2008 AMC 10A Problem 3

Question

For the positive integer

n,

let

\langle n \rangle

denote the sum of all the positive divisors of

n

with the exception of

n

itself. For example,

\langle 4 \rangle = 1 + 2 = 3

and

\langle 12 \rangle = 1 + 2 + 3 + 4 + 6 = 16.

What is

\langle\langle\langle 6 \rangle\rangle\rangle?

6

12

24

32

36

\langle 6 \rangle

is the sum of the divisors of

6

other than

6

Compute

\langle 6 \rangle

first; the result feeds back into the same operation

The positive divisors of

6

other than

6

are

1, 2,

and

3,

so

\langle 6 \rangle = 1 + 2 + 3 = 6.

Since applying the operation to

6

again returns

6,

we get

\langle\langle\langle 6 \rangle\rangle\rangle = 6.

(A number equal to the sum of its proper divisors is called a perfect number, and

6

is the smallest.)

Thus, the correct answer is A.

Problem 3 in other years