2021 Fall AMC 10A Problem 23

The value $12$ is fixed by the function, since $12$ has $6$ positive divisors and therefore $f_1(12)=12.$ First find all $n\le50$ with $f_1(n)=12,$ meaning $n$ has $6$ divisors. These are $$12,18,20,28,32,44,45,50.$$ Now check whether $f_1(n)$ can be one of these values before reaching $12.$ Since $f_1(n)$ is twice a divisor count, the only useful possibilities in that list are $18$ and $20,$ meaning $n$ has $9$ or $10$ divisors. For $n\le50,$ the additional possibilities are $36,$ which has $9$ divisors, and $48,$ which has $10$ divisors. Therefore there are $8+2=10$ values of $n.$ Thus, D is the correct answer.