After processing an infinite number of marbles, not once but twice in solving the puzzle Marbles Bonanza, you've grown rather tired of moving all these marbles around. Nevertheless, it is your duty to press on forward and try a third experiment. This time, though, you decide that you'll cut down on the amount of work by not removing any balls ever.

At the start of the minute, you put marbles 1-9 in the bag, and then add a 0 to the end of marble 1 (so that you now have duplicate marble 10s, one in the bag that you just modified, one out). Therefore you now have marbles numbered 2-10 in the bag, as in scenario B of the previous puzzle, and marble 10 outside the bag as in scenario A. 30 seconds later, you put marbles 11-19 in the bag, and add a 0 to marble 2, so that now you have two marbles numbered 20 - one in, one out. You continuously repeat this process, with each interval half as long as the one before. In general, for the nth operation, you put marbles 10n-9 to 10n-1 in the bag, and add a zero to marble n in the bag, so that it becomes marble 10n in the bag.

• How many marbles are in the bag at the end of the minute?
• What are the numbers on the marbles ?
• Is the situation inside the bag identical to either of the previous two problems after 31 seconds? 50 seconds? at the end of the minute? How about the situation outside the bag?

(In reply to re(8): uh oh, here I go again. by Ken Haley)

On this I can't say much more than what was in my previous comment.

Well, yes, it consists of two separate and distinct operations.  There is the first minute, and then the second.  Do you still consider them separate if the second minute begins a second before the first ends?  How about if it begins half a minute before the first ends?  And if the second minute takes half as long?  If you disagree, there's not much I can say.

I do understand the paradox of adding and removing a marble infinitely.  It is like the paradox of the infinite series involving 1s and -1s.

1-1+1-1+1-1+...
=(1-1)+(1-1)+(1-1)+...
=0
or
=1-(1-1)-(1-1)-(1-1)-...
=1

Posted by Tristan on 2005-07-06 21:12:46