1999 Bulgaria National Olympiad #2

-

Let $\{a_n\}$ be a sequence of integers satisfying $(n-1)a_{n+1}=(n+1)a_n-2(n-1) \forall n\ge 1$. If $2000|a_{1999}$, find the smallest $n\ge 2$ such that $2000|a_n$.




The answer is $\boxed{249}$. 

Define $a_n=b_n+2(n-1) $ . Using the given condition we have $$n(b_{n+1}-b_{n})=b_n+b_{n+1} \implies$$ $$\frac{b_{n+1}}{b_n}=\frac{n+1}{n-1}.$$Plugging in $n=2,3,.....,n$ and multiplying the terms results in a telescoping product from which we obtain $$b_n = \frac{n(n-1)}{2} \cdot b_2.$$ Hence, we plug into our original expression for $a_n$ to get $$a_n = \frac{n(n-1)}{2}b_2+2(n-1).$$It is given that $a_{2009} \equiv 0 \pmod{2000}$, so we know that $b_2 \equiv 4 \pmod{2000}$. Now, $b_2 \pmod{2000}$ combined with our above expression for $a_n$ gives us that the least possible $n$ such that $2000|a_n$ is $249$, as desired. $\square$

 

Comments

Post a Comment

Popular posts from this blog

1995 IMO #2

-
 Let $ a$, $ b$, $ c$ be positive real numbers such that $ abc = 1$. Prove that \[ \frac {1}{a^{3}\left(b + c\right)} + \frac {1}{b^{3}\left(c + a\right)} + \frac {1}{c^{3}\left(a + b\right)}\geq \frac {3}{2}. \]
Read more

2014 IMO SL #C2

-
  Problem : We have $2^m$ sheets of paper, with the number $1$ written on each of them. We perform the following operation. In every step we choose two distinct sheets; if the numbers on the two sheets are $a$ and $b$, then we erase these numbers and write the number $a + b$ on both sheets. Prove that after $m2^{m -1}$ steps, the sum of the numbers on all the sheets is at least $4^m$ .
Read more

2015 IMO SL #A1

-
 Suppose that a sequence $a_1,a_2,\ldots$ of positive real numbers satisfies\[a_{k+1}\geq\frac{ka_k}{a_k^2+(k-1)}\]for every positive integer $k$. Prove that $a_1+a_2+\ldots+a_n\geq n$ for every $n\geq2$.
Read more