Difference between revisions of "2010 AIME I Problems/Problem 1"
Tempaccount (talk | contribs) (Adding problem section) |
|||
| Line 1: | Line 1: | ||
| + | |||
| + | ==Problem== | ||
== Problem == | == Problem == | ||
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>. | ||
Revision as of 14:45, 9 August 2018
Contents
Problem
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
. 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)
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. 
