Difference between revisions of "2007 AMC 8 Problems/Problem 10"

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<math>\boxed{\boxed{11}} = \boxed{12} = 1 + 2 + 3 + 4 + 6 + 12 = 28</math>
 
<math>\boxed{\boxed{11}} = \boxed{12} = 1 + 2 + 3 + 4 + 6 + 12 = 28</math>
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<math>\boxed{D}</math>

Revision as of 16:20, 15 February 2010

Problem

For any positive integer $n$, $\boxed{n}$ to be the sum of the positive factors of $n$. For example, $\boxed{6} = 1 + 2 + 3 + 6 = 12$. Find $\boxed{\boxed{11}}$ .

$\mathrm{(A)}\ 13 \qquad \mathrm{(B)}\ 20 \qquad \mathrm{(C)}\ 24 \qquad \mathrm{(D)}\ 28 \qquad \mathrm{(E)}\ 30$

Solution

First we find $\boxed{11}$.

$\boxed{11} = 1 + 11 = 12$

Then we find $\boxed{12}$.

$\boxed{\boxed{11}} = \boxed{12} = 1 + 2 + 3 + 4 + 6 + 12 = 28$

$\boxed{D}$