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: $\textbf{Observation 1}$ $f_1 \left( 12 \right) = 12$
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 $\boxed{10}$

Answer: D 10