Wild About Math! Making Math fun and accessible

30Oct/094

MMM #38: We have a winner!

Random.org selected Ken Lareau as the winner for MMM #38. Congratulations, Ken! Here's what the problem was:

Prove or disprove: The product of any four consecutive integers is always one less than a perfect square.

Don’t assume the integers are all positive.

Here's Ken's solution:

The above conjecture is true. The domain can be simplified a bit by noting two things:

1) If there are both positive and negative integers in the product, then since they're consecutive, 0 has to be present, meaning the product is 0, which is one less than 1, a perfect square.

2) If all the numbers are negative, their product is positive and equal to the product of the negation of each number (in other words, (-a) * (-b) * (-c) * (-d) = a * b * c * d).

Therefore we can restrict our validation to consecutive positive integers. From this, the proof is easy to see since if the first number is 'n', then we have for the product:

n * (n + 1) * (n + 2) * (n + 3)

which expanded is:

n^4 + 6n^3 + 11n^2 + 6n

One more than this is:

n^4 + 6n^3 + 11n^2 + 6n + 1

which happens to factor down to:

(n^2 + 3n + 1)^2

which is a perfect square. Hence the conjecture holds for all possible products.

I have one comment on the problems and on the solutions submitted. Many of the submissions ignored the part about not assuming all the integers to be positive. While it may be trivial to many of you that the signs of the numbers doesn't affect the solution, if you ignore that part of the problem then you're not solving the whole problem. Ken did a superb job of showing the signs don't matter. Yes, I realize that the algebraic solutions don't assume anything about the signs of the numbers but it's important to make that point in your argument.

And, here's a solution at Watchmath.com

29Oct/091

How much would you pay for a 146 year old Math book?

Every now and then I like to buy old Math books. I buy them on Ebay. Most of the books I buy are from the early 1900's or the tail end of the 1800's. Yesterday, I happened upon this book:

Yes, it's from 1863. Yes, I paid $6.50 for it and there was no shipping charge. What's the book about? Beats me. I'm not quite sure what Analytical Arithmetic Antique College Math is. If you think you know, leave a comment. What condition is the book in? Dunno but for six and a half bucks I'm not complaining!

Filed under: Books 1 Comment
29Oct/090

‘Dismantling the calculus pyramid’ gets ton of views

A couple of days ago I posted a very brief article, Dismantling the calculus pyramid. The post has been seen 2,746 times, thanks mostly to being noticed by 'timwiseman' who posted a link to the article at Y Combinator Hacker News. The posting at Y Combinator has gotten 45 comments while, here, it's gotten four comments. The conversation at Y Combinator is quite lively. Check it out.

Filed under: Education No Comments
27Oct/096

Dismantling the calculus pyramid

Here's a quick 3 minute TED video by Mathemagician Arthur Benjamin. The topic: Benjamin's idea about how to change Math education. He makes the point that Math education is like a pyramid with all classes (e.g. algebra, geometry, trigonometry) building up to calculus. But, he argues, calculus is not very useful to many of us in our ordinary lives. So, what should the pinnacle of the Math pyramid be? Watch the video and leave your comments.

Filed under: Calculus 6 Comments
13Oct/091

Review: Mythematics

From time to time a publisher offers me a review copy of a book. I always say yes unless I don't think I'll find something positive to say about the book. If, when I receive the book, I can't find something good to say then I won't review it.

Princeton University Press offered me a copy of a new book, Mythematics: Solving the 12 Labors of Hercules by Michael Huber. Before getting into the Math of the book I should give a little bit of background on the mythology. What are the 12 labors of Hercules? Wikipedia comes to the rescue:

The Twelve Labours of Hercules (Greek: Δωδεκαθλος, dodekathlos) are a series of archaic episodes connected by a later continuous narrative, concerning a penance carried out by the greatest of the Greek heroes, Heracles, romanised as Hercules. The establishment of a fixed cycle of twelve labours was attributed by the Greeks to an epic poem, now lost, written by Peisander, dated about 600 BC (Burkert).

As they survive, the Labours of Hercules are not told in any single place, but must be reassembled from many sources. Ruck and Staples assert that there is no one way to interpret the labours, but that six were located in the Peloponnese, culminating with the rededication of Olympia. Six others took the hero farther afield. In each case, the pattern was the same: Hercules was sent to kill or subdue, or to fetch back for Hera's representative Eurystheus a magical animal or plant. "The sites selected were all previously strongholds of Hera or the 'Goddess' and were Entrances to the Netherworld".

The gist of Hercules' mission is in the last paragraph above. "Hercules was sent to kill or subdue, or to fetch back for Hera's representative Eurystheus a magical animal or plant."

So, what does Hercules' adventures have to do with Math? Mr. Huber uses the 12 labors as a backdrop with which to present inventive Math problems. How might Hercules have used mathematics to accomplish each of the labors? Labor 1, for example, involves trapping and killing a lion.

And when the lion took refuge in a cave with two mouths, Hercules built up the one entrance and came in upon the beast through the other, and putting his arm round its neck held it tight till he had choked it; so laying it on his shoulders he carried it to Cleonae.

Here's an inventive mathematical problem derived from the story.

To defeat the lion, Hercules must close up one cave entrance and attack the lion through the other. He finds several stacks of tiles nearby, each of which contains sets of regular polygons. There is one stack of equilateral triangles, one stack of squares, one stack of regular pentagons, one stack of regular hexagons, and one stack of regular octagons. Which stack(s) of polygons will allow Hercules to construct an edge-to-edge tiling in order to close up the mouth of the cave with no two tiles overlapping?

This problem is not too difficult for someone comfortable with the basics of angles within regular polygons.

Not all the problems in the book are this easy. Many are tougher and require a fair amount of engagement and mathematical sophistication to grasp and solve. Also, I wouldn't call this book a recreational Math book because I consider recreational Math to consist of problems which are accessible to students with a high school Math background. Mythematics has problems that span these branches of mathematics:

algebra, combinatorics, difference equations, differential calculus, differential equations, geometry, integral calculus, multivariable calculus, probability, simulations, statistics, and trigonometry

I do think that Mythematics would be enjoyable to college educated folks, or to high school students who learn and enjoy calculus. In particular, I think engineering students in college who like Math would appreciate the problems in this book.

Filed under: Book Review 1 Comment