Difference between revisions of "1998 OIM Problems/Problem 6"

 
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Find the remainder of the division of <math>x_{1998}</math> by <math>1998</math>.
 
Find the remainder of the division of <math>x_{1998}</math> by <math>1998</math>.
  
Note: The brackets indicate an integer part, that is, <math>\left\lfloor x \right\rfloor</math> is the only integer <math>k</math> such that <math>k le x < k+1</math>.  
+
Note: The brackets indicate an integer part, that is, <math>\left\lfloor x \right\rfloor</math> is the only integer <math>k</math> such that <math>k \le x < k+1</math>.  
  
 
~translated into English by Tomas Diaz. ~orders@tomasdiaz.com
 
~translated into English by Tomas Diaz. ~orders@tomasdiaz.com

Latest revision as of 14:57, 13 December 2023

Problem

Let $\lambda$ be the positive root of the equation $t^2 - 1998t - 1 = 0$. The sequence $x_0, x_1, x_2, \cdots , x_n, \cdots$ is defined by:

\[\begin{cases} x_0=1 \\ x_{n+1}=\left\lfloor \lambda x_n \right\rfloor\ , & \text{for }n=0,1,2,\cdots\end{cases}\]

Find the remainder of the division of $x_{1998}$ by $1998$.

Note: The brackets indicate an integer part, that is, $\left\lfloor x \right\rfloor$ is the only integer $k$ such that $k \le x < k+1$.

~translated into English by Tomas Diaz. ~orders@tomasdiaz.com

Solution

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See also

https://www.oma.org.ar/enunciados/ibe13.htm