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

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)

@@ Line 2: / Line 2: @@
 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 1==
-<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
+<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>
-== See also ==
+== 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>
+Thus, the probability is
+<cmath>\frac {\binom{16}{1}\binom{81-16}{1}}{\binom{81}{2}} = \frac {16\cdot65}{81\cdot40} = \frac {26}{81} \Longrightarrow 26+ 81 = \boxed{107}.</cmath>
+== Video Solution ==
+https://www.youtube.com/watch?v=YJeF9dLJZuw (Osman Nal)
+== See Also ==
 {{AIME box|year=2010|before=First Problem|num-a=2|n=I}}
 [[Category:Introductory Number Theory Problems]]
+{{MAA Notice}}

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

Page

Toolbox

Search