Difference between revisions of "Mock AIME 3 2006-2007 Problems/Problem 4"

(Created page with "== Problem == Let <math>x=(i-11)(i-10)....(i+8)(i+9)</math> where <math>i=\sqrt{-1}</math>. If <math>a</math> and <math>b</math> are positive integers such that <math>x(a-bi)</ma...")
 
(Solution)
 
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== Solution ==
 
== Solution ==
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Pairing the terms <math>(i-k)</math> with <math>(i+k)</math> for <math>k=1,2,...9</math> all result in integers, so we're left with making <math>i(i-10)(i-11)</math> an integer.
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<math>i(i-10)(i-11)=(21+109i)</math>. To make this an integer, we just multiply it by its conjugate <math>(21-109i)</math>, so <math>a+b=21+109=\boxed{130}</math>

Latest revision as of 23:35, 15 February 2015

Problem

Let $x=(i-11)(i-10)....(i+8)(i+9)$ where $i=\sqrt{-1}$. If $a$ and $b$ are positive integers such that $x(a-bi)$ is an integer, then find the minimum value of $a+b$


Solution

Pairing the terms $(i-k)$ with $(i+k)$ for $k=1,2,...9$ all result in integers, so we're left with making $i(i-10)(i-11)$ an integer.

$i(i-10)(i-11)=(21+109i)$. To make this an integer, we just multiply it by its conjugate $(21-109i)$, so $a+b=21+109=\boxed{130}$