Difference between revisions of "1987 AIME Problems/Problem 14"
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The original expression now becomes <cmath>\frac{\left[(7^2+9)(13^2+9)\right]\left[(19^2+9)(25^2+9)\right]\left[(31^2+9)(37^2+9)\right]\left[(43^2+9)(49^2+9)\right]\left[(55^2+9)(61^2+9)\right]}{\left[(1^2+9)(7^2+9)\right]\left[(13^2+9)(19^2+9)\right]\left[(25^2+9)(31^2+9)\right]\left[(37^2+9)(43^2+9)\right]\left[(49^2+9)(55^2+9)\right]}=\frac{61^2+9}{1^2+9}=\boxed{373}.</cmath> | The original expression now becomes <cmath>\frac{\left[(7^2+9)(13^2+9)\right]\left[(19^2+9)(25^2+9)\right]\left[(31^2+9)(37^2+9)\right]\left[(43^2+9)(49^2+9)\right]\left[(55^2+9)(61^2+9)\right]}{\left[(1^2+9)(7^2+9)\right]\left[(13^2+9)(19^2+9)\right]\left[(25^2+9)(31^2+9)\right]\left[(37^2+9)(43^2+9)\right]\left[(49^2+9)(55^2+9)\right]}=\frac{61^2+9}{1^2+9}=\boxed{373}.</cmath> | ||
~MRENTHUSIASM | ~MRENTHUSIASM | ||
| + | |||
| + | == Solution 3 (Complex Numbers) == | ||
| + | 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=0.</math> | ||
| + | |||
| + | <b>IN PROGRESS. NO EDIT PLEASE</b> | ||
== Video Solution == | == Video Solution == | ||
Revision as of 16:20, 1 July 2021
Contents
Problem
Compute
![\[\frac{(10^4+324)(22^4+324)(34^4+324)(46^4+324)(58^4+324)}{(4^4+324)(16^4+324)(28^4+324)(40^4+324)(52^4+324)}.\]](http://latex.artofproblemsolving.com/1/6/b/16b3624dd694603e72b8f0171eb31c63ad06597c.png)
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 ![\[x^4 + 4\cdot 3^4 = (x^2 + 2 \cdot 3^2 - 2\cdot 3\cdot x)(x^2 + 2 \cdot 3^2 + 2\cdot 3\cdot x) = (x(x-6) + 18)(x(x+6)+18),\]](http://latex.artofproblemsolving.com/0/4/5/045962c1e4a43ac7a5a749eacc748e8bdaea3438.png)
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)
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 ![\[\frac{\left[(7^2+9)(13^2+9)\right]\left[(19^2+9)(25^2+9)\right]\left[(31^2+9)(37^2+9)\right]\left[(43^2+9)(49^2+9)\right]\left[(55^2+9)(61^2+9)\right]}{\left[(1^2+9)(7^2+9)\right]\left[(13^2+9)(19^2+9)\right]\left[(25^2+9)(31^2+9)\right]\left[(37^2+9)(43^2+9)\right]\left[(49^2+9)(55^2+9)\right]}=\frac{61^2+9}{1^2+9}=\boxed{373}.\]](http://latex.artofproblemsolving.com/4/b/f/4bff1d2c80c44994e54758cdf9ba4aa2dbdd5e98.png)
~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 
IN PROGRESS. NO EDIT PLEASE
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. 
