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Preview Problem – Share the beer

Recently, I shared a problem on Facebook.  It’s typically called the “sharing wine” problem, but my friend, Cyrus, thought it’d be better with beer.  I agree, so I’ve modified these slightly.

It went something like this:

We have three containers of different sizes, 30L / 11L / 7L.  The 30L is filled with beer.  Empty exactly half of the 30L using only the 11L and 7L containers.

I’ll introduce some notation here so that we can talk about the answer.  Let’s say that we create a triple (a, b, c) indicating the amount of beer in each container at any one time.  Say that we order these largest to smallest for convenience.  Let a = the 30L container, b = the 11L container, c = the 7L container.

So, starting out we have (30,0,0).  Cyrus correctly pointed that for our first step we should pour 11L from the 30L into the 11L container.  Now, we have (19,11,0).  We should then fill the 7L container from the 11L giving us (19,4,7).  Then, pour the 7L back into the 30L so we now have (26,4,0).  If we then pour the remaining 4L into the 7L (26,0,4) bucket and then fill the 11L from the 30L, we arrive at (15,11,4).

If we use our new notation, I can put the five steps into a notation that looks like this:

I found that the problem was easy enough to do in my head.  It’s only five steps.  But, are there other versions that are more difficult?  Can we find a systematic way for solving these?

Let’s give it a try.  But, first – I want to pose a harder problem and see how you think about it.  Try this one, Cyrus!

Given 8L of beer in an 8L pitcher, use a 5L and 3L pitcher to equally divide the beer between us; 4L each.

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