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

$7$

$8$

$9$

$10$

$11$

Answer: D

Concepts:factor counting recursion

Difficulty rating: 2110

Solution:

Both $8$ and $12$ are fixed. For $n\le50,$ the first value $f_1(n)=2d(n)$ is at most $20.$ Checking the even values through $20$ shows that an orbit reaches $12$ exactly when its first value is $12,18,$ or $20.$ Thus we need $d(n)=6,9,$ or $10.$

The numbers at most $50$ with $6$ divisors are $12,18,20,28,32,44,45,50;$ the only one with $9$ divisors is $36,$ and the only one with $10$ divisors is $48.$ These $10$ values all reach the fixed point $12.$

Thus, the correct answer is D.

2021 AMC 12A Fall Problem 20

Answer: D

Difficulty rating: 2110

Solution:

Problem 20 in Other Years