T O P I C R E V I E W 
romangod5 
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?!?!?!?!?!?!?!? 
4 L A T E S T R E P L I E S (Newest First) 
the_hill1962 
Posted  08/20/2008 : 15:30:56 Thanks for the reply, pka. The "[60(60/2K)]" helps and is much better than my longwinded post. It has just been a long time since I have worked with limits. 
pka 
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(60/2^{K})] 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 (60/2^{K}) 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.

the_hill1962 
Posted  08/20/2008 : 09:02:04 pka, I also thought it was a problem dealing with a Zeno paradox. However, isn't it about the Zeno's dichotomy paradox and not the arrow paradox? I realize that wikipedia could have misinformation but it looks correct at http://en.wikipedia.org/wiki/Zeno's_paradoxes In the poster's (romangod5) problem, according to the paradox, the machine will never stop because it never adds enough time to get to 1 o'clock. 12:30>12:45 h:m:s format\/ 12:45>12:52.5 (12:52:30) 12:52.5>12:56.25 (12:56:15) 12:56.25>12:58.125 (12:58:15/2) 12:57.125>12:59.0625 (12:59:15/4) 12:59.0625>12:59.53125 (12:59:255/8) 12:59.53125>12:59.765625 (12:59:735/16) 12:59.765625>12:59.8828125 (12:59:1695/32) You see, the sequence 15/2, 15/4, 255/8, 735/16, 1695/32.... will never reach 60 seconds and therefore never will the time be 1:0:0. Of course, this is impossible because we know that you truly cannot keep just adding half the amount (as stated that the machine does be adding 30 minutes then 15 minutes then 7.5 minutes then 3.75 minutes). Even though the machine "runs at lightning speed", the branch of mathematics called Calculus shows that the limit of the above sequence is, in fact 60. However, in actuality, that value is never really arrived at. When I took Calculus, my teacher gave this description of the paradox: Imagine if you were going out on a date. Your girlfriend wants you to meet her father. While talking to her father, he stated that you can do anything on your date as long as you obey one rule. The rule is that you ARE allowed to keep getting closer to his daughter but by only half the distance you are now! Well, would you ever get to kiss her? 
pka 
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.


