Geek of The Week: Nick Hamblet

Not sure whether you have been following the comments on Prime Time? But, Nick, known to us in the comments as SumIdiot, came up with a nice proof showing that it's impossible to group the digits of a number n in primes so that you wind up with the prime factorization of that number. For example, the digits of 241 can be grouped as (2)(41), but 2 and 41 aren't the prime factors of 241. Another example, 573 could be grouped (5)(7)(3), but again the product (5)(7)(3) is not equal to 573. Enough.

What Nick actually showed was the larger result: any grouping of a number's digits, prime or not, will always have a product strictly less than the number itself. How did he do it? Here's his proof from the comments.

Given d+1 integers between 0 and 9, called a_0, ... a_d, let me write A=(a_d a_{d-1} ... a_0) for the number \sum_{i=0}^{d}a_i 10^i.

Consider the product obtained by splitting this digit string into (a_d ... a_s)(a_{s-1} ... a_0). So this is a product of a d-s+1 digit number and an s digit number (take s >= 1).

Now A=(a_d ... a_0) = 10^s(a_d ... a_s) + (a_{s-1} ... a_0) is at least as big as 10^s(a_d ... a_s). However, since 10^s is strictly bigger than (a_{s-1} ... a_0), A >= (a_d ... a_s)10^s > (a_d ... a_s)(a_{s-1} ... a_0).

So any integer is bigger than the product obtained by splitting it's decimal string into two pieces. It seems the general statement, splitting the decimal string into any number of pieces, follows by induction.

Clever! Nick has also given me the chance to break in a blogroll (Only on the new domain). He runs a real nice blog Sumidiot here.

**Reminder YofX will be moving to www.yofx.org in 11 days on 7/20**

Geek of the Week: Nerd Nook

I've seen a lot of math t-shirts. This one on Nerd Nook's photostream caught my attention. (might have to blow it up to read it.) In the 18 million times I've been asked, 'When will I ever use this in the real world?', I have yet to use this technique. It's kind of Dirty Harry meets underpaid math teacher. He also has a bunch of other t-shirts. I love this one.

Geek of the Week How To

How can I become Geek of the Week? I get the question all the time. I have to admit. There is no formula. There is a lot of luck, and not a little bit of skill involved. For aspirants, MaximumPC has compiled a litmus test for basic geekdom. As they see it, there are 50 fundamental skills. Sue will be glad to know that the recitation of pi to at least 23 digits is on there. Check and see how many of these must have geek-skills you wield. You also might feel that they have overlooked some fundamentals. I'd love to hear what you would add in the comments.

pic by koolbadges

Geek of the Week: Ken Colangelo

I have received my first ever geek-resumé. It came via email about a week ago. It almost knocked the slide-rule right out of my shirt pocket. Here are the contents of that message.

I’d like to nominate myself.

In my spare time at home (spare time mind you) I’ve been creating code to generate optimized 3D models of Menger sponges.

I used to use huge amounts of CPU time on the VAX at CCSU as a student exploring strange attractors, Julia sets, cellular automata and n-body gravitational simulation.

They even gave me a degree for it.

This rendering I found hangs in my bathroom. 4th level Menger sponges…for the bathroom…ha!
This, combined with my frequent and popular (?!) emails about when to see the ISS passing over and my geektastic panoramic photography I think puts me in the running.

Also, I got my first paying computer job when I was 9. Seriously. I used those monster 8-inch floppy disks and installed CPM on old 8088 computers.

I’ve got a goofy picture of me covered in electrodes and another wearing safety goggles and a breather if you need them for my geek-of-the-week portrait.

I’ve got my fingers crossed.

-Ken

He had me at Menger sponges, but he throws in a follow up email just to make it hurt.

I’m a rabid Scrabble player.
N.S.A.* member number 602852
*National Scrabble Association

Geek of the Week: Vadim Korf

Vadim wins Geek of the Week for solving one of the blog problems I wasn't able to solve. Not only solving it, but providing one of those elegant solutions that leaves you saying, "Of course, how come I couldn't think of that?" I hope you enjoy it as much as I do. Here's the problem:

A town creates a committee of six people to solve a new math problem. In the town, everyone is either a friend or an enemy. Prove that there are at least three friends or three enemies on the committee.

And here's Vadim's proof.

