Difference between revisions of "2013 Mock AIME I Problems/Problem 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>. |
| + | |||
| + | ==See also== | ||
| + | * [[2013 Mock AIME I Problems]] | ||
| + | * [[2013 Mock AIME I Problems/Problem 2|Preceded by Problem 2]] | ||
| + | * [[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 
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 
.
