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Mock AIME 1 2006-2007 Problems/Problem 5: Difference between revisions

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5. Let <math>p</math> be a prime and <math>f(n)</math> satisfy <math>0\le f(n) <p</math> for all integers <math>n</math>. <math>\lfloor x\rfloor</math> is the greatest integer less than or equal to <math>x</math>. If for fixed <math>n</math>, there exists an integer <math>0\le y < p</math> such that:
Let <math>p</math> be a prime and <math>f(n)</math> satisfy <math>0\le f(n) <p</math> for all integers <math>n</math>. <math>\lfloor x\rfloor</math> is the greatest integer less than or equal to <math>x</math>. If for fixed <math>n</math>, there exists an integer <math>0\le y < p</math> such that:




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then <math>f(n)=y</math>. If there is no such <math>y</math>, then <math>f(n)=0</math>. If <math>p=11</math>, find the sum: <math>f(1)+f(2)+...+f(p^{2}-1)+f(p^{2})</math>.
then <math>f(n)=y</math>. If there is no such <math>y</math>, then <math>f(n)=0</math>. If <math>p=11</math>, find the sum: <math>f(1)+f(2)+...+f(p^{2}-1)+f(p^{2})</math>.


[[Mock AIME 1 2006-2007]]
==Solution==
{{solution}}
 
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*[[Mock AIME 1 2006-2007/Problem 4 | Previous Problem]]
 
*[[Mock AIME 1 2006-2007/Problem 6 | Next Problem]]
 
*[[Mock AIME 1 2006-2007]]

Revision as of 16:27, 17 August 2006

Let $p$ be a prime and $f(n)$ satisfy $0\le f(n) <p$ for all integers $n$. $\lfloor x\rfloor$ is the greatest integer less than or equal to $x$. If for fixed $n$, there exists an integer $0\le y < p$ such that:


$ny-p\left\lfloor \frac{ny}{p}\right\rfloor=1$


then $f(n)=y$. If there is no such $y$, then $f(n)=0$. If $p=11$, find the sum: $f(1)+f(2)+...+f(p^{2}-1)+f(p^{2})$.

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