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

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== Solution 1 (Sophie Germain Identity) ==
 
== Solution 1 (Sophie Germain Identity) ==
−
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
+
The [[Sophie Germain Identity]] states that <math>a^4 + 4b^4</math> can be factored as <math>\left(a^2 + 2b^2 - 2ab\right)\left(a^2 + 2b^2 + 2ab\right).</math> Each of the terms is in the form of <math>x^4 + 324.</math> Using Sophie Germain, we get that
−
<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 />
+
<cmath>\begin{align*}
 +
x^4 + 324 &= x^4 + 4\cdot 3^4 \\
 +
&= \left(x^2 + 2 \cdot 3^2 - 2\cdot 3\cdot x\right)\left(x^2 + 2 \cdot 3^2 + 2\cdot 3\cdot x\right) \\
 +
&= (x(x-6) + 18)(x(x+6)+18),
 +
\end{align*}</cmath>
 +
so the original expression becomes
 +
<cmath>\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)]},</cmath> which simplifies to <cmath>\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)}.</cmath>
 +
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>
  
−
<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>
+
~Azjps (Solution)
  
−
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>.
+
~MRENTHUSIASM (Minor Reformatting)
  
 
== Solution 2 (Completing the Square and Difference of Squares) ==
 
== Solution 2 (Completing the Square and Difference of Squares) ==
Line 17: Line 24:
 
<cmath>\begin{align*}
 
<cmath>\begin{align*}
 
N^4+18^2&=\left(N^4+36N^2+18^2\right)-36N^2 \\
 
N^4+18^2&=\left(N^4+36N^2+18^2\right)-36N^2 \\
−
&=\left(N^2+18\right)-(6N)^2 \\
+
&=\left(N^2+18\right)^2-(6N)^2 \\
 
&=\left(N^2-6N+18\right)\left(N^2+6N+18\right) \\
 
&=\left(N^2-6N+18\right)\left(N^2+6N+18\right) \\
 
&=\left((N-3)^2+9\right)\left((N+3)^2+9\right).
 
&=\left((N-3)^2+9\right)\left((N+3)^2+9\right).
 
\end{align*}</cmath>
 
\end{align*}</cmath>
−
Now, the original expression becomes <cmath>\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}.</cmath>
+
The original expression now becomes <cmath>\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}.</cmath>
 +
~MRENTHUSIASM
 +
 
 +
== Solution 3 (Complex Numbers) ==
 +
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=0,</math> or <math>N^4=-18^2.</math>
 +
 
 +
Two solutions follow from here:
 +
 
 +
=== Solution 3.1 (Polar Form) ===
 +
We rewrite <math>N</math> to the polar form <cmath>N=r(\cos\theta+i\sin\theta)=r\operatorname{cis}\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>
 +
 
 +
By <b>De Moivre's Theorem</b>, we have <cmath>N^4=r^4\operatorname{cis}(4\theta)=18^2(-1),</cmath> from which
 +
<ol style="margin-left: 1.5em;">
 +
  <li><math>r^4=18^2,</math> so <math>r=3\sqrt2.</math></li><p>
 +
  <li><math>\begin{cases}
 +
\begin{aligned}
 +
\cos(4\theta) &= -1 \\
 +
\sin(4\theta) &= 0
 +
\end{aligned},
 +
\end{cases}</math> so <math>\theta=\frac{\pi}{4},\frac{3\pi}{4},\frac{5\pi}{4},\frac{7\pi}{4}.</math></li><p>
 +
</ol>
 +
By the <b>Factor Theorem</b>, we get
 +
<cmath>\begin{align*}
 +
N^4+18^2&=\biggl(N-3\sqrt2\operatorname{cis}\frac{\pi}{4}\biggr)\biggl(N-3\sqrt2\operatorname{cis}\frac{3\pi}{4}\biggr)\biggl(N-3\sqrt2\operatorname{cis}\frac{5\pi}{4}\biggr)\biggl(N-3\sqrt2\operatorname{cis}\frac{7\pi}{4}\biggr) \\
 +
&=\biggl[\biggl(N-3\sqrt2\operatorname{cis}\frac{\pi}{4}\biggr)\biggl(N-3\sqrt2\operatorname{cis}\frac{7\pi}{4}\biggr)\biggr]\biggl[\biggl(N-3\sqrt2\operatorname{cis}\frac{3\pi}{4}\biggr)\biggl(N-3\sqrt2\operatorname{cis}\frac{5\pi}{4}\biggr)\biggr] \\
 +
&=\left[(N-(3+3i))(N-(3-3i))\right]\left[(N-(-3+3i))(N-(-3-3i))\right] \\
 +
&=\left[((N-3)-3i)((N-3)+3i)\right]\left[((N+3)-3i)((N+3)+3i)\right] \\
 +
&=\left[(N-3)^2+9\right]\left[(N+3)^2+9\right].
 +
\end{align*}</cmath>
 +
We continue with the last paragraph of Solution 2 to get the answer <math>\boxed{373}.</math>
 +
 
