Difference between revisions of "2004 AIME I Problems/Problem 13"
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<math>\frac 3 {17}</math> and <math>\frac 4{19}</math> can both be seen to be larger than any of <math>\frac1{19}, \frac2{19}, \frac3{19}, \frac 1{17}, \frac2{17}</math>, so these latter five are the numbers we want to add. | <math>\frac 3 {17}</math> and <math>\frac 4{19}</math> can both be seen to be larger than any of <math>\frac1{19}, \frac2{19}, \frac3{19}, \frac 1{17}, \frac2{17}</math>, so these latter five are the numbers we want to add. | ||
| − | <math>\frac1{19}+ \frac2{19}+ \frac3{19}+ \frac 1{17}+ \frac2{17}= \frac6{19} + \frac 3{17} = \frac{6\cdot17 + 3\cdot19}{17\cdot19} = \frac{159}{323}</math> and so the answer is <math>159 + 323 = | + | <math>\frac1{19}+ \frac2{19}+ \frac3{19}+ \frac 1{17}+ \frac2{17}= \frac6{19} + \frac 3{17} = \frac{6\cdot17 + 3\cdot19}{17\cdot19} = \frac{159}{323}</math> and so the answer is <math>\displaystyle 159 + 323 = 482</math>. |
== See also == | == See also == | ||
* [[2004 AIME I Problems]] | * [[2004 AIME I Problems]] | ||
Revision as of 08:55, 28 July 2006
Problem
The polynomial 
has 34 complex roots of the form ![$z_k = r_k[\cos(2\pi a_k)+i\sin(2\pi a_k)], k=1, 2, 3,\ldots, 34,$](http://latex.artofproblemsolving.com/d/0/c/d0c96625746cab4901eb36724577280cf1515b7b.png)
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 
.
