Difference between revisions of "2008 AMC 10A Problems/Problem 3"
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==Solution== | ==Solution== | ||
| − | <math> | + | <math>\langle\langle\langle 6\rangle\rangle\rangle= \langle\langle 6\rangle\rangle = \langle 6\rangle=\ 6\quad\Longrightarrow\quad\mathrm{(A)}</math> |
==See also== | ==See also== | ||
{{AMC10 box|year=2008|ab=A|num-b=2|num-a=4}} | {{AMC10 box|year=2008|ab=A|num-b=2|num-a=4}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 09:30, 10 April 2014
Problem
For the positive integer 
, let 
denote the sum of all the positive divisors of with the exception of itself. For example, 
and 
. What is 
?
Solution
See also
| 2008 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 2 |
Followed by Problem 4 | |
| 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 AMC 10 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
