2021 AMC 10A Fall Problem 23

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 work backwards

Difficulty rating: 2130

Small hint:

f_1(n)=12

means

n

has

6

divisors

Big hint:

Also check

f_1(n)=18

or

20,

since those map to

12

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

Answer: D

Difficulty rating: 2130

Small hint:

Big hint:

Solution:

Problem 23 in other years