Colm Mulcahy, of Spelman College in Atlanta,  joins us to share his ice cream trick from his CardColm mathematical card trick column on the MAA website! You’re invited to explain how this works in the comments below.

Colm also shares a quick puzzle, tweeted on his What Would Martin Gardner Tweet feed @WWMGT. And finally we touch on the Gathering For Gardner and the Celebration of Mind, held all over the world around the time of Martin Gardner’s birthday, October 21.

And at last we get around to answering our quiz from a few weeks ago. There are indeed two solutions for correctly filling in the blanks in:

The number of 1′s in this paragraph is ___; the number of 2′s is ___; the number of 3′s is ____; and the number of 4′s is ___. 

Show Spoiler ▼

We can vary this puzzle at will, asking 

The number of 1′s in this paragraph is ___; the number of 2′s is ___; …..  and the number of N’s is ___. 

For N=2 or 3, there are no solutions (Asking that all the numbers we fill in are between 1 and N); for N=4 there are two. For N=5 there is just one, for N=6 there are none and beyond that there is just one. I think we’ll let the commenters explain that.

But here’s the cool thing. 

One way to approach the problem is to try filling in any answer at all, and then counting up what we have, filling that in, and repeating. Let’s illustrate, but first stipulate that we’ll stick with answers that are at least plausible– you can see that the total of all the numbers we fill in the blanks has to be 2N (since there are 2N total numbers in the paragraph). 

So here’s how this works. Suppose our puzzle is:

There are ___ 1′s;___ 2′s; ___ 3′s; ___ 4′s; ___ 5′s

Let’s pick a (bad) solution that totals 10, say, (2,4,1,2,1). So we fill in:

There are __2_ 1′s;   __4_ 2′s;    _1__ 3′s;      __2_ 4′s;     _1__ 5′s

That’s pretty wrong! There are actually three 1′s in that paragraph, three 2′s; at least there is just one 3, and two 4′s and one 5. In any case this gives us another purported solution to try: (3,3,1,2,1). Let’s fill that in:

There are __3_ 1′s;   __3_ 2′s;    _1__ 3′s;      __2_ 4′s;     _1__ 5′s

That attempt actually does have three 1′s; but has only two 2′s;  it does have three 3′s but only  one 4 and one 5. So let’s try (3,2,3,1,1):

There are __3_ 1′s;  __2_ 2′s;  _3__ 3′s;  __1_ 4′s;  _1__ 5′s

Lo and behold that works! We do in fact have three 1′s;  two 2′s; three 3′s and yes, one 4 and one 5.

So we can think of it this way: filling in a purported solution and reading off what we actually have gives another purported solution.

In this case (2,4,1,2,1) -> (3,3,1,2,1) -> (3,2,3,1,1) -> (3,2,3,1,1) etc,

We can keep following this process around, and if we ever reach a solution that gives back itself, we have a genuine answer, as we did here. 

So here’s an interesting thing to think about.

First, find, for N>=7, a correct solution; and a pair of purported solutions A,B  that cycle back and forth A->B->A->B etc.

Second, find a proof that this is all that can happen (unless I’m mistaken)–  any other purported solution eventually leads into  the correct one or that cycle.

I, like so many others, am deeply indebted to Tom; through his generosity (and a fair amount of his typical forceful prodding!) he enabled me and my collaborator Eugene Sargent to begin working together making large mathematical sculptures. Here are a few pictures of those creations. Until we next meet, Tom, thank you.

 

The number of 1′s in this quiz is ____

The number of 2′s in this quiz is ____

The number of 3′s in this quiz is ____

The number of 4′s in this quiz is ____

There are actually two solutions to this one, but more generally, what happens with more lines in the quiz?

Finally, here’s the link to the special issue of Nature with essays on the great Alan Turing.

We all know this feeling: someone’s in your seat, and now you’re the nutcase who’s going to take someone else’s seat. After all that what’s the probability the last person on the plane will be able to sit in the correct seat?

The three number trick is just a simple version of this one (but here it is quicker and simpler).

