This is the problems on the 46th International Mathematical Olympiad.

My friends got excellent result from it, some of them are still very young, the youngest is 2 years smaller than me. That could depress me much. [as I even could not get the qualification to join the competition]

http://gifted.hkedcity.net/Gifted/ActReview/imo2005Mexico/pdf/1dayenglish.pdf

http://gifted.hkedcity.net/Gifted/ActReview/imo2005Mexico/pdf/2dayenglish.pdf

I know I'm screwing up somewhere, but Q2 (day 1) looks trivial to me. Anyone else looked at this ?

Personally I'd say it looks the most managable (for me, at least). Don't know about trivial, but I've not doodled with it yet so I couldn't really say. I hate these kind of questions though - I just don't have the brain for them.

Okay, I must be making some really silly mistake here as it looks to me like a one-line proof.

 

Let me lay my head on the guillotine and actually write this down :

 

Assume [imath]a_p = a_q [/imath] for some q>p, then [imath]a_p \equiv a_q~ (mod~ q) [/imath]; but this is not possible since all of the first q terms leave different remainders with q. Hence the assumption was wrong. QED.

 

Feel free to let the blade drop...it won't hurt my feelings !

That certainly shows that any integer occuring in the sequence occurs at most once, but doesn't show that every integer is in the sequence.

Thanks matt. I misread the last line. This makes it non-trivial.