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Difference between revisions of "2010 AIME I Problems/Problem 1"

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== Problem ==
 
== Problem ==
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Maya lists all the positive divisors of <math>2010^2</math>. She then randomly selects two distinct divisors from this list. Let <math>p</math> be the probability that exactly one of the selected divisors is a perfect square. The probability <math>p</math> can be expressed in the form <math>\frac {m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m + n</math>.
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Maya lists all the positive divisors of <math>2010^2</math>. She then randomly selects two distinct divisors from this list. Let <math>p</math> be the [[probability]] that exactly one of the selected divisors is a [[perfect square]]. The probability <math>p</math> can be expressed in the form <math>\frac {m}{n}</math>, where <math>m</math> and <math>n</math> are [[relatively prime]] positive integers. Find <math>m + n</math>.
  
 
== Solution ==
 
== Solution ==
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{{AIME box|year=2010|before=First Problem|num-a=2|n=I}}
 
{{AIME box|year=2010|before=First Problem|num-a=2|n=I}}
  
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[[Category:Introductory Algebra Problems]]
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[[Category:Introductory Number Theory Problems]]

Revision as of 13:13, 17 March 2010

Problem

Maya lists all the positive divisors of $2010^2$. She then randomly selects two distinct divisors from this list. Let $p$ be the probability that exactly one of the selected divisors is a perfect square. The probability $p$ can be expressed in the form $\frac {m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m + n$.

Solution

$2010^2 = 2^2\cdot3^2\cdot5^2\cdot67^2$. Thus there are $(2+1)^4$ divisors, $2^4$ of which are squares (the exponent of each prime factor must either be $0$ or $2$). Therefore the probability is \[\frac {2\cdot2^4\cdot(3^4 - 2^4)}{3^4(3^4 - 1)} = \frac {26}{81} \Longrightarrow 26+ 81 = \boxed{107}.\]

See also

2010 AIME I (Problems • Answer Key • Resources)
Preceded by
First Problem 		Followed by
Problem 2
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions
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