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>. | ||
| − | == 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>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 | ||
<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> | ||
| − | == Solution 2 | + | == 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 22:25, 25 June 2021
Contents
Problem
Maya lists all the positive divisors of 
. She then randomly selects two distinct divisors from this list. Let 
be the probability that exactly one of the selected divisors is a perfect square. The probability can be expressed in the form 
, where 
and 
are relatively prime positive integers. Find 
.
Solution 1
. Thus there are 
divisors, 
of which are squares (the exponent of each prime factor must either be 
or 
). Therefore the probability is
![\[\frac {2\cdot2^4\cdot(3^4 - 2^4)}{3^4(3^4 - 1)} = \frac {26}{81} \Longrightarrow 26+ 81 = \boxed{107}.\]](http://latex.artofproblemsolving.com/2/6/d/26d4bcdbb00a5ce47108567249a4927492b0ad49.png)
Solution 2 (Using a Bit More Counting)
The prime factorization of is 
. Therefore, the number of divisors of is 
or 
, 
of which are perfect squares. The number of ways we can choose 
perfect square from the two distinct divisors is 
. The total number of ways to pick two divisors is 
Thus, the probability is
![\[\frac {\binom{16}{1}\binom{81-16}{1}}{\binom{81}{2}} = \frac {16\cdot65}{81\cdot40} = \frac {26}{81} \Longrightarrow 26+ 81 = \boxed{107}.\]](http://latex.artofproblemsolving.com/2/e/3/2e3379935124fd552165e887ddaaefa2de59ad16.png)
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
