Wild About Math! Making Math fun and accessible


New James Tanton video on sums of cubes

Have you ever wondered why the sum of the cubes of consecutive positive integers is always a square? The key to one visual proof lies in the humble multiplication table and in an array of square dots.

Here's a new video from James Tanton that shows in a remarkably elegant way that 1^3 + 2^3 + 3^3 + ... + n^3 = (1+2+3+...+n)^2.

And, you don't need to have very much of a background in Math to follow the proof. Absolutely amazing!


Filed under: Algebra, Beauty 4 Comments

12 cent Math trick

Here's a very interesting and simple trick to impress your friends. It's simple enough to do that even the little ones, if they can count to 12, can do this trick. There's no sleight of hand or any other difficult manipulation to do. In fact, once you know the very simple steps, the trick is pretty automatic.

The trick is based on some very simple algebra but I've found that even though I understand how the trick works it's still eerie to see it work.

I got this trick from Shecky's great Math-Frolic Blog. The trick was developed by Alfred Posamentier, a prolific writer of Math books. While Posamentier is not as well known as Martin Gardner or Cliff Pickover, his books are equally engaging.

I rewrote Posamentier's trick so that it could be performed as a bar trick so that maybe Scam School will pick up the idea and produce a video of it.

Here's my version of the trick:

The scamster hands the victim 12 pennies and then blindfolds himself. The victim is instructed to place the coins on the table such that exactly five of them are heads up. The scamster tells the victim that he can, without removing the blindfold, separate the 12 pennies into two groups, and turn over some pennies so that each group will have exactly the same number of coins heads up. Fumbling around because he can't see, the scamster moves the 12 pennies close together into a group and then somehow picks some of the pennies and moves them to another group. He then turns some coins over and, voila, both groups have the same number of heads up pennies.

How can the scamster know how to separate the coins into two piles? How does he know which coins to turn over? How can he do this all blind-folded? See if you can figure out the trick on your own then head on over to Shecky's blog for Posamentier's solution.

Filed under: Algebra, Trick 3 Comments

William Wallace creates a fun Math puzzle

William Wallace sent me the following email so I'm inspired to promote his puzzle:

I came across your math blog, and thought you might be interested in a puzzle I came up with.

To see how it works, go to (http://blog.coincidencetheories.com/?p=1522). You probably can figure it out first, then go verify. You can blog on it. I released it to the public domain so you don't need to give credit, but you may. Leave a pingback--I honor them. The above website has a pdf version.

How to play:

In order to calculate the number your child picked, you must be able to add two digit numbers in your head. If you can do this, think about the final sum, and start off by setting it to 0 in your mind. As you show each page, if your child picks a blue(cross) or green(square), find the lowest value on the page that is blue(cross), or green(square), whichever she picked, and add that to your sum. If your child picked red(circle), simply add zero (0) to your sum. Repeat this for each of the next three pages. After the forth page, the sum in your mind is the number that your child picked. Announce it. Your child will think you're a mind reader.

Filed under: Algebra No Comments

MMM #25: Monotonous Monday Math Madness

[ See below for MMM #25 ]

Blinkdagger has announced the winner for MMM #24 and I've got a new problem so keep reading.

Rich Berlin noticed the relationship to MMM #1 and submitted an interesting solution, based on Pat Ballew's Markov chain process which I wrote about in March.

The technique for solving this is the same as for MMM #1, except the numbers are a bit nastier to work with.

I solved MMM #1 using a decision tree, but as a result of reading Sol's blog I learned the Markov state approach and am happy to use that instead.


A simple but disturbing Math problem

My brother gave me a Math problem this morning. He and I both know how to solve it but we're both disturbed by the fact that the right answer seems unintuitive. I'm interested to see if one of you can explain the answer in a way that is intuitive.

Here's the problem: A car gets 50 miles per gallon for a 10 mile stretch of a trip and 40 miles per gallon for the next 10 mile stretch of the trip. What's the average miles per gallon for the two stretches of the trip?

Again, the point isn't to get the right answer. Go ahead and solve it, though, for your personal satisfaction. The point is how one can see that the answer isn't 45 miles per gallon.


