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 $f_{50}(n) = 12?$
对每个正整数 $n$，令 $f_1(n)$ 为 $n$ 的正整数因数个数的两倍；并且对 $j \ge 2$，令 $f_j(n)=f_1(f_{j-1}(n))$。问：满足 $n \le 50$ 且 $f_{50}(n)=12$ 的 $n$ 有多少个？
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 $f_1(x)=12$, possible values of $n$.
We also know that if $f_1(x)$ 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 $f_1(x)$ could only possibly be equal to $20$ and $18$, and still have $x$ less than or equal to $50$. This is because if $f_1(x)=20$ or $18$, $f_2 (x) = 12$. 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 $12,18,20,28,32,36,44,45,48,50$).
首先，我们可以测试哪些取值会使得 $f(x)=12$ 成立。要发生这种情况，$x$ 必须有 $6$ 个因数，这意味着它的质因数分解形如 $pq^2$ 或 $p^5$，其中 $p$ 和 $q$ 是质数。列出小于 $50$ 且具有这些质因数分解形式的数，我们得到 $pq^2$ 的情况为 $12,18,20,28,44,45,50$，而 $p^5$ 的情况只有 $32$。其中 $12$ 尤其引人注意，因为这意味着如果某个 $f_i(n)=12$，那么 $f_{i+1}(n), f_{i+2}(n), ...$ 都将是 $12$。这是因为 $f_{i+1}(n)=f(f_i(n))$（如题目所给），所以若 $f_i(n)=12$，代入得到 $f_{i+1}(n)=f(12)=12$，从而该模式重复。因此，只要存在某个 $i$ 满足 $i\leq 50$ 且 $f_{i}(n)=12$，就必有 $f_{50}(n)=12$，这也立刻使得我们先前列出的所有满足 $f_1(x)=12$ 的数都成为可能的 $n$ 值。
我们还知道，如果 $f_1(x)$ 等于这些数中的任意一个，那么 $x$ 也将满足 $f_{50}(n)$。逐一查看除 $12$ 之外的每种可能，我们发现当 $x\le 50$ 时，$f_1(x)$ 只有可能等于 $20$ 或 $18$。这是因为若 $f_1(x)=20$ 或 $18$，则 $f_2(x)=12$。这意味着 $x$ 必须分别有 $10$ 个或 $9$ 个因数；检验可知此时 $x$ 分别形如 $p^4q$ 或 $p^2q^2$。小于等于 $50$ 的仅有两个值，分别是 $48$ 和 $36$。从这里开始不再有更多可能值，因此统计所有可能，我们得到 $\boxed{\textbf{(D) }10}$ 个值（即 $12,18,20,28,32,36,44,45,48,50$）。