I’m doing something that I should have done a long time ago: collect past series of posts into a single, easy-to-reference post. The following posts formed my series on various lessons I’ve learned while trying to answer the questions posed by gifted elementary school students. (This is updated from my previous index.)

Part 1: A surprising pattern in some consecutive perfect squares.

Part 2: Calculating 2 to a very large exponent.

Part 3a: Calculating 2 to an even larger exponent.

Part 3b: An analysis of just how large this number actually is.

Part 4a: The chance of winning at BINGO in only four turns.

Part 4b: Pedagogical thoughts on one step of the calculation.

Part 4c: A complicated follow-up question.

Part 5a: Exponentiation is multiplication as multiplication is to addition. So, multiplication is to addition as addition is to what? (I offered the answer of incrementation, but it was rejected: addition requires two inputs, while incrementation only requires one.)

Part 5b: Why there is no binary operation that completes the above analogy.

Part 5c: Knuth’s up-arrow notation for writing very big numbers.

Part 5d: Graham’s number, reputed to be the largest number ever to appear in a mathematical proof.

Another poorly written word problem (Part 3)

Textbooks have included the occasional awful problem ever since Pebbles Flintstone and Bamm-Bamm Rubble chiseled their homework on slate tablets while attending Bedrock Elementary. But even with the understanding that there have been children have been doing awful homework problems since the dawn of time (and long before the advent of the Common Core), this one is a doozy.

There’s no sense having a debate about standards for elementary mathematics if textbook publishers can’t construct sentences that can be understood by elementary students.

Careers in Industry for Math Majors

I recently came across an excellent article promoting internships from math majors who would like to use their quantitative skills in an industrial setting (as opposed to an academic setting). The concluding paragraph:

Faculty will continue to train students for academic careers. Some will pursue tenure-track positions in the institutions of their choice, but an increasing number of our students will take positions very different from our own. Let’s learn about those options and share them with our students. Then, when a student takes a good job and enjoys a successful career, let’s call that a win.

Here’s the full article: http://www.americanscientist.org/blog/pub/internships-connect-math-students-to-new-career-paths

Teachers Who Are Funnier Than Their Students

I really enjoyed this, courtesy of BuzzFeed:

My Mathematical Magic Show: Part 8c

This mathematical trick was not part of my Pi Day magic show but probably should have been… I’ve performed this for my Precalculus classes in the past but flat forgot about it when organizing my Pi Day show. The next time I perform a magic show, I’ll do this one right after the 1089 trick. (I think I learned this trick from a Martin Gardner book when I was young, but I’m not sure about that.)

Here’s a description of the trick. I give my audience a deck of cards and ask them to select six cards between ace and nine (in other words, no tens, jacks, queens, or kings). The card are placed face up, side by side.

After about 5-10 seconds, I secretly write a pull out a card from the deck and place it face down above the others.

I then announce that we’re going to some addition together… with the understanding that I’ll never write down a number larger than 9. For example, the 4 and 6 of spades are next to each other. Obviously, $4+6 = 10$ , but my rule is that I’m going to write down a number larger than 9. So I’ll subtract 9 whenever necessary: $10-9 = 1$ . Since 1 corresponds to ace, I place an ace about the 4 and 6 of spades.

Continuing in this way (and having the audience participate in the arithmetic so that this doesn’t get boring), I eventually get to this position:

Finally, I add the two cards at the top (and, in this case, subtract 9) to get $6+9-9 = 6$ , and I dramatically turn over the last card to reveal a 6.

I’ll often perform this trick when teaching Precalculus, as the final answer involving Pascal’s triangle. As discussed yesterday, suppose that the six cards are $a$ , $b$ , $c$ , $d$ , $e$ , and $f$ . Forgetting for now about subtracting by 9, here’s how the triangle unfolds (turning the triangle upside down):

$a \qquad \qquad \qquad \quad b \qquad \qquad \qquad \quad c \qquad \qquad \qquad \quad d \qquad \qquad \qquad \quad e \qquad \qquad \qquad \quad f$

$a+b \qquad \qquad \qquad b+c \qquad \qquad \qquad c+d \qquad \qquad \qquad d+e \qquad \qquad \qquad e+f$

$a+2b+c \qquad \qquad b+2c+d \qquad \qquad c+2d+e \qquad \qquad d+2e+f$

$a+3b+3c+d \qquad \quad b+3c+3d+e \qquad \quad c+3d+3e+f$

$a+4b+6c+4d+e \qquad b+4c+6d+5e+f$

$a+5b+10c+10d+5e+f$

Not surprisingly, the coefficients in the above chart involve the numbers in Pascal’s triangle. Indeed, the reason that I chose to use 6 cards (as opposed to any other number of cards) is that the bottom row has only 1, 5, and 10 as coefficients, and $10 \equiv 1 (\mod 9)$ . Therefore, the only tricky part of the calculation is multiplying $b+e$ by $5$ , as the final answer can then be found by adding the remaining four numbers.

