^4$.

We have $f_1 \left( n \right) = 20$, $f_2 \left( n \right) = f_1 \left( 20 \right) = 12$. Hence, Observation 1 implies $f_{50} \left( n \right) = 12$.

In this case, the only $n$ is $48$.

$\textbf{Case 11}$: The prime factorization of $n$ takes the form $p_1^2 p_2^2$.

We have $f_1 \left( n \right) = 18$, $f_2 \left( n \right) = f_1 \left( 18 \right) = 12$. Hence, Observation 1 implies $f_{50} \left( n \right) = 12$.

In this case, the only $n$ is $36$.

$\textbf{Case 12}$: The prime factorization of $n$ takes the form $p_1 p_2 p_3$.

We have $f_1 \left( n \right) = 16$, $f_2 \left( n \right) = f_1 \left( 16 \right) = 10$, $f_3 \left( n \right) = f_2 \left( 10 \right) = 8$. Hence, Observation 2 implies $f_{50} \left( n \right) = 8$.

Putting all cases together, the number of feasible $n \leq 50$ is $\boxed{\textbf{(D) }10}$.

~Steven Chen (www.professorchenedu.com)

Video Solution by Mathematical Dexterity

https://www.youtube.com/watch?v=WQQVjCdoqWI

Video Solution by TheBeautyofMath

https://youtu.be/o2MAmtgBbKc

~IceMatrix

See Also

2021 Fall AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 19 		Followed by
Problem 21
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
2021 Fall AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 22 		Followed by
Problem 24
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 10 Problems and Solutions

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