# Difference between revisions of "2004 AIME I Problems/Problem 13"

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We see that the expression for the [[polynomial]] <math>P</math> is very difficult to work with directly, but there is one obvious transformation to make: sum the [[geometric series]]: | We see that the expression for the [[polynomial]] <math>P</math> is very difficult to work with directly, but there is one obvious transformation to make: sum the [[geometric series]]: | ||

− | <math>P(x) = (\frac{x^{18} - 1}{x - 1})^2 - x^{17} = \frac{x^{36} - 2x^{18} + 1}{x^2 - 2x + 1} - x^{17} = \frac{x^{36} - x^{19} - x^{17} + 1}{(x - 1)^2} = \frac{(x^{19} - 1)(x^{17} - 1)}{(x - 1)^2}</math>. | + | <math>P(x) = \left(\frac{x^{18} - 1}{x - 1}\right)^2 - x^{17} = \frac{x^{36} - 2x^{18} + 1}{x^2 - 2x + 1} - x^{17} = \frac{x^{36} - x^{19} - x^{17} + 1}{(x - 1)^2} = \frac{(x^{19} - 1)(x^{17} - 1)}{(x - 1)^2}</math>. |

This [[expression]] has [[root]]s at every 17th root and 19th root of unity, other than 1. Since 17 and 19 are [[relatively prime]], this means there are no duplicate roots. Thus, <math>a_1, a_2, a_3, a_4</math> and <math>a_5</math> are the five smallest fractions of the form <math>\frac m{19}</math> or <math>\frac n {17}</math> for <math>m, n > 0</math>. | This [[expression]] has [[root]]s at every 17th root and 19th root of unity, other than 1. Since 17 and 19 are [[relatively prime]], this means there are no duplicate roots. Thus, <math>a_1, a_2, a_3, a_4</math> and <math>a_5</math> are the five smallest fractions of the form <math>\frac m{19}</math> or <math>\frac n {17}</math> for <math>m, n > 0</math>. |

## Revision as of 17:52, 2 December 2006

## Problem

The polynomial has 34 complex roots of the form with and Given that where and are relatively prime positive integers, find

## Solution

We see that the expression for the polynomial is very difficult to work with directly, but there is one obvious transformation to make: sum the geometric series:

.

This expression has roots at every 17th root and 19th root of unity, other than 1. Since 17 and 19 are relatively prime, this means there are no duplicate roots. Thus, and are the five smallest fractions of the form or for .

and can both be seen to be larger than any of , so these latter five are the numbers we want to add.

and so the answer is .