Difference between revisions of "2006 Cyprus MO/Lyceum/Problem 24"

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==Problem==
 
==Problem==
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The number of divisors of the number <math>2006</math> is
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<math>\mathrm{(A)}\ 3\qquad\mathrm{(B)}\ 4\qquad\mathrm{(C)}\ 8\qquad\mathrm{(D)}\ 5\qquad\mathrm{(E)}\ 6</math>
  
 
==Solution==
 
==Solution==
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<math>2006=2\cdot17\cdot59</math>. A divisor of <math>2006</math> is therefore in the form <math>2^m\cdot 17^n\cdot 59^p</math>, where <math>m\leq 1</math>, <math>n\leq 1</math>, and <math>p\leq 1</math>.
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There are 2 choices for <math>m</math>, 2 choices for <math>n</math>, and 2 choices for <math>p</math>. Therefore, there are <math>2\cdot2\cdot2=\boxed{8}</math> divisors of <math>2006</math>.
  
 
==See also==
 
==See also==
 
{{CYMO box|year=2006|l=Lyceum|num-b=23|num-a=25}}
 
{{CYMO box|year=2006|l=Lyceum|num-b=23|num-a=25}}
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[[Category:Introductory Algebra Problems]]

Latest revision as of 13:15, 26 April 2008

Problem

The number of divisors of the number $2006$ is

$\mathrm{(A)}\ 3\qquad\mathrm{(B)}\ 4\qquad\mathrm{(C)}\ 8\qquad\mathrm{(D)}\ 5\qquad\mathrm{(E)}\ 6$

Solution

$2006=2\cdot17\cdot59$. A divisor of $2006$ is therefore in the form $2^m\cdot 17^n\cdot 59^p$, where $m\leq 1$, $n\leq 1$, and $p\leq 1$.

There are 2 choices for $m$, 2 choices for $n$, and 2 choices for $p$. Therefore, there are $2\cdot2\cdot2=\boxed{8}$ divisors of $2006$.

See also

2006 Cyprus MO, Lyceum (Problems)
Preceded by
Problem 23
Followed by
Problem 25
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