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The flooble question (Posted on 2003-02-18) Difficulty: 4 of 5
At flooble there are 40 problems in the queue. (this may not be true but lets just pretend it is.) A few crazy hackers somehow manage to promote themselves to scholars. On the first day the first hacker will vote thumbs up on all problems displayed.(The 10 most recent) On the second day the second hacker votes thumbs down on every second problem. On the third day the third hacker votes thumbs up on every third problem. And so on and so on. (When it gets to the eleventh day the eleventh hacker will do what the first hacker did)

How many days will it take for every problem in the queue to be live on the site?

Note: For those who don't know there are only 10 problems that can be voted thumbs up or thumbs down every day and these problems are the 10 least recent. Also a problem with three thumbs up will be posted to the site and taken out of queue. Only one problem can be posted to the site per day. Also if a problem gets 3 thumbs down it is deleted.

Btw: for those who like an extra challenge what if one problem is submitted every 3 days?

Also: A hacker will always vote before a problem becomes live.

See The Solution Submitted by Alan    
Rating: 4.0000 (10 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
re: crazy (for lack of a better subject) | Comment 19 of 52 |
(In reply to crazy (for lack of a better subject) by Cory Taylor)

The way I read the problem, puzzle 1 gets voted on on day 1 and then has to wait until day 10 to be voted on again, as only one hacker votes on a given day, and day 2 through day 10 hackers bypass 1, 2, 3, etc. potential votees in skipping over to every second, third, fourth, etc. problem. So at the end of day 5, the first 10 puzzles are still on the voting table with the following results:
10 11 20 12 20 21 10 12 20 21, indicating their positive votes in the first digit of each followed by their negative votes. I haven't tried assuming that "every second" means every odd rather than every even-numbered problem in the voting portion of the queue. Likewise, I have assumed "every third" means starting with number 3, then 6, then 9, rather than 1, 4, 7.
  Posted by Charlie on 2003-02-18 06:46:59

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