Original Post
Solve a problem, win over 1k(1000.01 tc)!
The following iterative sequence is defined for the set of positive integers:
n → n/2 (n is even)
n → 3n + 1 (n is odd)
Using the rule above and starting with 13, we generate the following sequence:
13 → 40 → 20 → 10 → 5 → 16 → 8 → 4 → 2 → 1
It can be seen that this sequence (starting at 13 and finishing at 1) contains 10 terms. Although it has not been proved yet (Collatz Problem), it is thought that all starting numbers finish at 1.
Which starting number, under two million, produces the longest chain?
NOTE: Once the chain starts the terms are allowed to go above two million.
First person to post the right answer gets 1000.01 tc
When you post the answer post how you got it, or you suck. And I'll post my c++ code. Btw thanks to wiirus for sending me tc to fund these things.
Previous thread of this was in wibbles and I'm pretty sure no one had posted the correct answer before wibbles was deleted. But post again, if you did.
edit: Man, the event sub forum sucks, hopefully people still participate.
edit 2: So this guy, he pms me and he's trying to convey some kind of information to me. I honestly can barely make out what he was saying but I guess it's along the lines that he's won or something. Then, to top it off, he calls me stupid, with his bad grammar and all.
So look here, boredlolz, 1. It's obvious you have to use an integer and 2. Learn some fucking English.
It says right up top to use an integer. Are you stupid or something?
Also, you used to be able to send decimal numbers, if they changed it then that's on them. http://img202.imageshack.us/img202/9201/scr5806513.jpg
EDIT: This will end on 5-13-11 at midnight -8 GMT. The winner will be either; the first person that posted the exact answer, or, if no one got the exact answer the winner will be the person that posted the starting number under 2 million that makes the longest chain posted.
EVENT DONE, THANKS FOR PARTICIPATING
n → n/2 (n is even)
n → 3n + 1 (n is odd)
Using the rule above and starting with 13, we generate the following sequence:
13 → 40 → 20 → 10 → 5 → 16 → 8 → 4 → 2 → 1
It can be seen that this sequence (starting at 13 and finishing at 1) contains 10 terms. Although it has not been proved yet (Collatz Problem), it is thought that all starting numbers finish at 1.
Which starting number, under two million, produces the longest chain?
NOTE: Once the chain starts the terms are allowed to go above two million.
First person to post the right answer gets 1000.01 tc
When you post the answer post how you got it, or you suck. And I'll post my c++ code. Btw thanks to wiirus for sending me tc to fund these things.
Previous thread of this was in wibbles and I'm pretty sure no one had posted the correct answer before wibbles was deleted. But post again, if you did.
edit: Man, the event sub forum sucks, hopefully people still participate.
edit 2: So this guy, he pms me and he's trying to convey some kind of information to me. I honestly can barely make out what he was saying but I guess it's along the lines that he's won or something. Then, to top it off, he calls me stupid, with his bad grammar and all.
Originally Posted by boredlolz
1999999.9999999999999999999999999999999999 (the nines continue forever) would because well... it's
1, not over 2000000, and
2, do you know how hard a fraction is to get it to one? sorry, i just exploded your theory pretty much, another solution is pi, since nobody knows how long it actually goes for, since it NEVER ENDING.
also, you cant give .01 of a tc, you stupid or something?
So look here, boredlolz, 1. It's obvious you have to use an integer and 2. Learn some fucking English.
It says right up top to use an integer. Are you stupid or something?
Also, you used to be able to send decimal numbers, if they changed it then that's on them. http://img202.imageshack.us/img202/9201/scr5806513.jpg
EDIT: This will end on 5-13-11 at midnight -8 GMT. The winner will be either; the first person that posted the exact answer, or, if no one got the exact answer the winner will be the person that posted the starting number under 2 million that makes the longest chain posted.
EVENT DONE, THANKS FOR PARTICIPATING
Last edited by isaac; May 11, 2011 at 11:24 PM.
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