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

Revision as of 14:55, 13 December 2023

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  such that .

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

This problem needs a solution. If you have a solution for it, please help us out by adding it.

@@ Line 4: / Line 4: @@
 <cmath>\begin{cases} x_0=1 \\ x_{n+1}=\left\lfloor \lambda x_n \right\rfloor\ , & \text{for }n=0,1,2,\cdots\end{cases}</cmath>
-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\$ is the only integer </math>k<math> such that </math>k le x < k+1$.
+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