 +
~MRENTHUSIASM
 +
 
 +
=== Solution 3.2 (Rectangular Form) ===
 +
We rewrite <math>N</math> to the rectangular form <cmath>N=a+bi</cmath> for some real numbers <math>a</math> and <math>b.</math>
 +
 
 +
Note that <math>N^2=\pm18i,</math> so there are two cases:
 +
<ol style="margin-left: 1.5em;">
 +
  <li><math>N^2=18i</math><p>
 +
We have
 +
<cmath>\begin{align*}
 +
(a+bi)^2&=18i \\
 +
a^2-b^2+2abi&=18i.
 +
\end{align*}</cmath>
 +
We need <math>\begin{cases}
 +
\begin{aligned}
 +
a^2-b^2 &= 0 \\
 +
2ab &= 18
 +
\end{aligned},
 +
\end{cases}</math> from which <math>(a,b)=(3,3),(-3,-3),</math> or <math>N=3+3i,-3-3i.</math></li><p>
 +
  <li><math>N^2=-18i</math><p>
 +
We have
 +
<cmath>\begin{align*}
 +
(a+bi)^2&=-18i \\
 +
a^2-b^2+2abi&=-18i.
 +
\end{align*}</cmath>
 +
We need <math>\begin{cases}
 +
\begin{aligned}
 +
a^2-b^2 &= 0 \\
 +
2ab &= -18
 +
\end{aligned},
 +
\end{cases}</math> from which <math>(a,b)=(3,-3),(-3,3),</math> or <math>N=3-3i,-3+3i.</math></li><p>
 +
</ol>
 +
By the <b>Factor Theorem</b>, we get
 +
<cmath>\begin{align*}
 +
N^4+18^2&=(N-(3+3i))(N-(-3-3i))(N-(3-3i))(N-(-3+3i)) \\
 +
&=\left[(N-(3+3i))(N-(3-3i))\right]\left[(N-(-3+3i))(N-(-3-3i))\right] \\
 +
&=\left[((N-3)-3i)((N-3)+3i)\right]\left[((N+3)-3i)((N+3)+3i)\right] \\
 +
&=\left[(N-3)^2+9\right]\left[(N+3)^2+9\right].
 +
\end{align*}</cmath>
 +
We continue with the last paragraph of Solution 2 to get the answer <math>\boxed{373}.</math>
 +
 
 
~MRENTHUSIASM
 
~MRENTHUSIASM
  
−
== Video Solution ==
+
== Solution 4 ==
 +
We use Sophie Germain's Identity to rewrite the first couple of multiplicands in the numerator and denominator. By Sophie Germain's: <cmath>(10^4 + 324) = (10^4 + \cdot 4 \cdot 3^4) = (10^2 + 2 \cdot 10 \cdot 3 + 2 \cdot 3^2)(10^2 - 2 \cdot 10 \cdot 3 + 2 \cdot 3^2) = (178)(58)</cmath> <cmath>(22^4 + 324) = (22^2 + 2 \cdot 3 \cdot 22 + 2 \cdot 3^2)(22^2 - 2 \cdot 3 \cdot 22 + 2 \cdot 3^2) = (634)(370)</cmath> <cmath>(4^4 + 324) = (4^2 + 2 \cdot 4 \cdot 3 + 2 \cdot 3^2)(4^2 - 2 \cdot 4 \cdot 3 + 2 \cdot 3^2) = (58)(10)</cmath> <cmath>(16^4 + 324) = (16^2 + 2 \cdot 16 \cdot 3 + 2 \cdot 3^2)(16^2 - 2 \cdot 16 \cdot 3 + 2 \cdot 3^2) = (370)(178)</cmath> If we only had these terms, then the fraction would rewrite to <math>\frac{(58)(178)(370)(634)}{(10)(58)(178)(370)}</math>. However, we notice most of the terms cancel, leaving us only with the largest term in the numerator and the smallest term in the denominator (<math>\frac{634}{10}</math>). We hypothesize that this will happen with the fraction as a whole. Then we will only be left with the largest term in the numerator, which is <math>(58^2 + 2 \cdot 58 \cdot 3 + 2 \cdot 3^2) = 3730</math>. The fraction simplifies to <math>\frac{3730}{10} = \boxed{373}</math>.
 +
 
