Difference between revisions of "2014 AIME I Problems/Problem 14"
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== Solution == | == 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 <math>3 | + | 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 <math>frac{3}{x-3}</math>, then the fraction becomes of the form <math>frac{x}{x - 3}</math>. 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. |
| − | <math>\ </math> | + | |
| + | <math>\frac{1}{x - 3} + \frac{1}{x - 5} + \frac{1}{x - 17} + \frac{1}{x - 19}</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 19:38, 14 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.
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. 
