In an increasing infinite sequence of positive integers, every term starting from the $2002$-th term divides the sum of all preceding terms. Prove that every term starting from some term is equal to the sum of all preceding terms.
Problem
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Tags: AMC, USA(J)MO, USAMO, Recursive Sequences
not_trig
17.03.2008 02:43
This seems suspiciously similar to USAMO 2007 #1. I think a similar argument works.
Peter
17.03.2008 02:47
Well, if it can be solved similarly, why don't you post a solution?
Rust
17.03.2008 08:00
Let $ x_n$ is suth sequence. Then \[ x_n|\sum_{i=1}^{n-1}x_i\] for big n, or \[ \sum_{i=1}^nx_i=y_ny_n\] for big n ($ y_n$ is natural). It give \[ \sum_{i=1}^nx_n=y_nx_n+x_n=y_{n+1}x_{n+1}\] for all big n. Because $ x_{n+1}>x_n$ we get $ y_{n+1}<y_n+1$ or $ y_{n+1}\le y_n$. Because $ y_n$ is natural, exist N, suth that $ y_n=a \ \forall n>N$ and $ (a+1)x_n=ax_{n+1} \ \forall n>N$. If $ a>1$, let $ z_n=ord_a(x_n),n>N \to z_{n+1}=z_n-1$ give contradition. Therefore $ a=1,x_{n+1}=2x_{n} \ \forall n>N$. It give $ x_n=2^{n-N}d, \ d=\sum_{i=1}^{N+1}x_n \ \forall n>N$.