Six people: A, B, C, D, E, F

Two kinds of relationships: X, Y
---- r(A,B)=X means A and B have relationship of type X.
---- r(A,B,C)=X means that r(A,B)=r(A,C)=r(B,C)=X

We have to prove that we have at least one r(P1,P2,P3)=X (or Y), where P1,P2,P3 could be anyone from {A, B, C, D, E, F}

Consider A (anyone). It has a relationship of type X (could be either friend or enemy, whichever has a bigger count for A) with at least three other guys, for example B, C, D (and possibly E and/or F as well), so r(A,B)=r(A,C)=r(A,D)=X

If r(B,C)=X then r(A,B,C)=X
If r(B,D)=X then r(A,B,D)=X
If r(C,D)=X then r(A,C,D)=X
If neither 1, 2, or 3 is true, then
r(B,C)=r(B,D)=r(C,D)=Y, then r(B,C,D)=Y

If you are not picking up on his notation, basically he's saying: consider any person in the group of six. That person has a relationship with 5 other people. He/She is either friends with 3 or more people or enemies with 3 or more people. That gives you two cases. The following proof will work in either case, so we just have to look at one. Assume that A is friends with B, C, and D. Vadim has given us a diagram to help visualize the argument. The red line between any two people means they're friends.
Consider the relationship between B and C, if they are friends, we have the picture. (Note: the green lines represent that the relation between them is unknown.) So A,B, and C are all friends. Therefore, there are three friends on the committee. In fact, the same is true if either BD or CD are friends. So what about if BC, CD, and BD are all enemies? Here's Vadim's picture. Remember the red lines represent friendship and the blue lines represent enmity. (Note: ignore the 4. Too lazy to edit it out.) But then B, C, and D are all enemies, so we have at least three enemies on the committee. All done.

Geek(s) of the Week: Academic Support Center

Every semester the tutors in the Academic Support Center miraculously save a good number of my students. I'm not sure how they do it but I know it involves a lot of work. Check out a couple of videos I made for an old course shell project. Who wouldn't want to talk math with either one of these guys? There are plenty of others in the ASC, too. I don't have videos of them, but that is just a matter of time. Keep up the great work!

Mohamed


Jeff

Geek of the Week: Laura

Laura is our resident Foxtrot expert, which is a surprisingly geeky comic. You might remember the Fibonachos. That was Laura. Below is her latest find. I like this one even better. I can't wait to give it to my students.


I am also happy to report that the caché of the Geek of the Week award has risen, evidenced by the fact that people are now starting to prepare acceptance speeches. Here is an excerpt from the full 20 minute version of Laura's speech.

...Wow! Geek of the Week! I never thought I'd win! It was an honor just to be nominated...I'd like to thank the math department for selecting me..my mom and dad for all their love and support and for giving me my first abacus and then my first slide rule...and let's not forget my Tunxis friends who love to remind me on a daily basis that I am a geek!

Geek of the Week: Robert Devaney

Last Friday I attended the Spring MATYCONN meeting with my infamous colleague Sue. I was treated to a really excellent tour of the Mandelbrot set by BU prof Robert Devaney. Where most presentations of fractals seem to die in the pictoral realm, Devaney gave the necessary background to appreciate the pictures and inner patterns of the set. He also had some cool visualizations. He's got some of them posted on his homepage.

pic by Fermion

Geek of the Week: Bertrand Russell

My youngest reader Lia sent me this quote from Bertrand Russell for the blog.

There is much pleasure to be gained from useless knowledge.

It made me think of another great quote Jean-Marc introduced me to from Russell's autobiography. (BTW, Russell had a very unhappy youth.)

There was a footpath leading across fields to New Southgate, and I used to go there alone to watch the sunset and contemplate suicide. I did not, however, commit suicide, because I wished to know more mathematics.

It suggests that all anyone seriously depressed needs is a little more math. I'm not sure geometry is ready to replace Prozac, but if it worked for Russell, it's worth a shot.

I googled Russell quotes because I forgot the exact wording on Jean-Marc's quote, and I found out that Russell was a master of the form. QuotationsPage had 63 entries for Russell. Here are some good ones.

Everything is vague to a degree you do not realize till you have tried to make it precise.

This beautifully echoes the wonderful failure of his project Principia Mathematica.

If there were in the world today any large number of people who desired their own happiness more than they desired the unhappiness of others, we could have paradise in a few years.

I have to believe he's exaggerating for effect, but let's hear it for a little enlightened self-interest! I could go on forever, but I'll end on this one about teaching.

Passive acceptance of the teacher's wisdom is easy to most boys and girls. It involves no effort of independent thought, and seems rational because the teacher knows more than his pupils; it is moreover the way to win the favour of the teacher unless he is a very exceptional man. Yet the habit of passive acceptance is a disastrous one in later life. It causes man to seek and to accept a leader, and to accept as a leader whoever is established in that position.

pic by Gisela Giardino

Geek of the Week: Colleen

Listen to this:


I think the geekiness explains itself. If you need more convincing...Colleen took time over vacation to memorize it, just so she could recite it for Sue and me after break. Not only is it geeky, it's sweet and geeky. Thanks for making my pi-day, Colleen!

