Difference between revisions of "1987 AIME Problems/Problem 14"
MRENTHUSIASM (talk | contribs) (Video solution should be at the end.) |
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== Problem == | == Problem == | ||
Compute | Compute | ||
− | < | + | <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> |
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+ | == Solution 1 (Generalized) == | ||
+ | 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 /> | ||
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Almost all of the terms cancel out! We are left with <math>\frac{58(64)+18}{4(-2)+18} = \frac{3730}{10} = \boxed{373}</math>. | Almost all of the terms cancel out! We are left with <math>\frac{58(64)+18}{4(-2)+18} = \frac{3730}{10} = \boxed{373}</math>. | ||
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+ | == Solution 2 (Specified) == | ||
== Video Solution == | == Video Solution == |
Revision as of 00:18, 1 July 2021
Problem
Compute
Solution 1 (Generalized)
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 (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.