In which we discuss mattress preservation, group theory, and the problem of the Wicked King.

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We also asked, on this week’s segment how to label the faces of some ordinary dice, with twelve different numbers (we did say different didn’t we?) so that every roll produces a prime number. This puzzle is from the fascinating site www.primepuzzles.net. Don’t peek!

We’ve never discussed the famous “Monty Hall Problem” here (though we did talk about it on the radio before we started podcasting). We recently got an interesting letter that highlights the difference between a game like “Let’s Make A Deal” and a game like “Deal or No Deal”.

Mark A. recently wrote us:
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I had a dream last night involving — (?) well I am not really sure, except that it left me wondering if there is a simple proof (if indeed it is true) that there must be a common factor of

m choose i = m!/(i! (m-i)!)
m choose j = m!/(j! (m-j)!)

for all counting numbers i,j,m with 1 < i,j < m Another way to state this same thing is: any pair of entries, on any row of Pascal's triangle (except for the 1's on the edges) will have a common factor. With facts of this sort, often there is a clever way to cast things in terms of counting something a couple of different ways which makes things clear.

Dennis Shasha, author of Puzzles for Programmers and Pros
joins us once again, posing a cake conundrum!

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How does this simple trick work?

Ask a friend to pick, silently, a three-digit number, then “double” it to make a six-digit number. For example, if she picks 412, the new number would be 412412. Then dividing by 7, then by 11, then by 13, presto! The original number!

Interestingly, there is no decent trick for two-digit numbers; and for four-digit numbers the trick is not so great. But for nine and fifteen digits (for the right kind of people only!!) there is a relatively simple variation.

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After explaining how the Princess escaped, we pose a simple puzzle from Dennis Shasha’s new book Puzzles for Programmers and Pros.

(In the next post we’ll say a little more about the princess.)

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As B Boom wrote, the first pirate can make a proposal that gives him all but 49 (about, depending on the rules) pieces of
the gold. Read the rest of this entry »

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This week we consider an odd number of odd people are milling about
with water pistols, on a large flat field. At a signal, everyone turns and squirts the closest
person (We may assume, since they are just milling about randomly, there is
a unique closest person to squirt.) Show that there will always be at least
person left dry!

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Last week’s puzzle on Perfectly Summing Sets can be solved in many different ways; here’s one!
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What numbers can 1,2,4,8,16,… etc “form”? Well, every number can be “formed” by summing various powers of 2. For example, 13 = 1 + 4 + 8.

In this way, we could say that a power of 2, say 64, is “full of divisors” since it has enough divisors to form any number up to 64. Its divisors are of course 1, 2, 4, 8, 16 and 32, and we can form any number from 1 to 63 by summing up these divisors as needed.

But what other numbers of “full of divisors”?

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