Difference between revisions of "2008 AMC 10A Problems/Problem 3"
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==Problem== | ==Problem== | ||
| − | For the positive integer <math>n</math>, let <math> | + | For the positive integer <math>n</math>, let <math>\langle n\rangle</math> denote the sum of all the positive divisors of <math>n</math> with the exception of <math>n</math> itself. For example, <math>\langle 4\rangle=1+2=3</math> and <math>\langle 12 \rangle =1+2+3+4+6=16</math>. What is <math>\langle\langle\langle 6\rangle\rangle\rangle</math>? |
<math>\mathrm{(A)}\ 6\qquad\mathrm{(B)}\ 12\qquad\mathrm{(C)}\ 24\qquad\mathrm{(D)}\ 32\qquad\mathrm{(E)}\ 36</math> | <math>\mathrm{(A)}\ 6\qquad\mathrm{(B)}\ 12\qquad\mathrm{(C)}\ 24\qquad\mathrm{(D)}\ 32\qquad\mathrm{(E)}\ 36</math> | ||
Revision as of 10:29, 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. 
