Here's to a prolific winter-break USAMO marathon
by shiningsunnyday, Dec 10, 2016, 10:52 AM
1997 USAMO P1 wrote:
Let 
be the prime numbers listed in increasing order, and let 
be a real number between 
and 
For positive integer 
, define
![\[ x_k = \begin{cases} 0 & \mbox{if} \; x_{k-1} = 0, \\[.1in] {\displaystyle \left\{ \frac{p_k}{x_{k-1}} \right\}} & \mbox{if} \; x_{k-1} \neq 0, \end{cases} \]](//latex.artofproblemsolving.com/e/1/7/e17b3fc5962fba938dee877dac266c3cd024a6e3.png)
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 
be the prime numbers listed in increasing order, and let 
be a real number between 
and 
For positive integer 
, define![\[ x_k = \begin{cases} 0 & \mbox{if} \; x_{k-1} = 0, \\[.1in] {\displaystyle \left\{ \frac{p_k}{x_{k-1}} \right\}} & \mbox{if} \; x_{k-1} \neq 0, \end{cases} \]](http://latex.artofproblemsolving.com/e/1/7/e17b3fc5962fba938dee877dac266c3cd024a6e3.png)
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 
Let 
where 
and 
Essentially the main observation is that the denominators 
monotonically decreases, since ![\[x_{k+1} = \left \{ \frac{p_{k+1} n_k}{m_k} \right \} = \frac{p_{k+1}n_k - \text{some integer}}{m_k}\]](http://latex.artofproblemsolving.com/e/0/6/e062b051c866af88279582d309827f438e77c169.png)
and 
Therefore, at some point we have 
and the following term ![\[x_{j+1} = \left \{ {p_{j+1} n_k} \right \} = 0,\]](http://latex.artofproblemsolving.com/d/5/b/d5bfc87b3f83cbade2078d2ddbf1689ac80335b3.png)
and the sequence becomes 
is irrational, no way on earth can 
or ![\[\frac{p_k}{x_{k-1}} - \left \lfloor \frac{p_k}{x_{k-1}} \right \rfloor\]](http://latex.artofproblemsolving.com/0/6/d/06d2287f57ad15c0d70f8acae83c896566a97b47.png)
be rational, and thus the sequence 
will remain irrational, finishing the problem.
at some point, 
is an integer, and thus by induction all the previous terms are rational as well.
May 2017
April 2017