2021 Fall AMC 12A Problem 20

Problem 20 of 25HarderAlgebra Number Theory

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

f_{50}(n) = 12?

Show solution

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,

20.

Thus we need

d(n)=6,9,

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.