consider the sequence mod 2004. it is equivalant to the following recurrence: x_0=1, x_1=2^1, x_2=2^2 ... x_2003=2^2003, x_n=x_n-1 ... + x_n-2004. so the recurrence is periodic. so the terms {1, 1, 2^1 ... 2^2003} occur again in that order somewhere on in the sequence. we can see that the 2004 terms before that are 1, followed by 2003 '0's. so we can have 2003 consecutive terms all divisible by 2004. We cannot have 2004 such terms, because if then all ensuing terms are divisible by 2004 - which then contradicts the periodicity of the recurrence mod 2004. And we are done.

vinoth_90_2004 wrote: we can see that the 2004 terms before that are 1, followed by 2003 '0's. Sorry, I think you've meant: We can see that the first term before that is 1, and the 2003 terms before this 1 are '0's. Darij

i think i have seen that problem before in the IMO 2003 short listed