2021 AMC 12A Fall Problem 20

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

f_1(12) = 2\cdot 6 = 12,

so

12

is a fixed point;

f_1(8) = 2\cdot 4 = 8

is another

After enough steps every

n

lands on

8

or

12;

count those whose orbit reaches

12

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.

Problem 20 in other years