Difference between revisions of "2015 AIME I Problems/Problem 12"
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At this point, we can see a pattern: our desired answer is always <math>\frac{p+1}{q+1}</math>. Plugging in <math>p = 2015</math> and <math>q = 1000</math>, the answer is <math>\frac{2016}{1001}=\frac{288}{143}</math>, so <math>288 + 143 = \boxed{431}</math>. | At this point, we can see a pattern: our desired answer is always <math>\frac{p+1}{q+1}</math>. Plugging in <math>p = 2015</math> and <math>q = 1000</math>, the answer is <math>\frac{2016}{1001}=\frac{288}{143}</math>, so <math>288 + 143 = \boxed{431}</math>. | ||
+ | |||
+ | ===Solution 4 (short)=== | ||
+ | In the "average case", the numbers evenly partition the interval [0,2016] into 1001 parts. Then because it asks for the expected value of the least element the answer is <math>\frac{2016}{1001}</math>. | ||
+ | |||
+ | -tigershark22 | ||
==Video Solution== | ==Video Solution== |
Latest revision as of 21:51, 1 March 2021
Contents
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 .
Hint
Use the Hockey Stick Identity in the form
(This is best proven by a combinatorial argument that coincidentally pertains to the problem: count two ways the number of subsets of the first numbers with elements whose least element is , for .)
Solution
Solution 1
Let be the desired mean. Then because subsets have 1000 elements and have as their least element, Using the definition of binomial coefficient and the identity , we deduce that The answer is
Solution 2
Each 1000-element subset of with contributes to the sum of the least element of each subset. Now, consider the set . There are ways to choose a positive integer such that ( can be anything from to inclusive). Thus, the number of ways to choose the set is equal to the sum. But choosing a set is the same as choosing a 1001-element subset from !
Thus, the average is . Our answer is .
Solution 3(NOT RIGOROUS)
Let be the size of the large set and be the size of the subset (i.e. in this problem, and ). We can easily find the answers for smaller values of and :
For and , the answer is .
For and , the answer is .
For and , the answer is .
For and , the answer is .
For and , the answer is .
For and , the answer is .
At this point, we can see a pattern: our desired answer is always . Plugging in and , the answer is , so .
Solution 4 (short)
In the "average case", the numbers evenly partition the interval [0,2016] into 1001 parts. Then because it asks for the expected value of the least element the answer is .
-tigershark22
Video Solution
https://www.youtube.com/watch?v=t-yHTkGkMK4 ~anellipticcurveoverq
Generalization
https://artofproblemsolving.com/wiki/index.php/1981_IMO_Problems/Problem_2
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.