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Difference between revisions of "2008 AMC 10A Problems/Problem 3"

(New page: ==Problem== For the positive integer <math>n</math>, let <math><n></math> denote the sum of all the positive divisors of <math>n</math> with the exception of <math>n</math> itself. For exa...)
 
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==Problem==
 
==Problem==
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For the positive integer <math>n</math>, let <math><n></math> denote the sum of all the positive divisors of <math>n</math> with the exception of <math>n</math> itself. For example, <math><4></math> and <math><12>=1+2+3+4+6=16</math>. What is <math><<<6>>></math>?
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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>
  
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==Solution==
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==Solution 1==
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<math><<<6>>>\ =\ <<6>>\ =\ <6>\ =\ 6\quad\Longrightarrow\quad\mathrm{(A)}</math>
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<math>\langle\langle\langle 6\rangle\rangle\rangle= \langle\langle 6\rangle\rangle = \langle 6\rangle=\ 6\quad\Longrightarrow\quad\mathrm{(A)}</math>
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==Solution 2==
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Since <math>6</math> is a perfect number, any such operation where <math>n=6</math> will yield <math>6</math> as the answer.
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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}}
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{{MAA Notice}}

Latest revision as of 14:12, 1 July 2021

Problem

For the positive integer $n$, let $\langle n\rangle$ denote the sum of all the positive divisors of $n$ with the exception of $n$ itself. For example, $\langle 4\rangle=1+2=3$ and $\langle 12 \rangle =1+2+3+4+6=16$. What is $\langle\langle\langle 6\rangle\rangle\rangle$?

$\mathrm{(A)}\ 6\qquad\mathrm{(B)}\ 12\qquad\mathrm{(C)}\ 24\qquad\mathrm{(D)}\ 32\qquad\mathrm{(E)}\ 36$

Solution 1

$\langle\langle\langle 6\rangle\rangle\rangle= \langle\langle 6\rangle\rangle = \langle 6\rangle=\ 6\quad\Longrightarrow\quad\mathrm{(A)}$

Solution 2

Since $6$ is a perfect number, any such operation where $n=6$ will yield $6$ 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)
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Problem 4
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All AMC 10 Problems and Solutions

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