Difference between revisions of "1997 USAMO Problems/Problem 1"

Latest revision as of 12:14, 13 February 2015

Problem

Let $p_1,p_2,p_3,...$ be the prime numbers listed in increasing order, and let $x_0$ be a real number between $0$ and $1$ . For positive integer $k$ , define

$x_{k}=\begin{cases}0&\text{ if }x_{k-1}=0\\ \left\{\frac{p_{k}}{x_{k-1}}\right\}&\text{ if }x_{k-1}\ne0\end{cases}$

where $\{x\}$ denotes the fractional part of $x$ . (The fractional part of $x$ is given by $x-\lfloor{x}\rfloor$ where $\lfloor{x}\rfloor$ is the greatest integer less than or equal to $x$ .) Find, with proof, all $x_0$ satisfying $0<x_0<1$ for which the sequence $x_0,x_1,x_2,...$ eventually becomes $0$ .

Solution

All rational numbers between 0 and 1 inclusive will eventually yield some $x_k = 0$ . To begin, note that by definition, all rational numbers can be written as a quotient of coprime integers. Let $x_0 = \frac{m}{n}$ , where $m,n$ are coprime positive integers. Since $0<x_0<1$ , $0<m<n$ . Now $x_1 = \left\{\frac{p_1}{\frac{m}{n}}\right\}=\left\{\frac{np_1}{m}\right\}.$ From this, we can see that applying the iterative process will decrease the value of the denominator, since $m<n$ . Moreover, the numerator is always smaller than the denominator, thanks to the fractional part operator. So we have a strictly decreasing denominator that bounds the numerator. Thus, the numerator will eventually become 0.

On the other hand, if $x_0$ is irrational, then a simple induction will show that $x_k$ will always be irrational. Indeed, the base case has been established, and, if $x_k$ is irrational, then $\dfrac{p_{k+1}}{x_k}$ must be too, and likewise for its fractional portion, which differs from it by an integer. Hence $x_{k+1}$ is irrational, completing the proof.

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 == Solution ==
-''This problem needs a solution. Help out by adding one.''
+All rational numbers between 0 and 1 inclusive will eventually yield some <math>x_k = 0</math>. To begin, note that by definition, all rational numbers can be written as a quotient of coprime integers. Let <math>x_0 = \frac{m}{n}</math>, where <math>m,n</math> are coprime positive integers. Since <math>0<x_0<1</math>, <math>0<m<n</math>. Now <cmath>x_1 = \left\{\frac{p_1}{\frac{m}{n}}\right\}=\left\{\frac{np_1}{m}\right\}.</cmath> From this, we can see that applying the iterative process will decrease the value of the denominator, since <math>m<n</math>. Moreover, the numerator is always smaller than the denominator, thanks to the fractional part operator. So we have a strictly decreasing denominator that bounds the numerator. Thus, the numerator will eventually become 0.
+On the other hand, if <math>x_0</math> is irrational, then a simple induction will show that <math>x_k</math> will always be irrational. Indeed, the base case has been established, and, if <math>x_k</math> is irrational, then <math>\dfrac{p_{k+1}}{x_k}</math> must be too, and likewise for its fractional portion, which differs from it by an integer. Hence <math>x_{k+1}</math> is irrational, completing the proof.
+== See Also ==
+{{USAMO newbox|year=1997|before=First Question|num-a=2}}
+[[Category:Olympiad Number Theory Problems]]
+{{MAA Notice}}

1997 USAMO (Problems • Resources)
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Difference between revisions of "1997 USAMO Problems/Problem 1"

Latest revision as of 12:14, 13 February 2015

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