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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 ___. 

[spoiler] namely (3,1,3,1) and (2,3,2,1) [/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.

Let’s see: First, the “Big News“, a discussion of Carlos May, and another puzzle (a pretty easy one)

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.

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Our band of intrepid pirates, having resolved previous squabbles over distributing booty amongst themselves and other issues have come across a treasure map fragment.  The picture has been destroyed, but the following text can be read:

Stand upon the gravesite and you’ll see two great palms towering above all others on the island.  Count paces to the tallest of them and turn 90 degrees clockwise and count the same number of paces and mark the spot with a flag.  Return to the gravesite and count paces to the second-tallest of the trees, turn 90 degrees counter-clockwise and count off that number of paces, marking the spot with a second flag.  You’ll find the treasure at the mid-point between the two flags.

Fortunately, our pirates knew which island the map referred to.  Sadly, upon arriving at the island, the pirates discovered that all evidence of a gravesite had faded.  The captain was preparing to order his men to dig up the entire island to find the fabled treasure when one of the more geometrically inclined pirates walked over to a particular spot and began to dig.  The treasure was quickly unearthed on that very spot.

How did the pirate know where to dig?

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So, I’m teaching a new course, Math 2033, Mathematical Thought, and it’s going great! I’d like to take a moment to write about it!

(This is one reason the MF has been kinda slow lately; another is that I’m chair) When it’s fully up and running, we’ll have about 150 students in one large section each semester (we’re starting with about 100). In a nutshell, it’s the Math Factor, as a course.

Read the rest of this entry »

This will be the last of my Lewis Carroll posts.  In Pillow Problems, Carroll writes:

   Three Points are taken at random on an infinite Plane.  Find the chance of their being the vertices of an obtuse-angled Triangle.

Note: An obtuse-angled triangle is one that has an angle measuring more than 90 degrees.

Strange appearance in the North Atlantic:

Here’s a puzzle that sounds a little like those, “A train leaves…” questions we were all prepared for but rarely saw on the SAT, but with a twist.

You are going to take a drive from City A to City B and back, but in a rather unusual car.  When travelling uphill, the car always moves at exactly 56 miles per hour.  On level ground, it travels at 63 miles per hour and finally when travelling downhill it travels at 72 miles per hour.  Assume that it transitions from one speed to another instantaneously and all of those other “mathematically perfect” qualities that make questions like this answerable.

You find that travelling from City A to City B takes exactly 4 hours of travel time.  On the return trip, driving time sums to 4 hours and 40 minutes.

How far apart are Cities A and B?

There has been a theme in some of the recent posts and problems. It’s a little buried but almost enough to say its another of those Mathfactor agendas when we try to sneak some knowledge to you buried in the fun.  Never one to miss such an opportunity I will jump in with a post, and a problem.  This is a slight change to a classic problem that comes out of the work of one of my mathematical heroes:  Leonardo of Pisa, also known as Fibonacci.  He is responsible for changing how we count! Not many people can claim that. He introduced the system base value, also known as Arabic numerals that we still use today into Europe.  He is more famous however for talking about rabbits:

Imagine that you have immortal rabbits, Bugs Bunny’s version of Olympus perhaps.  Even if they are immortal however rabbits are famous for one thing.  They breed like, well rabbits.  Some of the rabbits are children and some adults and are divided into pairs.  Each month any child pairs become adults and any adult pairs breed to produce a new child pair.  They are immortal so no pair ever dies.  These rabbits are also a little odd.  They live on a line (don’t complain, this is no more ludicrous than that they are immortal!), but can shuffle along.  Also if you are worried about inbreeding, the male rabbit leave the family hutch and shuffle along the line past others until they find a suitably unrelated mate.  Why we would be worried about inbreeding in immortal rabbits living on a line escapes me!

Anyway we start with one pair of children.  Lets put a c.  After a month they become adults, a.  Another month passes and they now have a pair of children, but are still there themselves.  We therfore have the original pair and a pair of children: ac.  Next month the adults have another pair of children and the children become adults: aca.  Can you see how this will work?  Each month the children become adults so we replace every c with and a, each pair of adults has a new pair of children but stays as adults, so we replace every a with ac.  We can continue to get longer and longer sequences of rabbits on this line:

aca  to acaac to acaacaca to acaacacaacaac….

Now some puzzles.  Given a line with 21 adult pairs and 13 child pairs, how many pairs of adults and children would there be after one month?

Given p adults and q children how many adults and children will there be after one month?

Finally a more difficult one.  How will the ratio of adults to children behave month on month?  Will it

a) Get closer and closer to a particular number?

b) Keep on changing without pattern?

In either case can you say more?

Here’s a quick puzzle that I don’t think has appeared here previously.

A man leaves at exactly 6:00 AM to climb a mountain.  He may not climb at a constant rate.  In fact, he may stop to rest, or even backtrack a bit to see something interesting.  He arrives at the summit at exactly 6:00 PM and camps for the night.

The next day, he starts down at exactly 6:00 AM, again meandering unevenly, but following exactly the same path and reaching the bottom at exactly 6:00 PM.

The question is, was there some point on the path he was sure to have visited at exactly the same time on the two trips?

I have just published a (rather long) article on mathematical surfaces, their models and links to art over at Maxwell’s Demon.  Here is a sneak preview.

Minimal Möbius, Benjamin Storch