- #1
fluidistic
Gold Member
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Hello,
I have posted a problem (on math stack exchange) I was given for fun by an uncle, who doesn't know whether the proof is possible to establish. I tried my best and failed so far. I don't think I can solve the problem with my current knowledge and I would love to know if you can find a solution or at least have some ideas.
Problem statement:
So far I have been able to show that there is no possible proof by induction, I believe. The reasoning is that if we start with a solution that works for ##n## steps, then we cannot make any further step because it would mean to withdraw ##n+1## balls from any box, making the number of balls negative in that box, which is not allowed.
So maybe number theory can offer a way to tackle the problem?
I have written a Python program that outputs the solutions for a given ##n##, and I have stared hard at them (their number grows quite quickly with ##n##. There are 36 or 72 solutions for ##n=9## for instance, starting from 1 or 2 solutions for n=5 depending on whether ##B## and ##C## are distinguishable). No luck, I couldn't notice obvious patterns or ways to create solutions. Some of them look almost trivial (i.e. withdraw balls from box A most of the time and place them into either ##B## or ##C## and then at the final step(s) withdraw from ##B## into ##C## or the opposite) but in general there is no such "trivial looking" solution. So I am not able to find an algorithmic or automatic way to build a solution from scratch.
I have posted a problem (on math stack exchange) I was given for fun by an uncle, who doesn't know whether the proof is possible to establish. I tried my best and failed so far. I don't think I can solve the problem with my current knowledge and I would love to know if you can find a solution or at least have some ideas.
Problem statement:
There are initially three boxes, labelled ##A##, ##B## and ##C##. ##A## contains ##3n## balls, with ##n\geq 5## a natural number. At step number ##k##, one withdraws ##k## balls from any box and place it into any other box. Show that it is always possible to end up with ##n## balls in each boxes after exactly ##n## steps.
So far I have been able to show that there is no possible proof by induction, I believe. The reasoning is that if we start with a solution that works for ##n## steps, then we cannot make any further step because it would mean to withdraw ##n+1## balls from any box, making the number of balls negative in that box, which is not allowed.
So maybe number theory can offer a way to tackle the problem?
I have written a Python program that outputs the solutions for a given ##n##, and I have stared hard at them (their number grows quite quickly with ##n##. There are 36 or 72 solutions for ##n=9## for instance, starting from 1 or 2 solutions for n=5 depending on whether ##B## and ##C## are distinguishable). No luck, I couldn't notice obvious patterns or ways to create solutions. Some of them look almost trivial (i.e. withdraw balls from box A most of the time and place them into either ##B## or ##C## and then at the final step(s) withdraw from ##B## into ##C## or the opposite) but in general there is no such "trivial looking" solution. So I am not able to find an algorithmic or automatic way to build a solution from scratch.