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

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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>.
 
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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== Solution ==
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== Solution 1==
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<math>2010^2 = 2^2\cdot3^2\cdot5^2\cdot67^2</math>. Thus there are <math>(2+1)^4</math> divisors, <math>2^4</math> of which are squares (the exponent of each prime factor must either be <math>0</math> or <math>2</math>). Therefore the probability is
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<math>2010^2 = 2^2\cdot3^2\cdot5^2\cdot67^2</math>. Thus there are <math>(2+1)^4</math> divisors, <math>(1+1)^4</math> of which are squares (the exponent of each prime factor must either be <math>0</math> or <math>2</math>). Therefore the probability is
 
<cmath>\frac {2\cdot2^4\cdot(3^4 - 2^4)}{3^4(3^4 - 1)} = \frac {26}{81} \Longrightarrow 26+ 81 = \boxed{107}.</cmath>
 
<cmath>\frac {2\cdot2^4\cdot(3^4 - 2^4)}{3^4(3^4 - 1)} = \frac {26}{81} \Longrightarrow 26+ 81 = \boxed{107}.</cmath>
  
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== Solution 2: Using a bit more counting==
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== Solution 2 (Using a Bit More Counting) ==
 
The prime factorization of <math>2010^2</math> is <math>67^2\cdot3^2\cdot2^2\cdot5^2</math>. Therefore, the number of divisors of <math>2010^2</math> is <math>3^4</math> or <math>81</math>, <math>16</math> of which are perfect squares. The number of ways we can choose <math>1</math> perfect square from the two distinct divisors is <math>\binom{16}{1}\binom{81-16}{1}</math>. The total number of ways to pick two divisors is <math>\binom{81}{2}</math>
 
The prime factorization of <math>2010^2</math> is <math>67^2\cdot3^2\cdot2^2\cdot5^2</math>. Therefore, the number of divisors of <math>2010^2</math> is <math>3^4</math> or <math>81</math>, <math>16</math> of which are perfect squares. The number of ways we can choose <math>1</math> perfect square from the two distinct divisors is <math>\binom{16}{1}\binom{81-16}{1}</math>. The total number of ways to pick two divisors is <math>\binom{81}{2}</math>
  

Latest revision as of 19:35, 17 December 2021

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 1

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

Solution 2 (Using a Bit More Counting)

The prime factorization of $2010^2$ is $67^2\cdot3^2\cdot2^2\cdot5^2$. Therefore, the number of divisors of $2010^2$ is $3^4$ or $81$, $16$ of which are perfect squares. The number of ways we can choose $1$ perfect square from the two distinct divisors is $\binom{16}{1}\binom{81-16}{1}$. The total number of ways to pick two divisors is $\binom{81}{2}$

Thus, the probability is \[\frac {\binom{16}{1}\binom{81-16}{1}}{\binom{81}{2}} = \frac {16\cdot65}{81\cdot40} = \frac {26}{81} \Longrightarrow 26+ 81 = \boxed{107}.\]

Video Solution

https://www.youtube.com/watch?v=YJeF9dLJZuw (Osman Nal)

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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