Difference between revisions of "Mock AIME 4 2006-2007 Problems/Problem 14"

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==Solution==
 
==Solution==
  
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We will assume that there is at least one solution, otherwise the answer would be undefined.
  
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Using the binomial theorem it is obvious that <math>(577-k)^4 \equiv k^4 \pmod {577}</math>. Thus the solutions come in pairs <math>\{k,577-k\}</math>, and hence their average is <math>\dfrac{577}2 = 288.5</math>, and the answer is <math>\boxed{288}</math>.
  
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(In this case, there are four solutions: <math>276</math>, <math>277</math>, <math>300</math>, and <math>301</math>.)
  
 
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Latest revision as of 01:52, 31 January 2009

Problem

Let $x$ be the arithmetic mean of all positive integers $k<577$ such that

$k^4\equiv 144\pmod {577}$.

Find the greatest integer less than or equal to $x$.

Solution

We will assume that there is at least one solution, otherwise the answer would be undefined.

Using the binomial theorem it is obvious that $(577-k)^4 \equiv k^4 \pmod {577}$. Thus the solutions come in pairs $\{k,577-k\}$, and hence their average is $\dfrac{577}2 = 288.5$, and the answer is $\boxed{288}$.

(In this case, there are four solutions: $276$, $277$, $300$, and $301$.)