Filed under: Algebra 23 Comments

Math doesn’t suck

When the Penguin Group publishing company contacted me to see if I was interested in reviewing Danica McKellar's popular books, Math Doesn't Suck, and Kiss My Math, I jumped at the chance. No, it wasn't for the free books. The time I spend reading and writing doesn't justify the cost savings. I review products I believe in. Period. For the sake of full disclosure, I received a free copy of each of the two books. That's it.

Danica McKellar is a well-known actress, a mathematician, and an advocate for Math education. I'm delighted to see people with a tremendous amount of influence use that influence to make Math more accessible.


MMM #15: We have a winner!

Richard Berlin, who has been participating in Monday Math Madness! for quite a while, was picked as the winner for this contest by Random.org. Congratulations, Richard! I'll be sending you your prize so send me your address. Stand by for another Monday Math Madness on Monday at Blinkdagger.


Monday Math Madness #11

MMM #11 is a variation on MMM #9. I promise I won't do any more variations on this problem after this one!

Consider all of the 6-digit numbers that one can construct using each of the digits between 1 and 6 inclusively exactly one time each. 123456 is such a number as is 346125. 112345 is not such a number since 1 is repeated and 6 is not used.

How many of these 6-digit numbers are divisible by 11?

While you may use a computer program to verify your answer, show how to solve the problem without use of a computer.

MMM #9 was interested in divisibility by 8. This contest is interested in divisibility by 11.

I have a Rubik’s Revolution, courtesy of Techno Source, (or $10 Amazon.com gift certificate) to give to the winner. I’ll give more than one prize if I get lots of correct submissions.

I've changed rule #9 to encourage original solutions, which I'm much more likely to acknowledge:

I may post names and website/blog links for people submitting timely correct well-explained solutions. I'm more likely to post your name if your solution is unique.

Here are the rules for the contest:

1. Email your answers with solutions to mondaymathmadness at gmail dot com.
2. Only one entry per person.
3. Each person may only win one prize per 12 month period. But, do submit your solutions even if you are not eligible.
3. Your answer must be explained. You must show your work! Wild About Math! and Blinkdagger will be the final judges on whether an answer was properly explained or not.
4. The deadline to submit answers is Tuesday, July 29, 2008 12:01AM, Pacific Time. (That’s Tuesday morning, not Tuesday night.) Do a Google search for "time California" to know what the current Pacific Time is.)
5. The winner will be chosen randomly from all timely well-explained and correct submissions, using a random number generator.
6. The winner will be announced Friday, August 1, 2008.
7. The winner (or winners) will receive a Rubik’s Revolution or a $10 gift certificate to Amazon.com. For those of you who don't want a prize I'll donate $10 to your favorite charity.
8. Comments for this post should only be used to clarify the problem. Please do not discuss ANY potential solutions.
9. I may post names and website/blog links for people submitting timely correct well-explained solutions. I'm more likely to post your name if your solution is unique.

Filed under: Algebra, Fun, Game, Puzzle 1 Comment

Ti-Nspire inspires Math students

A while ago I received an email out of the blue from Texas Instruments (TI). One of their marketing people discovered this blog and offered to send me a TI-Nspire calculator to review. I quickly accepted, after all, who would turn down a free fancy calculator, right? Once I received the calculator I realized that this was no ordinary calculator; it was a visual Math learning system. I did nothing with it for a couple of months until I finally realized that I was not the best person to review it as it would take me quite a bit of time and effort to learn and appreciate its power. Sure, I could read the manual and run some demos but I didn't think that would give me enough experience to write a very in-depth review.

In discussing my challenge with TI, I learned of some teachers who were successfully using the TI-Nspire in the classroom. One person in particular, Eric Butterbaugh, was teaching Math in Harlem, New York. It occurred to us in that conversation that readers of this blog would appreciate hearing about Mr. Butterbaugh's success with the Ti-Nspire system. I created some interview questions and received back the interview you're about to read.


Monday Math Madness #7

Monday Math Madness

It's time for Monday Math Madness #7. I love infinite series and I found today's infinite series problem on the web. This is one of the most interesting of these kinds of problems I have run into. It's challenging but not brutally difficult, so give it a try. I won't reveal the source until the contest ends because the answer is posted with the problem.

Thanks to the sponsors for this contest, I have one $25 gift certificate left for the Art of Problem Solving. I also have a couple of Rubik's Revolutions, courtesy of Techno Source. Depending on how many correct solutions I get I may give away two prizes.