In which we discuss still more 2012 facts—Matt Zinno points out that we are emerging from a spell of years with repeated digits, and in fact this is just about the longest run in the last 1000 years!  (So, folks, enjoy working out other long spells!)

Ben Anderman shares his online Princess-and-Suitor app.

And Kyle and I discuss some bar bets, including the great Barbette, shown here in a photo by Man Ray:

The challenge this week is to work out a strategy for the following game, that works 50% of the time on average:

The Victim believes you will lose twice as often as you win, so in order to make money, you should somehow get The Victim to bet a little bit more than you do, say $1.50 for each $1 you put up.

The Victim writes any three numbers on three pieces of paper, turns them over, and mixes them up. 

One by one you flip over a card, and then either stop, selecting that one, or discard it and move on. If you select the highest number overall, you win!

We discussed this in more generality a long time ago, but this version has the merit that it’s simple enough to demonstrate quickly, work out precisely why it works and on top of all that, of all the cases has the single highest probability of winning per round.

And still more 2012 facts! From Primepuzzles.net, we learn that

2012 =  (1+2-3+4)*(5-6+7*8*9)

and there’s still more amazing stuff there that we didn’t try to read on the air.

Many listeners will have heard about the hunter who walks one mile south, one mile east, then one mile north—and is right back where he started. But in fact there are infinitely many places on the Earth where he could be, and in at least a couple of different ways!

And what is the deal with Carlos May?

Three crazed, robotic professors (or, if you prefer, “spiders”) try to chase down a psychic, but slightly faster student (the “fly”) along the edges of a tetrahedron. It’s easier, perhaps, to draw it out in the view at right below.

Is there a strategy that allows the professors to catch their prey?

Wait a sec– did we say the prey is SLOWER? Obviously one fast hunter can chase down slower prey, even if the prey knows exactly what’s coming…

Wait another sec– isn’t the point that the professors are morons and robotic? Maybe we DID state the puzzle correctly.

Long time listeners know it wouldn’t be the first time we’d inadvertently added a “meta-puzzle”, namely, to figure out what it is we’d meant to say!! Thanks Byon for pointing that out!

(PS Don’t forget the spoiler tag if you post spoilers in the comments!)

(PPS Apologies: our policy of holding back solutions sometimes gets the order of the comments pretty scrambled)

In the meanwhile, here’s a little discussion about the glass of water problem.

Each time we add or subtract 50%, we are multiplying the quantity of water by 1/2 or 3/2. If we began with 1 glass’ worth, at each stage, we’ll have a quantity of the form 3^m/2ⁿ with m,n>0  Of course that can never equal 1, but we can get very close if m/n is very close to log₃ 2 = 0.63092975357145743710…

Unfortunately, there’s a serious problem: m/n has to hit the mark pretty closely in order for 3^m/2ⁿ to get really close to 1, and to get within “one molecule”s worth, m and n have to be huge indeed. 

How huge? Well, let’s see: an 8 oz. glass of water contains about 10²⁵ molecules; to get within 1/10²⁵ of 1, we need m=31150961018190238869556, n=49373105075258054570781 !!  One immediate problem is that if you make a switch about 100,000 times a second, this takes about  as long as the universe is old!

But there’s a more serious issue.

In a glass of water, there’s a real, specific number of molecules. Each time we add or subtract 50%, we are knocking out a factor of 2 from this number. Once we’re out of factors of 2, we can’t truly play the game any more, because we’d have to be taking fractions of water molecules. (For example, if we begin with, say, 100 molecules, after just two steps we’d be out of 2′s since 100=2*2*some other stuff.

But even though there are a huge number of water molecules in a glass of water, even if we arrange it so that there are as many 2′s as possible in that number, there just can’t be that many: 2⁸³ is about as good as we can do (of course, we won’t have precisely 8 ounces any more, but still.)

If we are only allowed 83 or so steps, the best we can do is only m= 53, n = 84 (Let’s just make the glass twice as big to accommodate that), and, as Byon noted, 3^53/2^84 is about 1.0021– not that close, really!