Difference between revisions of "2003 OIM Problems/Problem 6"
(Created page with "== Problem == Sequences <math>(a_n</math>)_{n \ge 0}, and <math>(b_n)_{n \ge 0}</math> with are defined by: <cmath>a_0=1 \text{, }b_0=4\text{, and}</cmath> <cmath>a_{n+1}=a_...") |
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== Problem == | == Problem == | ||
| − | Sequences <math>(a_n</math>)_{n \ge 0}, and <math>(b_n)_{n \ge 0} | + | Sequences <math>(a_n</math>)_{n \ge 0}<math>, and </math>(b_n)_{n \ge 0}$ with are defined by: |
<cmath>a_0=1 \text{, }b_0=4\text{, and}</cmath> | <cmath>a_0=1 \text{, }b_0=4\text{, and}</cmath> | ||
Revision as of 03:34, 14 December 2023
Problem
Sequences 
)_{n \ge 0}
(b_n)_{n \ge 0}$ with are defined by:
![\[a_0=1 \text{, }b_0=4\text{, and}\]](http://latex.artofproblemsolving.com/c/6/e/c6e6af268ab953f1fbc3de325f74019c3d5317fa.png)
![\[a_{n+1}=a_n^{2001}+b_n\text{, for }n \ge 0\text{.}\]](http://latex.artofproblemsolving.com/3/5/e/35e31f66225c73b04cc439fb8e2f77f0ae18fbc3.png)
![\[b_{n+1}=b_n^{2001}+a_n\text{, for }n \ge 0\text{.}\]](http://latex.artofproblemsolving.com/b/6/2/b62307a6806d57e18e7a905ac1f1e2b126621e88.png)
Show that 2003 does not divide any of the terms of these sequences.
~translated into English by Tomas Diaz. ~orders@tomasdiaz.com
Solution
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