Difference between revisions of "2005 AIME II Problems/Problem 3"

 
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== Problem ==
 
== Problem ==
How many positive integers have exactly three proper divisors, each of which is less than 50?
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An infinite geometric series has sum 2005. A new series, obtained by squaring each term of the original series, has 10 times the sum of the original series. The common ratio of the original series is <math> \frac mn </math> where <math> m </math> and <math> n </math> are relatively prime integers. Find <math> m+n. </math>
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== Solution ==
 
== Solution ==
 
== See Also ==
 
== See Also ==
 
*[[2005 AIME II Problems]]
 
*[[2005 AIME II Problems]]

Revision as of 23:19, 8 July 2006

Problem

An infinite geometric series has sum 2005. A new series, obtained by squaring each term of the original series, has 10 times the sum of the original series. The common ratio of the original series is $\frac mn$ where $m$ and $n$ are relatively prime integers. Find $m+n.$

Solution

See Also