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

(Created page with "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}...") |
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− | Problem | + | == 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 | + | == Solution == |

− | Notice that the | + | 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 be the greatest integer less than or equal to , and let . If , compute .

## Solution

Let . Notice that and that, by expanding using the binomial theorem, is an integer because the terms with radicals cancel. Thus, . The desired expression is .