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2008 AMC 10A Problem 3

2008 AMC 10A Problem 3AMC 10 Step by Step

Name: Date: September 13, 2026

Divisibility & Factors · difficulty 1 of 5 · about 1 min · position P1-10

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

A)
$6$
B)
$12$
C)
$24$
D)
$32$
E)
$36$

Answer

Key insight

The proper divisors of 6 are 1, 2, 3, which add to 6, so applying the operation any number of times returns 6.

Solution

Work from the inside out. The divisors of $6$ other than $6$ itself are $1, 2, 3$ , so
$\langle 6\rangle = 1 + 2 + 3 = 6.$

The operation sends $6$ back to $6$ , so the next two applications change nothing:
$\langle\langle\langle 6\rangle\rangle\rangle = \langle\langle 6\rangle\rangle = \langle 6\rangle = 6.$

The answer is $\boxed{\textbf{(A)}\ 6}$ .

Why this works

A nested operation is evaluated one layer at a time from the innermost. Here $6$ is a perfect number (its proper divisors sum to itself), so it is a fixed point of $\langle\cdot\rangle$ , and repeated application never leaves it. Always compute the first step before assuming the numbers will grow.

The trap

Including 6 itself in the divisor sum, getting 12, and then continuing with the divisors of 12.

Common mistakes

Including 6 itself in the divisor sum, getting 12, and then continuing with the divisors of 12.
Miscopying the definition from the examples and summing all divisors of the later values, drifting toward $24$ or $36$ .

Techniques

Set up the equation/formula and compute; no special trick needed