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!

Stephen Wolfram, creator of Mathematica and author of A New Kind of Science, discusses the Principle of Computational Equivalence and why even simple systems can give rise to irreducibly complex behavior. 

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Quick interviews with folks here at the Gathering For Gardner, including Stephen Wolfram, Will Shortz,  Dale Seymour, John Conway and many others. 

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We’re at The Gathering For Gardner, G4G9, and catch up with Ed Pegg of mathpuzzle.com and numerous other exploits; Ed gives us three quick puzzles to chew on– answers at his website. 

@MathFactor is tweeting pictures and more, and more short podcasts from the festivities!

(And Jeff didn’t make it! Rats!)

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You and other party-goers are lead to different places in a wood blind-folded.  You’re instructed to remove your blindfold at the sound of a horn and read a letter you’ve been provided.  When the bell sounds, you remove the blindfold and the letter reads thus:

You’ll notice around you that there are lengths of rope tied between hundreds of trees around you.  You may notice that any tree with any rope tied to it has exactly two ends tied to it, each stretching to a different tree.  In fact, these hundreds of treees form a giant concave polygon.  Some of you are inside that polyon and others are outside.  To win the game, you must be the first to reach the clubhouse at the top of the hill (which is outside of the polygon!) and report correctly whether you were initially inside the polygon or outside of it.  You can pass under the ropes, but please don’t change them in any way.  Good luck!

What strategy might you employ to determine your status and require as little time as possible to get back to the clubhouse?

First, a note from G.McN. (I’ll always assume , incidentally, that emails we get are fair game for reposting! Let me know if otherwise)

::: The problem :::
You are at a store that sells 6 items of interest, each item costs 1-6 cents, but you do not know which item costs what. How do you purchase these items all at once (so one bill) and figure out the cost of each item?

::: The original solution :::
The answer given was to purchase:
– 1 of first item
– 10 of the second item
…
 – 100,000 of the sixth item.

Then you can look at the columns and find the price of each item. That is all true but the answer is not universal nor efficient.

::: Efficient solution – 6 items :::
Let say you have 6 items with 6 costs, it is more efficient to do it this way. Purchase:
– 1 of the first item
– 7 of the second item
– 49 of the third item
– 343 of the fourth item
– 2,401 of the fifth item
– 16,807 of the sixth item
The way you figure out the cost of each is playing a remainder game. So divide the cost by 16,807 cents and split the answer between the whole number and the remainder. The whole number is the cost of the sixth item. Take the remainder and divide by 2,401. Split the answer into the whole number and the remainder. The whole number is the cost of the fifth item. You can see where this is going …

Or, if you want to be fancy, you can take the bill and use a computer (because, to be honest, I am lazy) and convert it to base 7 (as opposed to our base 10 decimal system) and look at the columns.

This maybe being picky of me, but it is much cheaper. If every item can be 6 cents, it is at most $1,176.48 instead of $6,666.66. If every item has to be different, it is at most $1,143.81 instead of $6,543.21. Either way, I would not want to pay that bill …

::: Solution – universal :::
To be universal, let there be n items and i = {1, 2, … n} where i is specific item numbers.Then you purchase n^(i-1) of each item i. Then you can covert the answer into base i+1 and look at the columns.
The most you would pay would turn out to be (if every item could be the same price) n^(i-1) if everything has to be different prices then the sum of i*n^(i-1) over all i, but I do not know how to do that off hand.

That’s it! Further  comments below…

We also got this amusing note from J. Kaivosoja

Hi,

and greetings from a snowy Otaniemi, a university campus next to Helsinki, Finland.

Your latest vintage puzzle was, as you predicted, quite easy. [[…]]

I’ve got some puzzles for you, too. Feel free to use them in your podcast (and radio show) and/or on your site.

First, an easy one. How are these numbers arranged?
8 5 4 9 1 7 6 3 2 0
[spoiler] Quite easy – alphabetically. [/spoiler]

How about these?
8 2 3 6 4 0 7 5 9 1
[spoiler]Again, alphabetically, but this time in my native Finnish. I can imagine that’d be a real head-scratcher for anyone who doesn’t speak the language.[/spoiler]

This one I like to give to my friends. How are these letters arranged?
Y S U L E A M I K N G T H R Z
[spoiler]Very few people know them all by heart – these are the commuter train routes in the Helsinki region (I’m a public transport enthusiast, myself).
http://www.vr.fi/eng/aikataulut/reittikartat/lahiliikenne.shtml [/spoiler]

I’m not sure if this last one would work in radio – it might be a bit too visual. How are these letters arranged into two groups?
A   EF HI KLMN
—————
 BCD  G  J    O
[spoiler]The ones below the dotted line are those with curvy lines.[spoiler]

I absolutely love your podcast – it’s kept me entertained for many a dull bus ride, and I’ve tried some of the coin and card tricks you’ve shown on my family, many of whom have since expressed interest in the podcast. Keep up the good work!

Juho Kaivosoja
Espoo, Finland
(the j’s are soft)

Wow, thanks for the great emails!

Chaim

We post last weeks encore presentation of GV. Mix Up At The Five And Dime, and then hit the road! Live updates from the Gathering For Gardner, G4G9 held this week in Atlanta! (Jeff learned he’s reporting when you did– apologies Jeff.)

Eugene Sargent and I will, among other things, be installing a 700 puzzle sculpture, 1 of N, shown here:

Unfortunately, I have not yet completely worked out just what the value of N is– has just been too busy for me.

Finally, we premiere James Greeson’s Canon and Chorale on Pi.

More soon!

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If LIFT = 17 and THIEF = 16 then what is LIFE?

Sorry, a silly one.

Puzzler Derrick Niederman tells us about his new book, Number Freak: From 1 to 200, the hidden language of numbers revealed, full of lore, mathematical amusements and numerical tidbits! 

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The world’s largest ever exhibit of Escher’s works is on display, right now, at the Boca Raton Musuem of Art If you can, this is a must see event! We talk with the collector, Rock J. Walker about his fascination with this amazing work.

And of course we answer last week’s puzzle, and hear from listeners!

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