Difference between revisions of "2002 AIME II Problems/Problem 11"
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So, the sum is <math>\frac{8x^2}{1-\frac{1}{8x}}=1</math>. Thus, <math>64x^3-8x+1 = (4x-1)(16x^2+4x-1) = 0 \Rightarrow x = \frac{1}{4}, \frac{-1 \pm \sqrt{5}}{8}</math>. | So, the sum is <math>\frac{8x^2}{1-\frac{1}{8x}}=1</math>. Thus, <math>64x^3-8x+1 = (4x-1)(16x^2+4x-1) = 0 \Rightarrow x = \frac{1}{4}, \frac{-1 \pm \sqrt{5}}{8}</math>. | ||
| − | The only solution in the appropriate form is <math>x = \frac{\sqrt{5}-1}{8}</math>. Therefore, <math>100m+10n+p = \boxed{518}</math>. | + | The only solution in the appropriate form is <math>x = \frac{\sqrt{5}-1}{8}</math>. Therefore, <math>100m+10n+p = \boxed{518}</math>. |
== See also == | == See also == | ||
{{AIME box|year=2002|n=II|num-b=10|num-a=12}} | {{AIME box|year=2002|n=II|num-b=10|num-a=12}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 19:49, 21 November 2015
Problem
Two distinct, real, infinite geometric series each have a sum of 
and have the same second term. The third term of one of the series is 
, and the second term of both series can be written in the form 
, where 
, 
, and 
are positive integers and is not divisible by the square of any prime. Find 
.
Solution
Let the second term of each series be 
. Then, the common ratio is 
, and the first term is 
.
So, the sum is 
. Thus, 
.
The only solution in the appropriate form is 
. Therefore, 
.
See also
| 2002 AIME II (Problems • Answer Key • Resources) | ||
| Preceded by Problem 10 |
Followed by Problem 12 | |
| 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. 
