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

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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Mock AIME 1 2006-2007

@@ Line 1: / Line 1: @@
-. 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:
@@ Line 7: / Line 7: @@
 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}}
+----
+*[[Mock AIME 1 2006-2007/Problem 4 | Previous Problem]]
+*[[Mock AIME 1 2006-2007/Problem 6 | Next Problem]]
+*[[Mock AIME 1 2006-2007]]

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

Revision as of 16:27, 17 August 2006

Solution