 +
~ cxsmi
 +
 
 +
== Video Solution by OmegaLearn ==
 
https://youtu.be/ZWqHxc0i7ro?t=1023
 
https://youtu.be/ZWqHxc0i7ro?t=1023
  
 
~ pi_is_3.14
 
~ pi_is_3.14
 +
 +
== Video Solution==
 +
https://www.youtube.com/watch?v=yoOWcx2Otcw
 +
 +
~Michael Penn
  
 
== See also ==
 
== See also ==

Latest revision as of 22:39, 11 September 2024

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 $\left(a^2 + 2b^2 - 2ab\right)\left(a^2 + 2b^2 + 2ab\right).$ Each of the terms is in the form of $x^4 + 324.$ Using Sophie Germain, we get that \begin{align*} x^4 + 324 &= x^4 + 4\cdot 3^4 \\ &= \left(x^2 + 2 \cdot 3^2 - 2\cdot 3\cdot x\right)\left(x^2 + 2 \cdot 3^2 + 2\cdot 3\cdot x\right) \\ &= (x(x-6) + 18)(x(x+6)+18), \end{align*} so the original expression becomes \[\frac{[(10(10-6)+18)(10(10+6)+18)][(22(22-6)+18)(22(22+6)+18)]\cdots[(58(58-6)+18)(58(58+6)+18)]}{[(4(4-6)+18)(4(4+6)+18)][(16(16-6)+18)(16(16+6)+18)]\cdots[(52(52-6)+18)(52(52+6)+18)]},\] which simplifies to \[\frac{(10(4)+18)(10(16)+18)(22(16)+18)(22(28)+18)\cdots(58(52)+18)(58(64)+18)}{(4(-2)+18)(4(10)+18)(16(10)+18)(16(22)+18)\cdots(52(46)+18)(52(58)+18)}.\] Almost all of the terms cancel out! We are left with $\frac{58(64)+18}{4(-2)+18} = \frac{3730}{10} = \boxed{373}.$

~Azjps (Solution)

~MRENTHUSIASM (Minor Reformatting)

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 (Polar Form)

We rewrite $N$ to the polar form \[N=r(\cos\theta+i\sin\theta)=r\operatorname{cis}\theta,\] where $r$ is the magnitude of $N$ such that $r\geq0,$ and $\theta$ is the argument of $N$ such that $0\leq\theta<2\pi.$

By De Moivre's Theorem, we have \[N^4=r^4\operatorname{cis}(4\theta)=18^2(-1),\] from which

  1. $r^4=18^2,$ so $r=3\sqrt2.$

  2. $\begin{cases} \begin{aligned} \cos(4\theta) &= -1 \\ \sin(4\theta) &= 0 \end{aligned}, \end{cases}$ so $\theta=\frac{\pi}{4},\frac{3\pi}{4},\frac{5\pi}{4},\frac{7\pi}{4}.$

By the Factor Theorem, we get \begin{align*} N^4+18^2&=\biggl(N-3\sqrt2\operatorname{cis}\frac{\pi}{4}\biggr)\biggl(N-3\sqrt2\operatorname{cis}\frac{3\pi}{4}\biggr)\biggl(N-3\sqrt2\operatorname{cis}\frac{5\pi}{4}\biggr)\biggl(N-3\sqrt2\operatorname{cis}\frac{7\pi}{4}\biggr) \\ &=\biggl[\biggl(N-3\sqrt2\operatorname{cis}\frac{\pi}{4}\biggr)\biggl(N-3\sqrt2\operatorname{cis}\frac{7\pi}{4}\biggr)\biggr]\biggl[\biggl(N-3\sqrt2\operatorname{cis}\frac{3\pi}{4}\biggr)\biggl(N-3\sqrt2\operatorname{cis}\frac{5\pi}{4}\biggr)\biggr] \\ &=\left[(N-(3+3i))(N-(3-3i))\right]\left[(N-(-3+3i))(N-(-3-3i))\right] \\ &=\left[((N-3)-3i)((N-3)+3i)\right]\left[((N+3)-3i)((N+3)+3i)\right] \\ &=\left[(N-3)^2+9\right]\left[(N+3)^2+9\right]. \end{align*} We continue with the last paragraph of Solution 2 to get the answer $\boxed{373}.$

