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

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<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>
  

Revision as of 21:55, 18 August 2019

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 \[\frac {2\cdot2^4\cdot(3^4 - 2^4)}{3^4(3^4 - 1)} = \frac {26}{81} \Longrightarrow 26+ 81 = \boxed{107}.\]

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

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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