Difference between revisions of "1997 USAMO Problems/Problem 1"
(added solution) |
|||
Line 7: | Line 7: | ||
== Solution == | == Solution == | ||
− | {{ | + | |
+ | All rational numbers between 0 and 1 will eventually become 0 under this iterative process. 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, irrational <math>x_0</math> will never become 0, because <math>x_k</math> will always be irrational. | ||
+ | |||
== See Also == | == See Also == | ||
{{USAMO newbox|year=1997|before=First Question|num-a=2}} | {{USAMO newbox|year=1997|before=First Question|num-a=2}} |
Revision as of 05:12, 3 August 2013
Problem
Let be the prime numbers listed in increasing order, and let be a real number between and . For positive integer , define
where denotes the fractional part of . (The fractional part of is given by where is the greatest integer less than or equal to .) Find, with proof, all satisfying for which the sequence eventually becomes .
Solution
All rational numbers between 0 and 1 will eventually become 0 under this iterative process. To begin, note that by definition, all rational numbers can be written as a quotient of coprime integers. Let , where are coprime positive integers. Since , . Now From this, we can see that applying the iterative process will decrease the value of the denominator, since . 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, irrational will never become 0, because will always be irrational.
See Also
1997 USAMO (Problems • Resources) | ||
Preceded by First Question |
Followed by Problem 2 | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAMO Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.