# Best of Math Mole Volume 1

A selection of interesting problems with some commentary and solutions.

From 2000-2010 (less 2002) I taught a math class in a summer program for high school students.  I always included some content beyond the base content, starting with Mathematicians of the Day (MotD) and in 2005 adding an official daily newsletter, the Math Mole.  Edited by me and later the students, included several features, but the most popular were the puzzles.  I thought it would be fun to revisit some of them on a continuing basis, and so here we go (solutions at the bottom, questions and comments are welcome):

(the original logo)

1.1: A Hershey’s Chocolate Bar comes in a solid rectangle subdivided into a grid of 4 by 3 rectangles. Counting a ‘break’ as any time you take any solid piece and break it all the way through using a given subdividing line, what is the minimum number of breaks needed to completely divide a Hershey’s bar into 12 pieces?

This was the first puzzle from the first edition of the Math Mole.  From this I learned that no matter how clear I think the problem is, students will have their own interpretations and will try to stretch the rules to fit their answers and understanding. (Most important is that you can’t stack the pieces as one and count it as one break.) Most students tried to figure out schemes for breaking the bar and came up with the correct answer, but they usually weren’t able to convince me that it was the minimum. As one emphasis of the class was on proofs, it was important that they could justify their conclusions.

This was not the first logic puzzle, but the first with a back story.  I like it because it brings back some good memories from when I was growing up.  My mom was a big fan of crossword and other word-type puzzles and always bought the Dell magazine to do their puzzles.  She didn’t do the other problems and so I’d get to go through the magazine and do the number and logic problems.  The thing is that I could do the number puzzles (like CrossSums)  pretty regularly, but usually would make a mistake or miss some connection and I could never do the logic puzzles.  Even for this one, I had to look up the answer on the internet just to make sure I had the right answer.  For the students, most were aware of how you can create a grid to solve these types of problems and were able to get the solution.

1.3: Four prisoners are placed in a line so that #1 can not see the others, #2 can only see #1, #3 can see #2 and #1, and #4 can see all the others. The warden then secretly places a hat on each prisoner and then makes the following announcement. I have placed either a black hat or a white hat on each of your heads. There is at least one hat of each color. The first to tell me the color of his or her hat will be set free.” (No guessing is allowed)

The original question is if the order of hats (from #1 to #4) is BBWB, then who gets set free?

A more challenging question is: If you were one of the prisoners, which one would you want to be and why?

This represents a class of puzzles that I love, the ones where a lack of information is useful information. There are lots of great puzzles which use this approach. The challenge question brings in strategies and probabilities, which is always a nice twist.

### Solutions:

Soln1.1: You need 11 breaks.  Actually, if you follow the rules, you will always take exactly 11 breaks, and it doesn’t matter about the geometry. The basic idea is that it takes a break to make one piece, except when you have a piece consisting of 2 pieces, then a break produces 2 pieces.  Add a little induction and you have a proof.

Soln1.2: The German keeps the Fish. As I mentioned in the commentary, I struggle with the details on these, so I suggest looking up a solution, or just forming the grid and working it out.

Soln1.3: For the first question, #3 is the only one that can truly know the color of his hat.  #1 never has a chance as he can’t see anything and #2 is also as helpless.  Initially #3 has no idea, but when #4 doesn’t say his color (because he sees at least one black and one white), then #3 can deduce what #4 sees, and since #3 sees just 2 blacks, he know he must have a white hat.  For the challenge question, #1 can never figure out his hat, #2 can only figure it out when #1 and #2 have different colors, #3 wins when #1 and #2 have them same color and #3 is different, and #4 can only figure it out when #1, #2 and #3 have the same color.  To figure out the best, we have to divide up all the 14 possible configurations among who can know their hat (X is either B or W):

#1 – (none)

#2 – BWXX(4), WBXX(4)

#3 – BBWX(2), WWBX(2)

#4 – BBBW, WWWB

So #2 has the best chance (most opportunities) to identify his or her own hat.

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