Difference between revisions of "2008 AMC 10A Problems/Problem 3"
Math-passion (talk | contribs) |
(→Solution 2) |
||
| Line 7: | Line 7: | ||
<math>\langle\langle\langle 6\rangle\rangle\rangle= \langle\langle 6\rangle\rangle = \langle 6\rangle=\ 6\quad\Longrightarrow\quad\mathrm{(A)}</math> | <math>\langle\langle\langle 6\rangle\rangle\rangle= \langle\langle 6\rangle\rangle = \langle 6\rangle=\ 6\quad\Longrightarrow\quad\mathrm{(A)}</math> | ||
==Solution 2== | ==Solution 2== | ||
| − | Since <math>6</math> is a perfect number, any such operation will yield <math>6</math> as the answer. | + | Since <math>6</math> is a perfect number, any such operation will yield <math>6</math> as the answer. |
| + | Note: A perfect number is a number that equals the sum of its positive divisors excluding itself. | ||
| + | |||
==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 16:21, 4 June 2021
Contents
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 1
Solution 2
Since 
is a perfect number, any such operation will yield as the answer.
Note: A perfect number is a number that equals the sum of its positive divisors excluding itself.
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. 
