2008 AMC 10A Problems/Problem 3

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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 where $n=6$ will yield $6$ as the answer.

Note: A perfect number is defined as a number that equals the sum of its positive divisors excluding itself.

See also

2008 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 2
Followed by
Problem 4
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All AMC 10 Problems and Solutions

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