Dec 14

Did you know?

Here is a fun fact to demonstrate to your students:

Think of two positive integers.
Add them to get a third number.
Add the second number and the third number to get a fourth number.
Continue in this way until you have 10 numbers.
The sum of the 10 numbers is 11 times the seventh number.

found via Futility Closet

Oct 13

Infinite Chocolate

chocolate bar

This gif (double click to play) came up in my google+ feed this morning and was bugging me out. Luckily I found this post to explain the trick. The area of the missing piece is sufficiently spread out to make the the smaller area seem visually similar to the original area.

Aug 12

Math + Mind Tickle

I place $20 in a box.
So do you.
Now the box contains $40, and we both know it.
I sell the box to you for $30.
And we both walk away with a $10 profit.

Jay sent over a link this great problem. It’s from one of my favorite blogs Futility Closet. This is the perfect kind of problem to throw out in class…easy to remember and easy to think through in a few minutes. What’s lovely is that initially it seems so plausible and yet impossible that they both profit $10. It reminds me of a Nova show (embedded below) I was watching on neuroscientists studying magic. Most tricks rely on our visual system’s strong bias toward detecting and anticipating motion. I feel like this problem and ones like it trick our reason, possibly with the momentum of language? Whoa, that got deep! Better stop right there.

Watch Magic and the Brain on PBS. See more from NOVA scienceNOW.

Apr 12

Miura Fold

miura fold
I came across the Miura Fold on Lifehacker. It basically allows one to fold/unfold something with a single motion. The easiest way to think about the value of the Miura fold is in terms of a map. I can’t count the number of maps I’ve screwed up by not folding it back up the right way. Unfortunately it’s fairly hard to learn the fold. I have to build up those origami skills.

Nov 11

Math + Unblievable

A couple of my students told me about this guy–the human calculator. There are a lot of tricks to doing mental calculations quickly. This guy is unbelievable though.

Thanks Dan and Matt!

Oct 11

Pets + Math

math dogBeau the math labrador. Too weird to be true, or can dogs really do arithmetic?

Mar 11

Pi/Pie Day

pi day puz

Jay sent this over to me from The Puzzler. It took me a second to get it, but ended up giving me a big smile. Happy Pi Day everyone!

Feb 10

Cuttin’ Cards

One of my old students asked me to explain this card trick. If you go to the comments on the dailymotion page, you’ll see a couple of other explanations. The trick breaks down in to two sections: stacking the deck and sifting cards.

Stacking the deck
Though the aces look semi-randomly inserted into the deck, it is only an illusion. The aces will always wind up in positions 16, 32, and 48 in the stack no matter how one cuts the cards. Follow the bouncing cards…
Here’s the original setup:
(pile 1)(pile 2)(pile 3) (left overs)
Let’s insert the aces. We put the first ace on pile 1:
(1st ace)
(pile 1)     (pile 2)(pile 3) (left overs)
Now we cut pile 2 and make a space for the 2nd ace in pile 2:
(top of pile 2)
(1st ace)           (2nd ace)
(pile 1)              (bottom of pile 2)(pile 3) (left overs)
Now we do the same thing to pile 3, cut it and insert the third ace.
(top of pile 2)    (top of pile 3)
(1st ace)              (2nd ace)                (3rd ace)
(pile 1)                 (bottom of pile 2)(bottom of pile 3) (left overs)
Now they all get stacked on top of each other in this way:
(left overs)
(3rd ace)
(bottom of pile 3)
(top of pile 3)
(2nd ace)
(bottom of pile 2)
(top of pile 2)
(1st ace)
Then the first 4 left overs are put on the bottom (remembering there are 9 left overs all together):
(last 5 left overs)
(3rd ace)
(bottom of pile 3)
(top of pile 3)
(2nd ace)
(bottom of pile 2)
(top of pile 2)
(1st ace)
(pile 1)
(1st four left overs)
It is easy to see that the 3rd ace is the 6th card in the stack. It’s a little harder to see that the 2nd ace is 22nd and the 1st ace is 38. But remember, pile 2 and pile 3 each had 15 cards, so if the bottom of pile 3 sits on top of the top of pile 3 that would still 15 cards. The same is true for pile 2. We also know that pile 1 had 10 cards in it. So we can simplify the stack.
(last 5 left overs)
(3rd ace)
(15 cards)
(2nd ace)
(15 cards)
(1st ace)
(10 cards)
(1st four left overs)
Now it is easy to see our aces are in the 6th, 22nd, and 38th position. So the cutting of decks is really just slight of hand.

Sifting the cards
In the second part of the trick, the man goes through the deck and throws out every other card four times. We need to show the result of the process will always yield the 6th, 22nd, and 38th cards.
The first sifting out is easy to follow. The man turns the first card up, the second card down, the third card up… The result is all the odd cards are thrown out. We have just the even cards left, but they’re stacked from high to low, so…
Now the process is repeated, 52 is up, 50 is down…. So that we end up throwing out all the multiples of 4 and reordering the deck from low to high.
Repeating the process again will throw out all those numbers that have a remainder of 2 when divided by 8. (BTW, there is an interesting thing going on with modular arithmetic here.)
We do it once more and throw out everything that has a remainder of 14 when divided by 16. We get the desired:

Sorry it took so long, Jay!

Nov 09

Cocktail Party Math

If there is anyone reading this blog thinking about teaching math, there are papers to grade, emails to answers, and meetings to go to. It isn’t all as glamorous as it looks. But you occasionally get a day when a colleague, perhaps Paul Argazzi, drops in and gives you a great new math trick. That’s when you know you’ve made the right choice. Here’s the trick. My choices are indicated in parentheses.

Paul walks in and says, give me a book. I hand him one from the shelf. He leaves through a couple of pages, gives it back to me, and says, pick a three digit number. (439) I do. He says to make sure the 100’s and 1’s digit are two or more apart. I got lucky and don’t have to re-pick. Okay, he says, take that number and flip it around so that the 100’s becomes the 1’s and vice versa. (934) Now subtract the smaller (439) from the bigger number (934), and call the result A (495). Then, he said, take A (495), spin it again, call that number B (594), now add A to B (1089). Then he told me to ignore the ones place and take the digits in front of the ones place (108). He said find that page in the book. I did. Then he told me to take the number in the ones place (9) and find that number word in the first line of the page. (So I looked for the ninth word in the first line of the page. It read “meaning”.) Then he asked me, is the word “meaning”?

Pretty cool, huh? So here’s the question. How does it work?