2021 AMC Fall 10A Problem 23

Write $d(n)$ for the number of divisors, so $f_1(n) = 2d(n)$.

Where do iterates go? For $n \le 50$, $d(n) \le 10$ (the maximum is $d(48) = 10$), so $f_1(n)$ is an even number at most $20$. Now track every even number up to $20$:
$$2 \to 4 \to 6 \to 8 \to 8, \qquad 10 \to 8, \qquad 14 \to 8, \qquad 16 \to 10 \to 8,$$
$$12 \to 12, \qquad 18 \to 12, \qquad 20 \to 12.$$
So $8$ and $12$ are the only fixed points, and after one step every $n \le 50$ is at some even number $\le 20$, which then falls into $8$ or $12$ within a few steps. Hence $f_{50}(n) = 12$ exactly when $f_1(n) \in \{12, 18, 20\}$, i.e. when $d(n) \in \{6, 9, 10\}$.

Count. Using $d(p^a q^b) = (a+1)(b+1)$:

• $d(n) = 6$: $n = p^5$ or $p^2 q$. Up to $50$: $32$; $12, 18, 20, 28, 44, 45, 50$. That is $8$ numbers.
• $d(n) = 9$: $n = p^8$ or $p^2 q^2$. Only $36$.
• $d(n) = 10$: $n = p^9$ or $p^4 q$. Only $48$.

Total $8 + 1 + 1 = 10$.

The answer is $\boxed{\textbf{(D)}\ 10}$.

Iterating a function on a bounded set always ends in a cycle, and here the range shrinks immediately: $f_1$ maps everything below $50$ into $\{2, 4, \dots, 20\}$, a set small enough to trace by hand. Once the two fixed points $8$ and $12$ are known, work backwards one step to see which first values feed $12$, and the problem reduces to the standard task of counting integers with a prescribed number of divisors.

• Counting only the n with d(n) = 6 (eight of them) and missing 36 and 48, whose f_1 values 18 and 20 also feed into the fixed point 12.
• Forgetting $32 = 2^5$ (six divisors) or including $52 = 4 \cdot 13$, which exceeds $50$.