Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

m (see also box)
m (added link to all problems)
 
(One intermediate revision by the same user not shown)
Line 8: Line 8:
  
 
==See also==
 
==See also==
 +
* [[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]]

Latest revision as of 08:01, 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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2013_Mock_AIME_I_Problems/Problem_3&oldid=223958"