2013 Mock AIME I Problems/Problem 3

Revision as of 00:14, 5 March 2017 by Rocketscience (talk | contribs)
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

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