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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.

We’ll have some pursuit puzzles over the next couple of weeks; this segment’s puzzle has a simple and elegant solution, but it might take a while to work it out!

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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!

In Golden Earring – Radar Love I had a big problem.

This the solution so please read the problem first and have a go at solving it yourself.

My wife has lost a golden earring, I can buy a bunch of similar earrings but I know one is a fake.

I can weigh two groups of earrings, each weighing can give one of three results: they weigh the same; the left group is heavier, the right group is heavier.

The question is – how many earrings can be in the bunch I buy and leave me confident that I can find the fake with just three weighings?

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The Math Factor podcast is taking a rest for a while — we’ll be back with new podcasts at some point (probably, we think) so check back every once in a while!

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In the meantime, from time to time the Math Factor crew will still be posting here, on our traditional highly irregular schedule.

I’m really proud of the bookends to the pieces so far: Cantor’s Theorem, in the segments leading up through AH. QED and now the discussion of undecidability in the last two podcasts. Along the way, we’ve managed to get in quite a bit of sophisticated stuff — not bad for local radio! 

I’ve really enjoyed all the conversations the Math Factor has initiated, between me and Kyle, with mathematicians and those using mathematics to do really interesting stuff, with book authors, and especially all those who have written in — and even become active collaborators! (Hi Jeff and Stephen) The tremendously supportive feedback we’ve gotten really means a lot.

I think it’s time, though, to take an extended break from the podcast. Kyle is now incredibly busy producing five hours of original magazine format radio journalism a week. He’s always been a dynamo, but lately the man’s a blur! And much of my energy has been directed elsewhere too (check out math2033.uark.edu!) I’ve started a couple of books that I hope you’ll check out when the time comes, and in the meantime, please read my articles Can’t Decide? Undecide! and another on tilings and computation.

I’ll be hanging out in Marseille, Mexico City and Oaxaca in June, with a lot of neat people, so might get all inspired to make some new posts soon, but on the whole, I feel like I’ve said what I needed to say for a while. The Theory of Computation is really an astounding and important perspective, and I’m delighted to have helped spread the word a bit more. It’s a great resting spot!

This is the solution to Morris: Trial/Trual/Whatever.  Please look there before reading the solution.

It turns out the right word is truel, first coined in 1954 by Martin Shubik.

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{ Hi, Steve here. Jeff asked me to post a solution and I’m more than happy to oblige. It’s a fun puzzle with some nice maths to explore. I learnt a lot about graph theory and a new theorem (new to me), Turan’s theorem. More on that later. }

In Yoak: Batteries, and the Problem-of-the Week Jeff posed a great problem from Stan Wagon’s Problem of the Week.

You have eight batteries, four good and four dead. You need two good batteries to work the device; if either battery is dead then the device shows no sign of life. How many tests using two batteries do you need to make the device work?

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In Living With Crazy Buttocks  I posed a problem where 20 party guests were each given an unusual book.  These books were placed in identical boxes.  The guests enter the room with the boxes one at a time and are allowed to open half of the boxes.  They leave by a different door and cannot communicate with the other guests.  The room is put back identically before the next guest enters.

If every guest finds their book then the whole group win a trip to Paris.

What is their best strategy?

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In working out the proof for World of Britain I came across a paradox.  Maybe smarter Math Factorites can help me out?  My sanity could depend on it.

In the puzzle you have five different tasks.  On each day one of these tasks is given at random.  How long do you expect it to take to get all five tasks?

First consider a simple case.  Suppose some event has a probability, p, of happening on any one day.  Let’s say that E(p) is the expected number of days we have to wait for the event to happen.  For example if p=1 then the event is guaranteed to happen every day and so E(p)=1.

How can we calculate E(p)? 

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This is a followup to my earlier post, A Rather Odd Car Trip.  It provides a solution so if you haven’t read that yet, you should do so first as this won’t make much sense without it.

[spoiler]

Though czarrandy and others provided the distance between the cities, I’d like to show a way that you can work the problem.

We want to work out an equation that explains the amount of time traveled each  downhill, level and uphill in terms of the distances involved.

How long does it take to travel a particular distance at a particular rate?  Take the uphill portion as an example.  At 56 mph, you travel 1 hour / 56 miles or n hours / x miles for any given distance x.  So:

1/56 = n / x

Solve for n (hours) and you get:

x/56 hours traveled for distance x.

Do something similar for the level distance (call that y) and the downhill distance (call that z) and you get this equation for the time traveled in terms of the three distances traveling from A to B in 4 hours:

x/56 + y/63 + z/72 = 4

For the trip back, where uphill becomes downhill and vice versa, we get:

z/56 + y/63 + x/72 = 4 2/3 or 14/3

If we could solve for x, y and z, we would get our answer, but generally there will not be a unique solution for two equations with three unknowns.  But in this case, what we need is the value of x+y+z , which we can attempt to extract that from the equations.

First, add the two equations together and get:

x/56 + x/72 + y/63 + y/63 + z/56 + z/72 = 26/3

To collect the terms, multiply both sides by the least common multiple of the denominators, which is 504, and we get:

9x + 7x + 8y + 8y + 9z + 7z = 4368
16x + 16y + 16z = 4368
16(x+y+z) = 4368
x+y+z = 273

for 273 miles each way.  Notice that what we did was to add the equations up to get a value for the round trip, which might have been a first intuitive step if you happened to think of it.

Now… it was very lucky that we were able to nicely factor out that 16 in order to solve for x+y+z.  This won’t always be the case!  The three speeds were chosen carefully so that this would work out.  Consider if we used some other randomly chosen values:

Uphill: 50 mph
Level: 60 mph
Downhill: 70 mph

Leave the times the same, so:

x/50 + y/60 + z/70 = 4
x/70 + y/60 + z/50 = 14/3

or x/50 + x/70 + y/60 + y/60 + z/70 + z/50 = 26/3

Multiply by the least common multiple of the denominators, 2100, and get:

42x + 30x + 35y + 35y + 30z + 42z = 18200
72x + 70y + 72z = 18200

As you can see, there isn’t going to be any way to factor out x+y+z to get a unique total distance.  You could experiment to demonstrate that you can pick values for x, y and z that satisfy the equation and for which x+y+z will differ.

The final question that interested me is how to identify the cases where there does turn out to be a solution.  From the process we just followed, you can see that you want the final coefficient of the x and z terms, or the uphill and downhill time traveled, which will always be equal to each other, to also be equal to the coefficient of the y term or the time traveled level.

Since we’ve added the two equations from the round trip, the meaning of these terms is the time taken to travel round-trip over a slanted piece of road.  So in English, in order to have a unique solution to the problem, the speeds must be specially selected such that the time it takes to travel round-trip over a slanted piece of road must be the same as the time it takes to travel over a level piece of road.

I got this problem originally from Nick’s Mathematical Puzzles. There are a lot of neat puzzles on the site to enjoy.

[/spoiler]

I took one of Peter Winkler’s puzzle books on holiday recently.  After dinner each night I intended to impress my friend with an amazing math puzzle.  I had done this before.

The book dissapeared on the flight out.  After dinner each night my friend impressed me with an amazing math puzzle.  I haven’t seen the book since.

Serves me right!

This is one of those puzzles, you will understand why I have to do it from memory.

I really like Jeff’s post  A Fun Trick – Guess the Polynomial.  You might want to look at it first.

If you relax the conditions a bit you have a similar sounding puzzle with a very different solution.

So my puzzle is this:

I am thinking of a polynomial.  All of the co-efficients are fractions.   You may use any number as your test number.  When you give me a test number I will tell you the result.

How many test numbers do you need to identify the polynomial?