Difference between revisions of "2015 AIME I Problems/Problem 12"
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Consider all 1000-element subsets of the set {1, 2, 3, ... , 2015}. From each such subset choose the least element. The arithmetic mean of all of these least elements is <math> \frac{p}{q} </math>, where <math>p</math> and <math>q</math> are relatively prime positive integers. Find <math>p + q</math>. | Consider all 1000-element subsets of the set {1, 2, 3, ... , 2015}. From each such subset choose the least element. The arithmetic mean of all of these least elements is <math> \frac{p}{q} </math>, where <math>p</math> and <math>q</math> are relatively prime positive integers. Find <math>p + q</math>. | ||
| + | |||
| + | == See also == | ||
| + | {{AIME box|year=2015|n=I|num-b=11|num-a=13}} | ||
| + | {{MAA Notice}} | ||
| + | [[Category:Introductory Geometry Problems]] | ||
Revision as of 18:01, 20 March 2015
Problem
Consider all 1000-element subsets of the set {1, 2, 3, ... , 2015}. From each such subset choose the least element. The arithmetic mean of all of these least elements is 
, where 
and 
are relatively prime positive integers. Find 
.
See also
| 2015 AIME I (Problems • Answer Key • Resources) | ||
| Preceded by Problem 11 |
Followed by Problem 13 | |
| 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. 