~MRENTHUSIASM

Solution 3.2 (Rectangular Form)

We rewrite $N$ to the rectangular form \[N=a+bi\] for some real numbers $a$ and $b.$

Note that $N^2=\pm18i,$ so there are two cases:

  1. $N^2=18i$

    We have \begin{align*} (a+bi)^2&=18i \\ a^2-b^2+2abi&=18i. \end{align*} We need $\begin{cases} \begin{aligned} a^2-b^2 &= 0 \\ 2ab &= 18 \end{aligned}, \end{cases}$ from which $(a,b)=(3,3),(-3,-3),$ or $N=3+3i,-3-3i.$

  2. $N^2=-18i$

    We have \begin{align*} (a+bi)^2&=-18i \\ a^2-b^2+2abi&=-18i. \end{align*} We need $\begin{cases} \begin{aligned} a^2-b^2 &= 0 \\ 2ab &= -18 \end{aligned}, \end{cases}$ from which $(a,b)=(3,-3),(-3,3),$ or $N=3-3i,-3+3i.$

By the Factor Theorem, we get \begin{align*} N^4+18^2&=(N-(3+3i))(N-(-3-3i))(N-(3-3i))(N-(-3+3i)) \\ &=\left[(N-(3+3i))(N-(3-3i))\right]\left[(N-(-3+3i))(N-(-3-3i))\right] \\ &=\left[((N-3)-3i)((N-3)+3i)\right]\left[((N+3)-3i)((N+3)+3i)\right] \\ &=\left[(N-3)^2+9\right]\left[(N+3)^2+9\right]. \end{align*} We continue with the last paragraph of Solution 2 to get the answer $\boxed{373}.$

~MRENTHUSIASM

Solution 4

We use Sophie Germain's Identity to rewrite the first couple of multiplicands in the numerator and denominator. By Sophie Germain's: \[(10^4 + 324) = (10^4 + \cdot 4 \cdot 3^4) = (10^2 + 2 \cdot 10 \cdot 3 + 2 \cdot 3^2)(10^2 - 2 \cdot 10 \cdot 3 + 2 \cdot 3^2) = (178)(58)\] \[(22^4 + 324) = (22^2 + 2 \cdot 3 \cdot 22 + 2 \cdot 3^2)(22^2 - 2 \cdot 3 \cdot 22 + 2 \cdot 3^2) = (634)(370)\] \[(4^4 + 324) = (4^2 + 2 \cdot 4 \cdot 3 + 2 \cdot 3^2)(4^2 - 2 \cdot 4 \cdot 3 + 2 \cdot 3^2) = (58)(10)\] \[(16^4 + 324) = (16^2 + 2 \cdot 16 \cdot 3 + 2 \cdot 3^2)(16^2 - 2 \cdot 16 \cdot 3 + 2 \cdot 3^2) = (370)(178)\] If we only had these terms, then the fraction would rewrite to $\frac{(58)(178)(370)(634)}{(10)(58)(178)(370)}$. However, we notice most of the terms cancel, leaving us only with the largest term in the numerator and the smallest term in the denominator ($\frac{634}{10}$). We hypothesize that this will happen with the fraction as a whole. Then we will only be left with the largest term in the numerator, which is $(58^2 + 2 \cdot 58 \cdot 3 + 2 \cdot 3^2) = 3730$. The fraction simplifies to $\frac{3730}{10} = \boxed{373}$.

~ cxsmi

Video Solution by OmegaLearn

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

~ pi_is_3.14

Video Solution

https://www.youtube.com/watch?v=yoOWcx2Otcw

~Michael Penn

See also

1987 AIME (Problems • Answer Key • Resources)
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