Difference between revisions of "1987 AIME Problems/Problem 14"
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== Problem == | == Problem == | ||
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 == | ||
+ | |||
The [[Sophie Germain Identity]] states that <math>a^4 + 4b^4</math> can be [[factor]]ized 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 <math>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)</math>.<br /><br /> | The [[Sophie Germain Identity]] states that <math>a^4 + 4b^4</math> can be [[factor]]ized 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 <math>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)</math>.<br /><br /> | ||
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<math>= \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)}</math></div> | <math>= \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)}</math></div> | ||
− | Almost all of the terms cancel out! We are left with <math>\frac{58(64)+18}{4(-2)+18} = \frac{3730}{10} = 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>. |
== See also == | == See also == | ||
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[[Category:Intermediate Algebra Problems]] | [[Category:Intermediate Algebra Problems]] | ||
+ | {{MAA Notice}} |
Revision as of 19:17, 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 .
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.