Difference between revisions of "2014 AIME I Problems/Problem 14"
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<math>z^2 - 104z + 2504 = 0</math> | <math>z^2 - 104z + 2504 = 0</math> | ||
− | Using the quadratic formula, we get that the largest solution for z is <math>z = 52 + 10\sqrt{2}</math>. Then, repeatedly substituting backwards, we find that the largest value of x is <math>11 + \sqrt{52 + \sqrt{200}}</math>. The answer is 11 + 52 + 200 = 263 | + | Using the quadratic formula, we get that the largest solution for z is <math>z = 52 + 10\sqrt{2}</math>. Then, repeatedly substituting backwards, we find that the largest value of x is <math>11 + \sqrt{52 + \sqrt{200}}</math>. The answer is thus <math>11 + 52 + 200 = \boxed{263}</math> |
== See also == | == See also == | ||
{{AIME box|year=2014|n=I|num-b=13|num-a=15}} | {{AIME box|year=2014|n=I|num-b=13|num-a=15}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 14:59, 15 March 2014
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 x = 0 is not the highest solution (if we realize it is, we can always backtrack) we can cancel the common factor of x from both sides of the equation.
Then, if we make the substitution y = x - 11, 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 y from both sides, knowing that is a possible solution. After we do that, we can make the final substitution z = y^2.
Using the quadratic formula, we get that the largest solution for z is . Then, repeatedly substituting backwards, we find that the largest value of x 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.