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 == | ||
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{{AIME box|year=2010|before=First Problem|num-a=2|n=I}} | {{AIME box|year=2010|before=First Problem|num-a=2|n=I}} | ||
− | [[Category:Introductory | + | [[Category:Introductory Number Theory Problems]] |
Revision as of 12:13, 17 March 2010
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
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 |