My students usually find this to be a clever application of Pascal’s triangle for impressing their friends after class.

P.S. After typing this series, it hit me that it’s really easy to do this trick mod 10 (which means getting rids of only the face cards prior to the trick). All the magician has to do is subtly ensure that the second and fifth cards are both even or both odd, so that $b+e$ is even and hence $5(b+e)$ is a multiple of 10. Therefore, since $10c+10d$ is also a multiple of 10, the answer will be just $a+f$ or $a+f-10$ .

(If the magician can’t control the placement of the second and fifth cards so that one is even and one is odd, the answer will be just $a+f+5$ or $a+f-5$ .)

Henceforth, I’ll be doing this trick mod 10 instead of mod 9.

My Mathematical Magic Show: Part 8b

After about 5-10 seconds, I secretly write a pull out a card from the deck and place it face down above the others.

Continuing in this way (and having the audience participate in the arithmetic so that this doesn’t get boring), I eventually get to this position:

Finally, I add the two cards at the top (and, in this case, subtract 9) to get $6+9-9 = 6$ , and I dramatically turn over the last card to reveal a 6.

How does this trick work? This is an exercise in modular arithmetic (see also Wikipedia). Suppose that the six cards are $a$ , $b$ , $c$ , $d$ , $e$ , and $f$ . Forgetting for now about subtracting by 9, here’s how the triangle unfolds (turning the triangle upside down):

$a \qquad \qquad \qquad \quad b \qquad \qquad \qquad \quad c \qquad \qquad \qquad \quad d \qquad \qquad \qquad \quad e \qquad \qquad \qquad \quad f$

$a+b \qquad \qquad \qquad b+c \qquad \qquad \qquad c+d \qquad \qquad \qquad d+e \qquad \qquad \qquad e+f$

$a+2b+c \qquad \qquad b+2c+d \qquad \qquad c+2d+e \qquad \qquad d+2e+f$

$a+3b+3c+d \qquad \quad b+3c+3d+e \qquad \quad c+3d+3e+f$

$a+4b+6c+4d+e \qquad b+4c+6d+5e+f$

$a+5b+10c+10d+5e+f$

Therefore, the top card will simply be $a+5b+10c+10d+5e+f$ minus a multiple of 9.

That’s a pretty big calculation for the magician to do on the spot. Fortunately, $9c + 9d$ is also a multiple of 9, and so the top card will be

$a+5b+10c+10d+5e+f - (9c + 9d)$ minus a multiple of 9, or

$5(b+e) + a + c + d + f$ minus a multiple of 9.

For the case at hand, $b = 6$ and $e =8$ , so $5(b+e) = 70$ . That’s still a big number to keep straight when performing the trick. However, since I’m going to be subtracting 9’s anyway, I can do this faster by replacing the 8 by $8 - 9 = -1$ . So, for the purposes of the trick, $5(b+e) = 5 \times (6-1) = 25$ , and I subtract $18$ to get $7$ .

I now add the rest of the cards, subtracting 9 as I go along. For this example, I’d add the 2 first to get 9, which is 0 after subtracting another 9. I then add the remaining cards of 4, 3, and 8 (remembering that the 8 is basically $8-9 = -1$ , yielding $4+3-1 = 6$ . So the top card has to be 6.

The key point of this calculation is to subtract 9 whenever possible to keep the numbers small, making it easier to do in your head when performing the trick.

My Mathematical Magic Show: Part 8a

After about 5-10 seconds, I secretly write a pull out a card from the deck and place it face down above the others.

Next, I consider the 6 of spades and 2 of diamonds. Adding, I get 8. That’s less than 9, so I pull an 8 out of the deck.

Next, $2+3 = 5$ , so I pull out a 5 from the deck.

Next, $8+8=16$ , and $16-9=7$ . So I pull out a 7.

(To keep this from getting dry, I have the audience perform the arithmetic with me.)

