Difference between revisions of "2013 Mock AIME I Problems/Problem 3"

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Problem
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== Problem ==
  
 
Let <math>\lfloor x\rfloor</math> be the greatest integer less than or equal to <math>x</math>, and let <math>\{x\}=x-\lfloor x\rfloor</math>. If <math>x=(7+4\sqrt{3})^{2^{2013}}</math>, compute <math>x\left(1-\{x\}\right)</math>.
 
Let <math>\lfloor x\rfloor</math> be the greatest integer less than or equal to <math>x</math>, and let <math>\{x\}=x-\lfloor x\rfloor</math>. If <math>x=(7+4\sqrt{3})^{2^{2013}}</math>, compute <math>x\left(1-\{x\}\right)</math>.
  
Solution
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== Solution ==
  
Notice that the radical conjugate of x is a positive integer from 0-1. Since the sum of powers of two radical conjugates is an integer, <math>1-{x}</math> is just the conjugate of x to the <math>2^{2013}</math> power. Therefore, the desired expression is just 001.
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Let <math>y=(7-4\sqrt{3})^{2^{2013}}</math>. Notice that <math>y<<1</math> and that, by expanding using the binomial theorem, <math>x+y</math> is an integer because the terms with radicals cancel. Thus, <math>y=1-\{x\}</math>. The desired expression is <math>x\left(1-\{x\}\right)=xy=((7+4\sqrt{3})(7-4\sqrt{3}))^{2^{2013}}=\boxed{1}</math>.

Latest revision as of 00:14, 5 March 2017

Problem

Let $\lfloor x\rfloor$ be the greatest integer less than or equal to $x$, and let $\{x\}=x-\lfloor x\rfloor$. If $x=(7+4\sqrt{3})^{2^{2013}}$, compute $x\left(1-\{x\}\right)$.

Solution

Let $y=(7-4\sqrt{3})^{2^{2013}}$. Notice that $y<<1$ and that, by expanding using the binomial theorem, $x+y$ is an integer because the terms with radicals cancel. Thus, $y=1-\{x\}$. The desired expression is $x\left(1-\{x\}\right)=xy=((7+4\sqrt{3})(7-4\sqrt{3}))^{2^{2013}}=\boxed{1}$.

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