Difference between revisions of "1987 AIME Problems/Problem 14"
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In both the numerator and the denominator, each factor is of the form <math>N^4+324=N^4+18^2</math> for some positive integer <math>N.</math> | In both the numerator and the denominator, each factor is of the form <math>N^4+324=N^4+18^2</math> for some positive integer <math>N.</math> | ||
− | We factor <math>N^4+18^2</math> by solving the equation <math>N^4+18^2</math> <math>= | + | We factor <math>N^4+18^2</math> by solving the equation <math>N^4+18^2=0,</math> or <math>N^4=-18^2.</math> |
− | + | Two solutions follow from here: | |
− | <math>N^ | + | === Solution 3.1 (Exponential Form) === |
+ | We rewrite <math>N</math> to the exponential form <cmath>N=re^{i\theta},</cmath> where <math>r</math> is the magnitude of <math>N</math> such that <math>r\geq0,</math> and <math>\theta</math> is the argument of <math>N</math> such that <math>0\leq\theta<2\pi.</math> | ||
− | + | === Solution 3.2 (Rectangular Form) === | |
== Video Solution == | == Video Solution == |
Revision as of 14:08, 2 July 2021
Contents
Problem
Compute
Solution 1 (Sophie Germain Identity)
The Sophie Germain Identity states that can be factored as
. Each of the terms is in the form of
. Using Sophie Germain, we get that
so the original expression becomes
![$\frac{[(10(10-6)+18)(10(10+6)+18)][(22(22-6)+18)(22(22+6)+18)]\cdots[(58(58-6)+18)(58(58+6)+18)]}{[(4(4-6)+18)(4(4+6)+18)][(16(16-6)+18)(16(16+6)+18)]\cdots[(52(52-6)+18)(52(52+6)+18)]}$](http://latex.artofproblemsolving.com/4/d/f/4df1c85169d79080a1e4ad7cd3cb96f9c11ec0c7.png)
![$= \frac{(10(4)+18)(10(16)+18)(22(16)+18)(22(28)+18)\cdots(58(52)+18)(58(64)+18)}{(4(-2)+18)(4(10)+18)(16(10)+18)(16(22)+18)\cdots(52(46)+18)(52(58)+18)}$](http://latex.artofproblemsolving.com/4/b/d/4bdb1dff907fef36c98d06fdb044a05f81004d30.png)
Almost all of the terms cancel out! We are left with .
Solution 2 (Completing the Square and Difference of Squares)
In both the numerator and the denominator, each factor is of the form for some positive integer
We factor by completing the square, then applying the difference of squares:
The original expression now becomes
~MRENTHUSIASM
Solution 3 (Complex Numbers)
In both the numerator and the denominator, each factor is of the form for some positive integer
We factor by solving the equation
or
Two solutions follow from here:
Solution 3.1 (Exponential Form)
We rewrite to the exponential form
where
is the magnitude of
such that
and
is the argument of
such that
Solution 3.2 (Rectangular Form)
Video Solution
https://youtu.be/ZWqHxc0i7ro?t=1023
~ pi_is_3.14
See also
1987 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 13 |
Followed by Problem 15 | |
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.