2008 AMC 10A Problems/Problem 3

Problem

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

$\mathrm{(A)}\ 6\qquad\mathrm{(B)}\ 12\qquad\mathrm{(C)}\ 24\qquad\mathrm{(D)}\ 32\qquad\mathrm{(E)}\ 36$

Solution 1

$\langle\langle\langle 6\rangle\rangle\rangle= \langle\langle 6\rangle\rangle = \langle 6\rangle=\ 6\quad\Longrightarrow\quad\mathrm{(A)}$

Solution 2

Since $6$ is a perfect number, any such operation will yield $6$ as the answer. Note: A perfect number is a number that equals the sum of its positive divisors excluding itself.

2008 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 2	Followed by Problem 4
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2008 AMC 10A Problems/Problem 3

Contents

Problem

Solution 1

Solution 2

See also