For each positive integer n, let f_1(n) be twice the number of positive integer divisors of n, and for j ≥ 2, let f_j(n) = f_1(f_j-1(n)). For how many values of n ≤ 50 is f_50(n) = 12?

2021 AMC 12A Fall Problem 20 — Solution

2021 AMC 12A Fall Problem 20

Below is the professionally curated solution for Problem 20 of the 2021 AMC 12A Fall, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2021 AMC 12A Fall solutions, or check the answer key.

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Concepts:factor counting recursion

Difficulty rating: 2110

20.

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:

Both $8$ and $12$ are fixed: $f_1(8) = 2\cdot 4 = 8$ and $f_1(12) = 2\cdot 6 = 12.$ The small chain $2 \to 4 \to 6 \to 8$ funnels most numbers to $8;$ to reach $12$ the orbit must hit $12,$ $18$ (since $f_1(18) = 12$ ), or $20.$

Tracing each $n \le 50,$ the ones reaching $12$ are $12, 18, 20, 28, 32, 36, 44, 45, 48, 50$ — for instance $36 \to 18 \to 12$ and $48 \to 20 \to 12.$ That is $10$ values.

Thus, the correct answer is D.

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