Difference between revisions of "2008 AMC 10A Problems/Problem 3"
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<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> | ||
− | ==Solution== | + | ==Solution 1== |
<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== | ||
+ | Since <math>6</math> is a perfect number, any such operation where <math>n=6</math> will yield <math>6</math> as the answer. | ||
+ | |||
+ | Note: A perfect number is defined as 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}} |
Latest revision as of 14:12, 1 July 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 where will yield as the answer.
Note: A perfect number is defined as 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 |
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