Geek of the Week: Lee Bradley

Welcoming the newest member to our math community Lee Bradely. Lee has just been hired on as a full time math tutor in the Academic Support Center. He gave me a great problem. You have to create 5 rectangles, such that the length of each side of one of the rectangles is an integer 1-10. You have to use all 10 numbers, so no sides of the same length. Furthermore, the resulting rectangles have to tile together to form a square. He has sworn there are 4 solutions and has given me one. Stop over in the ACS to say hello and get a hint from him. Also check out Lee's website. He uses mathematics to make beautiful puzzles.

Geek of the Week: Jesse Abbot

Jesse pulled off another great evening in his Proof & Possibility series. Barry Loewer was incredibly engaging as he walked us through the potential philisophic implications of quantum physics. I know nothing about physics and was riveted. When's the next one? (btw, it was packed make sure to get there early next time if you want a good seat.)

If you missed it, they should have a podcast of the lecture and Q&A pretty soon.

(Sorry about the pic Jesse. I have to get a better camera. I'm still at 4 Megapixels. That is nowhere near geeky enough.)

Geek of the Week: Anna and Will

This post is going to warm your heart. One, because we haven't had a geek of the week in a while, and two, it's a post about families sharing their geekiness...about the passing of primal geek energies from one generation to another. It's a post about continuity, tradition, and history. All of this in time for the holidays.

I ask you first to consider Sandy's son Will, decked out in full on geek regalia for Haloween.



This one is going straight to the head of the class. He has all the style he needs. (He also schooled me in Guitar Hero.)

Now check out this other Geek prodigy. You're about to listen to a voicemail Anna left her mother, Sue, at work. Anna gets so excited doing her math homework that she has to call her mother to sing the digits of pi. Have you ever heard such irrepressible joy at the discovery of irrational numbers?

Geek of the Week: Barack Obama

This seemed really timely. The Freaking News is running a photoshop contest for the best photoshop doctored photo's in a geek-wise manner. There are some really funny ones. Unfortunately, there' no McCain photo. Here's an old friend though.

Geek of the Week: Sue

Sue released her inner geek in an all out effort to win this year's Halloween costume contest. This is our best chance ever. I know she'll bring top prize back to the department. Great job squared, Sue!

I know a story that is sad but true
about a geek that I once knew.
She took my calculator and ran around
with every single theorem in town.

Geek of the Week-Charles Hartman

Pi mnemonics have been around forever to help people remember the irrational number pi to a desired number of places. Here's an example:
"How I wish I could calculate pi."
The first word has three letters, the second, one, the third, four...giving us the digits of pi 3.141592 to 6 decimal places. The idea has been appropriated as a poetic form by Conn College professor/poet Charles Hartman, who starts his collection Island off with a pi mnemonic, possibly the longest in the world. Island is available from Ahsahta Press for the curious.

Geek of the Week: Sam Loyd

It is really hard to come up with a good math problem. I'm not talking about the routine problems that constitute the homework questions in math classes. I mean the kind of puzzle that piques your curiosity/keeps you up at night/won't let you think about anything else. Sam Loyd didn't further mathematical theory and won't go down in history with names like Gauss, Cantor, and Newton, but he had a special talent for creating this type problem. Try the one above, his famous Donkey Puzzle. The instructions are

To solve it, cut out the three rectangles and reassemble the pieces so that the two jockeys are riding the two donkeys.


Care to try anymore: scans of Sam Loyd's Cyclopedia are available online.

Geek of the Week: Christian

This smug young man holding the cup of coffee is our new geek of the week. Everybody wants to use shortcuts in math. The dilemma for math teachers is often short cuts are used improperly resulting in tons of errors. Christian had one such short cut picked out for finding the y-intercept of any line from the slope and a point on the line. I assured him the technique was sound, but warned him that he was likely to screw up if the numbers given weren't nice. He didn't look worried at all, which piqued me, so I threw up a difficult example on the board and bet him a cup of coffee he couldn't find the y-intercept. Of course, he did. This is a picture of him enjoying my humiliation. By the way, if any of you want to congratulate him on his geek eminence, he takes hazelnut, cream, and sugar.

Geek of the Week: GIMPS

The mathematical news of the week is that a new Mersenne prime of over 10 million digits has been found. It was found by an organization called GIMPS (The Great Internet Mersenne Prime Search). It is really the collective effort of people donating spare computing power to search for primes. The hunt for large primes has been a long-time fascination of mathematicians, and recently has become important to government and commerce because of RSA public key encryption. There is no known algorithm for generating ever higher primes, so we have to keep checking numbers. This gets pretty tedious as numbers get large, but you can donate your spare computing power to help out. Just visit GIMPS and see how your computer can help.

Geek of the Week: Danica McKellar

Bringing glam back (or maybe "to") math. Not only does she have geek cred with the The Chayes-McKellar-Winn Theorem, but she is spreading the math love with her book, Math Doesn't Suck. I couldn't agree more.