Difference between revisions of "1987 AIME Problems/Problem 14"
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Pi is 3.14 (talk | contribs) (→Solution) |
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Compute | Compute | ||
<div style="text-align:center;"><math>\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)}.</math></div> | <div style="text-align:center;"><math>\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)}.</math></div> | ||
| + | |||
| + | == Video Solution == | ||
| + | https://youtu.be/ZWqHxc0i7ro?t=1023 | ||
| + | |||
| + | ~ pi_is_3.14 | ||
== Solution == | == Solution == | ||
Revision as of 18:17, 15 January 2021
Contents
Problem
Compute
Video Solution
https://youtu.be/ZWqHxc0i7ro?t=1023
~ pi_is_3.14
Solution
The Sophie Germain Identity states that 
can be factorized as 
. Each of the terms is in the form of 
. Using Sophie-Germain, we get that 
.
![$\frac{[(10(10-6)+18)(10(10+6)+18)][(22(22-6)+18)(22(22+6)+18)]\cdots[(58(58-6)+18)(58(58+6)+18)]}{[(4(4-6)+18)(4(4+6)+18)][(16(16-6)+18)(16(16+6)+18)]\cdots[(52(52-6)+18)(52(52+6)+18)]}$](http://latex.artofproblemsolving.com/4/d/f/4df1c85169d79080a1e4ad7cd3cb96f9c11ec0c7.png)
Almost all of the terms cancel out! We are left with 
.
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. 
