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Difference between revisions of "1979 AHSME Problems/Problem 19"

(Created page with "== Problem 19 == Find the sum of the squares of all real numbers satisfying the equation <math>x^{256}-256^{32}=0</math>. <math>\textbf{(A) }8\qquad \textbf{(B) }128\qquad...")
 
 
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Solution by e_power_pi_times_i
 
Solution by e_power_pi_times_i
  
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Notice that the solutions to the equation <math>x^{256}-1=0</math> are the <math>256</math> roots of unity. Then the solutions to the equation <math>x^{256}-256^{32}=0</math> are the <math>256</math> roots of unity dilated by <math>\sqrt[256]{256^{32}} = \sqrt[256]{2^{256}} = 2</math>. However, the only real solutions to the equation are the first root of unity and the root of unity opposite of it, as both are on the real axis in the complex plane. These two roots of unity are <math>\pm1</math>, and dilating by <math>2</math> gives <math>\pm2</math>. The sum of the squares is <math>(2)^2+(-2)^2 = \boxed{\textbf{(A) } 8}</math>.
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Notice that the solutions to the equation <math>x^{256}-1=0</math> are the <math>256</math> roots of unity. Then the solutions to the equation <math>x^{256}-256^{32}=0</math> are the <math>256</math> roots of unity dilated by <math>\sqrt[256]{256^{32}} = \sqrt[256]{2^{256}} = 2</math>. However, the only real solutions to the equation are the first root of unity and the root of unity opposite of it, as both are on the real axis in the complex plane. These two roots of unity are <math>\pm1</math>, and dilating by <math>2</math> gives <math>\pm2</math>. The sum of the squares is <math>(2)^2+(-2)^2 = \boxed{\textbf{(A) } 8}</math>
 +
 
 +
==Solution 2==
 +
Notice that we can change the equation to be <math>x^{256}=256^{32}</math>.
 +
 
 +
The RHS can be simplified to <math>x^{256}=2^{256}</math>
 +
 
 +
Hence, the only real solutions are <math>x=-2,2</math>
 +
 
 +
Hence, <math>(-2)^2+(2)^2=8=\boxed{A}</math>.
  
 
== See also ==
 
== See also ==

Latest revision as of 15:09, 8 June 2017

Problem 19

Find the sum of the squares of all real numbers satisfying the equation $x^{256}-256^{32}=0$.

$\textbf{(A) }8\qquad \textbf{(B) }128\qquad \textbf{(C) }512\qquad \textbf{(D) }65,536\qquad \textbf{(E) }2(256^{32})$

Solution

Solution by e_power_pi_times_i

Notice that the solutions to the equation $x^{256}-1=0$ are the $256$ roots of unity. Then the solutions to the equation $x^{256}-256^{32}=0$ are the $256$ roots of unity dilated by $\sqrt[256]{256^{32}} = \sqrt[256]{2^{256}} = 2$. However, the only real solutions to the equation are the first root of unity and the root of unity opposite of it, as both are on the real axis in the complex plane. These two roots of unity are $\pm1$, and dilating by $2$ gives $\pm2$. The sum of the squares is $(2)^2+(-2)^2 = \boxed{\textbf{(A) } 8}$

Solution 2

Notice that we can change the equation to be $x^{256}=256^{32}$.

The RHS can be simplified to $x^{256}=2^{256}$

Hence, the only real solutions are $x=-2,2$

Hence, $(-2)^2+(2)^2=8=\boxed{A}$.

See also

1979 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 18 		Followed by
Problem 20
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All AHSME Problems and Solutions

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