ags@pucc-i (Seaman) (02/18/84)
I am reposting the original balls-in-the-bowl question, since no one has
answered it correctly and there has been no discussion of it on the net.
What makes this puzzle hard is that it has an answer which is simple,
obvious, and WRONG. I received mail responses from several people
giving the wrong answer. If you read the puzzle and dismissed it as
trivial, you should think about it again. If there is no response,
I will point out later why the obvious answer is incorrect.
Some tried to read more into the problem than is really there. Notice
that there is nothing about dividing any number of balls by N, nor was
that my intention. The original posting follows...
------------------------------------------------------------------------
You see before you a large empty bowl, two machines marked A and B, and a
*huge* supply of balls.
1. At one minute before noon, machine A places 100 balls into the bowl.
Machine B removes one ball from the bowl.
2. At 1/2 minute before noon, machine A places 100 balls into the bowl.
Machine B removes one ball from the bowl.
.
.
.
N. At 1/N minute before noon, machine A places 100 balls into the bowl.
Machine B removes one ball from the bowl.
.
.
.
It is now noon. How many balls are there in the bowl?
--
Dave Seaman
..!pur-ee!pucc-i:ags
"Against people who give vent to their loquacity
by extraneous bombastic circumlocution."kaufman@uiucdcs.UUCP (kaufman ) (02/20/84)
#R:pucc-i:-21000:uiucdcs:40700006:000:1074
uiucdcs!kaufman Feb 19 16:43:00 1984
There are two ways to look at this problem, each of which guarantees a
different answer. First assume all balls can be numbered, with balls 1-100
going in the first time, then balls 101-200, then balls 201-300, and so on.
Solution 1: The first ball removed is ball 1; the second one removed is ball
2; and so on. Ball n will be removed at 1/n before noon ==>
all balls put in will be removed before noon.
Therefore, no balls in the bowl at noon.
Solution 2: The first ball removed is ball 100; the second one is ball 200,
and so on. 99 out of every 100 balls will never be removed and
since we put balls in an infinite number of times, an infinite
number of balls will be in the bowl at noon.
Those are the "normal" ways to look at the problem. Of course, for any
k < 100 we could say the first ball removed is ball k+1, the second is k+2,
etc. At noon, only k balls: balls 1-k will be in the bowl. TAKE YOUR PICK!
Ken Kaufman (uiucdcs!kaufman)cjp@vax135.UUCP (CharlesPoirier) (02/20/84)
It is now noon. There are no balls in the bowl. There is no bowl, machine, or observer either. Because, acouple of seconds ago, the machines tried to move at infinite speed as the time approached noon; they became relativistically massive, and eventually dragged everything nearby into itself, forming a small lump of neutronium. C'mon, guys, if you want to give a math problem, *state it* as a math problem. I can't deal with infinitely fast machines that have to move mass around. Especially if we are not allowed to add some assumptions. Like, suppose that both machines have the same (fixed, arbitrary) speed at which they break down, but that the 100-ball machine has to run 100 times as fast as the 1-ball machine. Is there a fastest such break-down speed for which the 1-ball machine just manages to empty the bowl before breaking down itself? (Add assumptions if you want.) (I haven't solved this one myself ....) Charles Poirier
csc@watmath.UUCP (Computer Sci Club) (02/20/84)
The problem is similar to finding the sum of the series
100 - 1 + 100 - 1 + 100 - 1 ...
As written the sum of the above series is infinity (ie the limit
of the sequence of partial sums is infinity). However the subseries
consisting of all negative terms is divergent. Hence we cannot
rearange more than a finite number of terms and expect to have
a limit for the sequence of partial sums. This is what is involved
in such arguments as "number the balls sequentialy and say that
machine B removes ball N at time noon - 1/N".
If we assume the problem is find the sum of
100 - 1 + 100 -1 .....
and use the standard definition of the sum of a series, then the
problem has a well defined answer, there will be an infinite number of
balls in the bowl at noon. Else the argument "machine B removes ball
N + K at time noon - 1/N" allows an arbitrary answer. (note also that
infinity = limit as e goes to zero of the number of balls in the
bowl at time noon - e)
William Hughesmcewan@uiucdcs.UUCP (mcewan ) (02/24/84)
#R:pucc-i:-21000:uiucdcs:40700009:000:396 uiucdcs!mcewan Feb 23 23:15:00 1984 It seems quite obvious to me that at each point 99 balls are being added to the bowl -> infinite number of balls at noon. You could also think of this as the sum 100 - 1 + 100 - 1 + 100 - 1 + ... which gives the same result. I would be very interested in seeing why this is incorrect (or is there some other obvious answer that never occured to me?). Scott McEwan pur-ee!uiucdcs!mcewan
johna@haddock.UUCP (02/28/84)
#R:pucc-i:-21000:haddock:25400001:000:70 haddock!johna Feb 27 15:03:00 1984 It appears there would be 297 balls left. johna @ ima!haddock
andree@uokvax.UUCP (03/01/84)
#R:pucc-i:-21000:uokvax:13500008:000:351 uokvax!andree Feb 27 19:44:00 1984 You can actually do better than that. You are summing a series of the form 100 -1 +100 -1 .... Here we have the sum of an alternating series which is not absolutely convergent. Hence, if you hand me a number, I can make the series sum to that number by rearranging terms. In other words, the answer is any number - or just plain ill defined. <mike