Difference between revisions of "2003 AIME II Problems/Problem 15"
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Therefore the answer is <math>8+4+3 = \boxed{015}</math>. | Therefore the answer is <math>8+4+3 = \boxed{015}</math>. | ||
+ | ==Solution 2== | ||
+ | |||
+ | Note that <math>x^k + x^{k-1} + \dots + x + 1 = \frac{x^{k+1} - 1}{x - 1}</math>. Our sum can be reformed as <cmath>\frac{x(x^{47} - 1) + x^2(x^{45} - 1) + \dots + x^{24}(x - 1)}{x-1}</cmath> | ||
+ | |||
+ | So <cmath>\frac{x^{48} + x^{47} + x^{46} + \dots + x^{25} - x^{24} - x^{23} - \dots - x}{x-1} = 0</cmath> | ||
+ | |||
+ | <math>x(x^{47} + x^{46} + \dots - x - 1) = 0</math> | ||
+ | |||
+ | <math>x^{47} + x^{46} + \dots - x - 1 = 0</math> | ||
+ | |||
+ | <math>x^{47} + x^{46} + \dots + x + 1 = 2(x^{23} + x^{22} + \dots + x + 1)</math> | ||
+ | |||
+ | <math>\frac{x^{48} - 1}{x - 1} = 2\frac{x^{24} - 1}{x - 1}</math> | ||
+ | |||
+ | <math>x^{48} - 1 - 2x^{24} + 2 = 0</math> | ||
+ | |||
+ | <math>(x^{24} - 1)^2 = 0</math> | ||
+ | |||
+ | And we can proceed as above. | ||
== See also == | == See also == | ||
{{AIME box|year=2003|n=II|num-b=14|after=Last Question}} | {{AIME box|year=2003|n=II|num-b=14|after=Last Question}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 18:37, 16 February 2014
Contents
Problem
Let
Let be the distinct zeros of and let for where and and are real numbers. Let
where and are integers and is not divisible by the square of any prime. Find
Solution
We can rewrite the definition of as follows:
This can quite obviously be factored as:
Note that . So the roots of are exactly all -th complex roots of , except for the root .
Let . Then the distinct zeros of are .
We can clearly ignore the root as it does not contribute to the value that we need to compute.
The squares of the other roots are .
Hence we need to compute the following sum:
Using basic properties of the sine function, we can simplify this to
The five-element sum is just . We know that , , and . Hence our sum evaluates to:
Therefore the answer is .
Solution 2
Note that . Our sum can be reformed as
So
And we can proceed as above.
See also
2003 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Last Question | |
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.