# 2021 Fall AMC 12A Problems/Problem 20

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.

## 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)

~IceMatrix