Question

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 $f_{50}(n) = 12?$

$7$

$8$

$9$

$10$

$11$

Solution:

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.

2021 AMC 10A Fall Problem 23

Solution:

Problem 23 in Other Years