let ${a_{n}}$ be a sequence of integers,$a_{1}$ is odd,and for any positive integer $n$,we have $n(a_{n+1}-a_{n}+3)=a_{n+1}+a_{n}+3$,in addition,we have $2010$ divides $a_{2009}$ find the smallest $n\ge\ 2$,so that $2010$ divides $a_{n}$
Problem
Source: 2009 HongKong mathematic Olympiad
Tags: algebra proposed, algebra
tenniskidperson3
11.02.2012 20:32
Strange... Letting $n=1$ gives $a_2-a_1+3=a_2+a_1+3$ or $a_1=0$ which isn't odd...
sayantanchakraborty
05.04.2014 21:44
Ya....I think the problem is wrong somewhere.....Please see too it.....
nawaites
13.07.2014 10:57
Anyone can fix it?
Tan
05.05.2015 04:18
Sorry for the bump....this might trivialize the problem, but perhaps it was: let ${a_{n}}$ be a sequence of integers,$a_{1}$ is odd,and for any positive integer $n$,we have $n(a_{n+1}-a_{n}+3)=a_{n+1}-a_{n}+3$,in addition,we have $2010$ divides $a_{2009}$ find the smallest $n\ge\ 2$,so that $2010$ divides $a_{n}$. If so, here's my solution: Let $b_n=a_{n+1}-a_n+3$. We have $(n-1)b_n=0$, so $a_{n+1}=a_n-3$. Since $2010 | a_{2009}$, we can get the smallest value is $n=671$. (i think?)