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 = \boxed{\textbf{(D)}\ 28}</math> | <math>\boxed{\boxed{11}} = \boxed{12} = 1 + 2 + 3 + 4 + 6 + 12 = \boxed{\textbf{(D)}\ 28}</math> | ||
| + | |||
| + | ==Video Solution by WhyMath== | ||
| + | https://youtu.be/Ih8lEBwPqEY | ||
| + | |||
| + | ~savannahsolver | ||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2007|num-b=9|num-a=11}} | {{AMC8 box|year=2007|num-b=9|num-a=11}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 17:08, 26 February 2021
Problem
For any positive integer 
, 
to be the sum of the positive factors of .
For example, 
. Find 
.
Solution
First we find 
.
Then we find 
.
Video Solution by WhyMath
~savannahsolver
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
