Difference between revisions of "1987 AIME Problems/Problem 14"
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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 /> |
Revision as of 14:28, 24 March 2018
Problem
Compute

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 |
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