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romangod5
New Member

USA
1 Posts

 Posted - 08/14/2008 :  22:18:49 ok. i have been working on this problem for about two months. but i can't figure it out.There are two buckets one on the left and one on the right and a machine in the middle that transfers the balls at lightning speed. Each bucket can hold infinite amount of balls. The bucket on the left has infinite amount of balls and each ball has a number on it(1,2,3,4,...) and so on. The machine comes on at 12 and stops forever at 1 o'clock. It comes on at twelve o'clock and transfers balls 1 and 2 to the bucket on the right then takes ball number 1 back to the bucket on the left. the next time it comes on is at 12:30 and it takes balls 3,4, and 5 to the bucket on the right and takes ball number two to the bucket on the left. so u have balls 3,4, and 5 in the bucket on the right. the next time it came on was 12:45 and it took balls 6,7,8, and 9 to the bucket on the right and put ball number 3 in the left. So it keeps taking an extra ball and takes the next number ball in the right bucket to the left. The machine keeps coming on in half the amount of time added last time.12 to 12:30 then 30 by 2=15 so 12:45 then 15 by 2=7.5 so 12:52.5 and so on till it gets to 1:00. In the end all the balls end up in the bucket on the left. WHY?!?!?!?!?!?!?!?

pka

USA
2731 Posts

 Posted - 08/15/2008 :  19:30:58 romangod5, Surely if you are capable of working on this problem, then you would have studied Zeno Paradox. If not, do a web search for that classical paradox. Essentially it says that because an arrow in flight covers only half the distance in each of successive countable time intervals can never reach its target.This problem is just a slight variation of that famous paradox.

the_hill1962

USA
1438 Posts

pka

USA
2731 Posts

 Posted - 08/20/2008 :  11:58:25 I did not reply to a MP from the poster after he said that he was only in first year algebra. However, because there seems to be some interest I will address this.We can think of the machine acting at [60-(60/2K)] intervals is seconds between 12 and 1. Note that after the first action at 12:30 ball 1 remains in the first box. After 12:45 balls 1 & 2 remain in the first box. After Kth action balls 1, 2, …,K remain in the first box. As noted above the term (60/2K) approaches 0 as K increases but in fact never is 0. But note that for any N, ball N remains in the first box for all K > N. In effect all the balls are in the first box at 1 o’clock. Edited by - pka on 08/20/2008 12:00:46

the_hill1962

USA
1438 Posts

 Posted - 08/20/2008 :  15:30:56 Thanks for the reply, pka. The "[60-(60/2K)]" helps and is much better than my long-winded post.It has just been a long time since I have worked with limits.
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