Difference between revisions of "1987 AIME Problems/Problem 14"
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<cmath>\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)}.</cmath> | <cmath>\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)}.</cmath> | ||
| − | == Solution 1 ( | + | == Solution 1 (Sophie Germain Identity) == |
The [[Sophie Germain Identity]] states that <math>a^4 + 4b^4</math> can be factored as <math>(a^2 + 2b^2 - 2ab)(a^2 + 2b^2 + 2ab)</math>. Each of the terms is in the form of <math>x^4 + 324</math>. Using Sophie Germain, we get that <cmath>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),</cmath> so the original expression becomes | The [[Sophie Germain Identity]] states that <math>a^4 + 4b^4</math> can be factored as <math>(a^2 + 2b^2 - 2ab)(a^2 + 2b^2 + 2ab)</math>. Each of the terms is in the form of <math>x^4 + 324</math>. Using Sophie Germain, we get that <cmath>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),</cmath> so the original expression becomes | ||
<div style="text-align:center;"><math>\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)]}</math><br /><br /> | <div style="text-align:center;"><math>\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)]}</math><br /><br /> | ||
Revision as of 23:20, 30 June 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 (Specified)
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. 
