mmm. where have i been mistaken? i forgot to remove 18 numbers?

...

oh i know. i forgot to substract the 220,221,223-228, and the 1220,1221,1223. thats a total of 18 exactly, my bad.

now cls, since i see you're a programmer, care to try my problem?

also, what languages do you know?

A few.


a(n) = a(n-1) + gcd(n, n-1)
gcd(n, n-1) = 1
a(n) = a(n-1) + 1
a(n+1) - a(n) = (a(n) + 1) - a(n) = 1

It was you who invited me here btw, shos.

19 years old...'sthat you Marz?! this is so awsome XD
why cls, though?

edit: nvm got it XD
but you didn't give an answer to my riddle heh =] the fact that your program will make sure my theory is correct, is obvious since it is correct, but suppose you had a paper and pencil, and then you'd be given that question.

The program is the function you described (in tail-recursive Scheme). Below that is the proof, which was done without any code. I'm not very good at mathematics, but I seem good at recursive functions, at least.

oh boy, you are completely correct, i have written it wrong. it is the gcd of n and a(n-1) that is added, not the gcd of n and n-1. my bad.

2's? Just... count them, lol. In every 10, there's 1. In every 100, there's 9 of those 1's, plus 10 for every all the number's in the 20's = 19 total. Every thousand, there's 9 of those 19's, plus 100 for all the numbers in the 200's, = 271. We're dealing with two thousands, to * 2 = 542. Add 1 for all the numbers in the 2000's. +10 = 525.

It took me longer to read shos' post than to calculate this.

....but you were wrong =\

edit: nvm, you made a typo. :X

now think about what i posted. i still don't understand that one's solution >_>;

a(n) = a(n-1) + gcd(n, a(n-1)) ?
That's somewhat harder.

somewhat?! XD
that competition had 6 questions; 2 worth 4 points, 2 worth 6, and 2 worth 9. i solved 1 4, 2 6s, and 1 9, so i tried my hardest on this one which was a 9 pointer too. so i wasted the rest of the time i had on this one, which was around 2.5 hours, thinking. i didn't manage to solve it :X but then 2 weeks later they published the solutions, so...if you have time and are willing to face challanges, there ya go, and i have plenty more.

I have good news for all of you...
I just qualified for the National Mathematical Olympiad Special Round!

I did not know whether it was luck or what, but it is two weeks away...

Don't believe me? Check this  out. [url]http://www.catonamug.com/coppermine/displayimage.php?pid=55&fullsize=1[/url]

of dear, WHY do you have six windows of the Task Manager?!?!

and congrats, what does that mean, thou?

means that i am one of the top ten percent of all participants in singapore. And btw shos, my com is a laggy one,

oh.
and if your computer is laggy you only neeed one of those; opening more will make it lag more.

I added poll for you guys. Tough question. There may be more than one correct answer here...

i think livio=6.

there's something wrong with the poll. there's no option for 69

so you chose the starting '6'?...i'd say that 222+444=666. now stop it! this is a serious topic! answer my question or die!

I won a silver for the mo

& why can't I edit the poll?

congrats =] and cuz you're not a moderator.

uh, lately i've been into the international math olympiad. here's a question that i couldn't solve there:

Problem 3. Suppose that s(1), s(2), s(3), . . . is a strictly increasing sequence of positive integers such
that the subsequences
s(s(1)) , s(s(2)) , s(s(3)) , . . . and s(s(1)+1), s(s(2)+1), s(s(3)+1), . . .
are both arithmetic progressions. Prove that the sequence s1, s2, s3, . . . is itself an arithmetic progression.

but this is my own topic!

But this poll is awesome, so it cancels you out.

anyone wants some hard stuff?

lol, -karma for this?
I was serious, you know - I have some 6 IMO level questions, if you're interested.

also, dare you:
http://www.gamecabinet.com/rules/Rithmomachia.html