December 23, 2011
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The Mathcast

How much does my True Love love me truly?

Kyle and Chaim ponder the question…

[spoiler]

This is, of course, the on air version of last year’s post, The Nth Day of Christmas, when the podcast was quiet. But don’t peek!

[/spoiler]

Quick answer soon, and in the New Year, a longer discussion on the merits of mixing cream with your coffee. In the meantime,

Peace on Earth, for all living things,

Chaim and Kyle

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December 12, 2011
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answers, calculusey stuff, guests, math puzzles, The Mathcast

Harry Kaplan joins us for discussion of cake and coffee– and leaves us with a counter-intuitive puzzle…

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December 4, 2011
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game theory, math puzzles, The Mathcast

A quick hello from Chaim and Kyle as the Math Factor returns!

We’ll be the first to say our Coffee Pot Question isn’t our deepest puzzle ever, but it sure did make a difference in Chaim’s life!

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November 17, 2011
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math puzzles, paradoxes

When I became chairman of my department a few years ago, I moved from my office far down at the end of the hall to one much closer to the center of action: the tea room! I make a lot more visits there than I used to, and began to notice a frustrating pattern:

Far more often than seems reasonable, there’s not even a full cup of coffee in the coffee pot! Once again, someone has left a nearly empty pot with no regard to the next person (me, whine)!

This seems to happen so often I began to wonder what kind of boors I’ve been working with all these years. They *seem* like nice people and all, but…?

And then I realized: there’s a perfectly logical reason, a mathfactor puzzle, if you will, that explains this phenomenon perfectly, no boors required, no special tricks, just sensible activity by all. My faith in my colleagues has been restored.

**Why is it that on average I see an emptier rather than fuller coffee pot?**

*PS let us know what works for when we return…*

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November 15, 2011
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The Mathcast

Well, it’s been quite a while, and I guess the important news is that the M.F. is coming back soon! Look for new segments starting around the end of the year!

All those encouraging emails, prodding emails, whining emails have had their effect—but I have to say, in the end, it was the Juniors Fabulous cheesecake that probably made the difference.

Meanwhile, a steady accumulation of great ideas and on top of that, when we do come back, we’ll be listed on the National Science Foundation’s app/site, Science 360. Thanks to all those that have hung around waiting—your patience is valued.

One question: Over the years we’ve experimented with a lot of formats (short, long, interviews, deep stuff, light stuff, book reviews, puzzles, Kyle and me shooting the breeze.) What format works best for you?

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November 13, 2011
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Authors, math puzzles, Yoak

It’s been a while since I posted a puzzle and happened upon a nice one today.

You find yourself babysitting a friend’s marble collection while he’s away at a conference of collectors. The housekeeper, being an honest type, admits to having disturbed one of the displays. She assures you that she returned all the marbles and shows you ten marbles laid out thus:

* * * * *

* * * * *

That looks fine, but the note to the side of it says five lines of four marbles each. Since this is two lines of five marbles, must there be marbles missing or can they be laid out in a manner consistent with the note?

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July 14, 2011
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guests, Podcasts

Samuel Hansen’s Strongly Connected Components podcast features interviews with all kinds of mathematical luminaries (that sounds familiar!) If you’ve been missing the Math Factor, be sure to check it out!

Here, we discuss, well, Chaim Goodman-Strauss—the tables are turned!

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July 4, 2011
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Morris, The Mathcast

25 horses have entered your race meeting. You have to award ribbons to the three fastest, in order.

Only five horses can compete in one race.

Assuming each horse always runs at the same pace, and that all 25 horses have different abilities, what is the fewest number of races you can use?

You can’t use stopwatches, you can only use the finishing order of each race.

UPDATE: Thanks for all the great comments. At the time of writing we’ve had three correct answers; Mike and Blaise in the comments and Jim via email.

I have a follow up question: can you prove that this answer is minimal? There is a neat spot, can you find it?

Apparantly this was an interview question for Facebook. I found it in this list of impossible interview questions. Math Factor afficionados will recognise some of these problems and not find them so impossible. Have fun!

http://skymcelroy.tumblr.com/post/6244020081/impossible-interview-questions-from-facebook-goldman

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April 9, 2011
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The Mathcast

By an amazing coincidence, world-class puzzle creators Ed Pegg (mathpuzzle.com) and James Stephens (puzzlebeast.com) were in Fayetteville Ark. on the very same day! We sit down and discuss the art of puzzle making, their own wonderful puzzles, and their personal favorites.

Here are some links to some of the things they talk about:

I don’t know how often we’ll be posting new podcasts, but stay tuned!

vortex

rush hour

the gordian knot

smart games think fun

pentamonos, hexamonoes

burr tools

sliding block puzzleworld

click mazes

logic mazes

nikoli

games magazine

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January 4, 2011
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The Mathcast

Today’s the Nth day of Christmas (The tenth day, to be precise) — as a function of N, about how many gifts total has my true love given me since the first day of Christmas?

It’s cooler to have a quick estimate of the rate at which the number of gifts grows, rather than the exact formula; there’s a quick way to get a rough sense of the right answer.

Does the total number of gifts grow exponentially? Factorially? As the square of N? Something else?

Don’t peek until you want the answer!

[spoiler]

Ok: on the Nth day, my true love gives to me 1+2+3+…+N gifts; you might know the formula for that sum, and it’s a good exercise to work it out again, but the really important thing is that when you sum up consecutive integers, the total grows about like N^2 (ignoring pesky constants)

In fact, if we sum up consecutive kth powers, the total grows about like N^(k+1). This is really a kind of counting version of integration (and in fact, is exactly one of the tools ancient mathematicians, such as Archimedes, used to work out certain integral calculus problems 2000 years before Newton and Leibnitz invented calculus!) So if on day n, we receive about n^2 gifts, by day N, we’ve received about 1^2 + 2^2 + 3^2 + … + N^2 gifts, for a total of roughly N^3.

Thus the total grows about like N^3, ignoring pesky constants and lower level terms. The exact formula, should you need it for checking your true love’s love, is going to be N^3/6 + N^2/2 + N/3; on day twelve you should be expecting a total of 364 gifts!

[/spoiler]

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