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2007 AMC 8 Problems/Problem 10: Difference between revisions

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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>


<math>\boxed{D}</math>
<math>\boxed{\textbf{(D)}\ 28}</math>


==See Also==
==See Also==
{{AMC8 box|year=2007|num-b=9|num-a=11}}
{{AMC8 box|year=2007|num-b=9|num-a=11}}

Revision as of 12:18, 9 December 2012

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 $\boxed{12}$.

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

$\boxed{\textbf{(D)}\ 28}$

See Also

2007 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 9 		Followed by
Problem 11
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AJHSME/AMC 8 Problems and Solutions
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