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Difference between revisions of "2006 USAMO Problems/Problem 1"
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Thus, choose <math>m=j+1</math> and <math>n=x</math> to complete the proof.
 
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== Resources ==
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== See also ==
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* [[2006 USAMO Problems]]
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* [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=490569#p490569 Discussion on AoPS/MathLinks]
 
  
 
[[Category:Olympiad Number Theory Problems]]
 
[[Category:Olympiad Number Theory Problems]]
 
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Revision as of 20:49, 27 July 2014

Problem

Let $p$ be a prime number and let $s$ be an integer with $0 < s < p$. Prove that there exist integers $m$ and $n$ with $0 < m < n < p$ and

$\left\{ \frac{sm}{p} \right\} < \left\{ \frac{sn}{p} \right\} < \frac{s}{p}$

if and only if $s$ is not a divisor of $p-1$.

Note: For $x$ a real number, let $\lfloor x \rfloor$ denote the greatest integer less than or equal to $x$, and let $\{x\} = x - \lfloor x \rfloor$ denote the fractional part of $x$.

Solution

We proceed by contradiction. Assume that $s|(p-1)$. Then for some positive integer $k$, $sk=p-1$. The conditions given are equivalent to stating that $sm \mod p < sn \mod p< s\mod p$. Now consider the following array modulo p:

$\mbox{row 1}\qquad s\qquad 2s\qquad \hdots\qquad ks\\ \mbox{row 2}\qquad s-1\qquad 2s-1\qquad \hdots\qquad ks-1\\ \vdots\\ \mbox{row s}\qquad 1\qquad s+1\qquad \hdots\qquad (k-1)s+1$

Obviously, there are $s$ rows and $k$ columns. The first entry of each row is simply $((r-1)k+1)s\mod p$. Since we wish for $sm\mod p$ and $sn\mod p$ to both be in the first column, while also satisfying the given conditions, we can easily see that $sm$ must be in a row $m_r$ with $m_r>n_r$, where $n_r$ denotes the row $sn \mod p$ is in. However, since the values of each entry decreases while $((r-1)k+1)s$ keeps increasing, we can see that the condition can never be satisfied and thus, $s\not|(p-1).$


To prove the other direction, let $sj+r=p-1$ for positive integers $j$ and $r$ with $j$ being the largest integer such that $sj<p-1$ and $r<s$. Since $s\not|(p-1)$, $1<s<p-1$.

Note that for $0<l<j$, $ls\ge s\mod p$. Thus, the first integer multiplied by s modulo p that will be less than s is $j+1.$ $(j+1)s\equiv s-r-1\mod p$. Since $\lbrace s,2s,\hdots,js,(j+1)s,\hdots,xs,\hdots,(p-1)s \rbrace$ is a complete residue system mod p with the exception of the 0 term, we can find an $x>j+1$ such that $xs \equiv s-1 \mod p$.

Thus, choose $m=j+1$ and $n=x$ to complete the proof.

See also

  • <url>viewtopic.php?t=84548 Discussion on AoPS/MathLinks</url>
2006 USAMO (Problems • Resources)
Preceded by
First Question 		Followed by
Problem 2
1 • 2 • 3 • 4 • 5 • 6
All USAMO Problems and Solutions

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