2002 AIME II Problems/Problem 11

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Problem

Two distinct, real, infinite geometric series each have a sum of $1$ and have the same second term. The third term of one of the series is $1/8$ , and the second term of both series can be written in the form $\frac{\sqrt{m}-n}p$ , where $m$ , $n$ , and $p$ are positive integers and $m$ is not divisible by the square of any prime. Find $100m+10n+p$ .

Solution

Let the second term of each series be $x$ . Then, the common ratio is $\frac{1}{8x}$ , and the first term is $8x^2$ .

So, the sum is $\frac{8x^2}{1-\frac{1}{8x}}=1$ . Thus, $64x^3-8x+1 = (4x-1)(16x^2+4x-1) = 0 \Rightarrow x = \frac{1}{4}, \frac{-1 \pm \sqrt{5}}{8}$ .

The only solution in the appropriate form is $x = \frac{\sqrt{5}-1}{8}$ . Therefore, $100m+10n+p = \boxed{518}$ .

2002 AIME II (Problems • Resources)
Preceded by Problem 10	Followed by Problem 12
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

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2002 AIME II Problems/Problem 11

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Problem

Solution

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