Difference between revisions of "1987 AIME Problems/Problem 14"

Revision as of 13:08, 2 July 2021

Problem

Compute $\frac{(10^4+324)(22^4+324)(34^4+324)(46^4+324)(58^4+324)}{(4^4+324)(16^4+324)(28^4+324)(40^4+324)(52^4+324)}.$

Solution 1 (Sophie Germain Identity)

The Sophie Germain Identity states that $a^4 + 4b^4$ can be factored as $(a^2 + 2b^2 - 2ab)(a^2 + 2b^2 + 2ab)$ . Each of the terms is in the form of $x^4 + 324$ . Using Sophie Germain, we get that $x^4 + 4\cdot 3^4 = (x^2 + 2 \cdot 3^2 - 2\cdot 3\cdot x)(x^2 + 2 \cdot 3^2 + 2\cdot 3\cdot x) = (x(x-6) + 18)(x(x+6)+18),$ so the original expression becomes

$\frac{[(10(10-6)+18)(10(10+6)+18)][(22(22-6)+18)(22(22+6)+18)]\cdots[(58(58-6)+18)(58(58+6)+18)]}{[(4(4-6)+18)(4(4+6)+18)][(16(16-6)+18)(16(16+6)+18)]\cdots[(52(52-6)+18)(52(52+6)+18)]}$

$= \frac{(10(4)+18)(10(16)+18)(22(16)+18)(22(28)+18)\cdots(58(52)+18)(58(64)+18)}{(4(-2)+18)(4(10)+18)(16(10)+18)(16(22)+18)\cdots(52(46)+18)(52(58)+18)}$

Almost all of the terms cancel out! We are left with $\frac{58(64)+18}{4(-2)+18} = \frac{3730}{10} = \boxed{373}$ .

Solution 2 (Completing the Square and Difference of Squares)

In both the numerator and the denominator, each factor is of the form $N^4+324=N^4+18^2$ for some positive integer $N.$

We factor $N^4+18^2$ by completing the square, then applying the difference of squares: $\begin{align*} N^4+18^2&=\left(N^4+36N^2+18^2\right)-36N^2 \\ &=\left(N^2+18\right)^2-(6N)^2 \\ &=\left(N^2-6N+18\right)\left(N^2+6N+18\right) \\ &=\left((N-3)^2+9\right)\left((N+3)^2+9\right). \end{align*}$ The original expression now becomes $\frac{\left[(7^2+9)(13^2+9)\right]\left[(19^2+9)(25^2+9)\right]\left[(31^2+9)(37^2+9)\right]\left[(43^2+9)(49^2+9)\right]\left[(55^2+9)(61^2+9)\right]}{\left[(1^2+9)(7^2+9)\right]\left[(13^2+9)(19^2+9)\right]\left[(25^2+9)(31^2+9)\right]\left[(37^2+9)(43^2+9)\right]\left[(49^2+9)(55^2+9)\right]}=\frac{61^2+9}{1^2+9}=\boxed{373}.$ ~MRENTHUSIASM

Solution 3 (Complex Numbers)

In both the numerator and the denominator, each factor is of the form $N^4+324=N^4+18^2$ for some positive integer $N.$

We factor $N^4+18^2$ by solving the equation $N^4+18^2=0,$ or $N^4=-18^2.$

Two solutions follow from here:

Solution 3.1 (Exponential Form)

We rewrite $N$ to the exponential form $N=re^{i\theta},$ where $r$ is the magnitude of $N$ such that $r\geq0,$ and $\theta$ is the argument of $N$ such that $0\leq\theta<2\pi.$

Solution 3.2 (Rectangular Form)

Video Solution

https://youtu.be/ZWqHxc0i7ro?t=1023

~ pi_is_3.14

@@ Line 27: / Line 27: @@
 In both the numerator and the denominator, each factor is of the form <math>N^4+324=N^4+18^2</math> for some positive integer <math>N.</math>
-We factor <math>N^4+18^2</math> by solving the equation <math>N^4+18^2</math> <math>=0</math>.
+We factor <math>N^4+18^2</math> by solving the equation <math>N^4+18^2=0,</math> or <math>N^4=-18^2.</math>
-<b>SOLUTION IN PROGRESS. WAITING FOR THE API ISSUE TO BE FIXED.</b>
+Two solutions follow from here:
-<math>N^4+18^2=0^2</math> (code testing--this should be rendered correctly).
+=== Solution 3.1 (Exponential Form) ===
+We rewrite <math>N</math> to the exponential form <cmath>N=re^{i\theta},</cmath> where <math>r</math> is the magnitude of <math>N</math> such that <math>r\geq0,</math> and <math>\theta</math> is the argument of <math>N</math> such that <math>0\leq\theta<2\pi.</math>
-TEST
+=== Solution 3.2 (Rectangular Form) ===
 == Video Solution ==

1987 AIME (Problems • Answer Key • Resources)
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