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

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==See also==
 
==See also==
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* [[2013 Mock AIME I Problems]]
 
* [[2013 Mock AIME I Problems/Problem 2|Preceded by Problem 2]]
 
* [[2013 Mock AIME I Problems/Problem 2|Preceded by Problem 2]]
 
* [[2013 Mock AIME I Problems/Problem 4|Followed by Problem 4]]
 
* [[2013 Mock AIME I Problems/Problem 4|Followed by Problem 4]]
 
[[Category:Intermediate Algebra Problems]]
 
[[Category:Intermediate Algebra Problems]]

Latest revision as of 09:01, 30 July 2024

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 $x\left(1-\{x\}\right)=xy=((7+4\sqrt{3})(7-4\sqrt{3}))^{2^{2013}}=\boxed{1}$.

See also

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