Difference between revisions of "2014 AIME I Problems/Problem 14"
m (added latex tags) |
Tempaccount (talk | contribs) (Adding problem section) |
||
Line 1: | Line 1: | ||
+ | |||
+ | ==Problem== | ||
== Problem 14 == | == Problem 14 == | ||
Revision as of 14:41, 9 August 2018
Contents
Problem
Problem 14
Let be the largest real solution to the equation
There are positive integers , , and such that . Find .
Solution
The first step is to notice that the 4 on the right hand side can simplify the terms on the left hand side. If we distribute 1 to , then the fraction becomes of the form . A similar cancellation happens with the other four terms. If we assume is not the highest solution (if we realize it is, we can always backtrack) we can cancel the common factor of from both sides of the equation.
Then, if we make the substitution , we can further simplify.
If we group and combine the terms of the form and , we get this equation:
Then, we can cancel out a from both sides, knowing that is a possible solution. After we do that, we can make the final substitution .
Using the quadratic formula, we get that the largest solution for is . Then, repeatedly substituting backwards, we find that the largest value of is . The answer is thus
See also
2014 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 13 |
Followed by Problem 15 | |
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.