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

m (see also box)
m (category)
Line 10: Line 10:
 
* [[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]]

Revision as of 08:54, 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