On the the next row. The next cards are $1+8 = 9$ , $8+5-9 = 4$ , $5+2 =7$ , and $2+7 = 9$ .

On the the next row. The next cards are $9+4-9=4$ , $latex $4+7-9 = 2$, and $7+9-9 = 7$ .

Almost there: $4+2 = 6$ and $2+7= 9$ .

Finally, $6+9-9 = 6$ , and I dramatically turn over the last card to reveal a 6.

Naturally, everyone wonders how I knew what the last card would be without first getting all of the cards in the middle. I’ll discuss this in tomorrow’s post.

My Mathematical Magic Show: Part 7

This mathematical trick, which may well be the best mathematical magic trick ever devised, was not part of my Pi Day magic show. However, it should have been. Here’s a description of the trick, modified from the description at http://mathoverflow.net/questions/20667/generalization-of-finch-cheneys-5-card-trick:

The magician walks out of the room. A volunteer from the crowd chooses any five cards at random from a deck, and hands them to your assistant so that nobody else can see them. The assistant glances at them briefly and hands one card back, which the volunteer then places face down on the table to one side. The assistant quickly place the remaining four cards face up on the table, in a row from left to right. After all of this is completed, the magician re-enters the room, inspects the faces of the four cards, and promptly names the hidden fifth card.

In turns out that the trick is a clever application of permutations (there are $3! = 6$ possible ways of ordering 3 objects) and the pigeon-hole principle (if each object belongs to one of four categories and there are five objects, then at least two objects must belong to the same category). These principles from discrete mathematics (specifically, combinatorics) make possible the Fitch-Cheney 5-Card Trick.

Unlike the other tricks in this series, the Fitch-Cheney 5-Card Trick requires a well-trained assistant (or a smartphone app that plays the role of the assistant).

A great description of how this trick works can be found at Math With Bad Drawings. For a deeper look at some of the mathematics behind this trick, I give the following references:

My Mathematical Magic Show: Part 6

Last March, on Pi Day (March 14, 2015), I put together a mathematical magic show for the Pi Day festivities at our local library, compiling various tricks that I teach to our future secondary teachers. I was expecting an audience of junior-high and high school students but ended up with an audience of elementary school students (and their parents). Still, I thought that this might be of general interest, and so I’ll present these tricks as well as the explanations for these tricks in this series. From start to finish, this mathematical magic show took me about 50-55 minutes to complete. None of the tricks in this routine are original to me; I learned each of these tricks from somebody else.

After presenting four different but thoroughly impressive mathematical magic tricks (including the explanations for each trick as well as a chance for a child in the audience to present the trick for themselves) over the past 50-55 minutes, I have now reached the climatic end of my routine.

Here’s the patter for my grand finale. I got this trick from a book of card tricks that my parents bought me when I was a boy. (The book remains one of my prized possessions from my childhood.) This trick requires an ordinary deck of playing cards — preferably a newly purchased, sealed, and unused deck of cards.

It’s now time for the final magic trick of the show. And every magic act has to include a magic trick involving… a deck of cards.

Now, when most magicians perform a magic trick, they have a deck of cards that are pre-arranged in a certain order. But I am not like most magicians. (I hand the deck to someone.) Here, please take the deck of cards and shuffle it a few times. (He shuffles the deck and gives it back to me.)

Now, when most magicians perform a magic trick, they have a confederate in the audience who helps me cheat by shuffling the deck in a certain way. (Laughter.) But I am not like most magicians. (I hand the deck to someone else.) Here, please take the deck and shuffle it a few more times. (She shuffles the deck some more and gives it back to me.)

Now, when most magicians perform a magic trick, they ask someone in the audience to take a card from a certain spot in the deck. But I am not like most magicians. (I hand the deck to a third person.) Here, please choose a card from anywhere in the deck, and show it to everyone. (The third person picks a card and shows it to everyone except me.)

Now, when most magicians perform a magic trick, they have the card returned to a certain spot in the deck. But…

(And the audience invariably says, “You are not like most magicians!”)

Oh, you’ve seen this trick before?

That’s right, I’m not like most magicians. So don’t return that card to a certain spot in the deck. Instead, place it anywhere in the middle you want. (The third person places the card back in the deck.)

Now, when most magicians perform a magic trick, they’ll do something special to the deck so that the card pops out. But I am not like most magicians. Take the deck, and shuffle it again, and then give me back the deck. (The third person shuffles the deck and returns it to me.)

Now, when most magicians perform a magic trick, they search through the deck to find the selected card. But I am not like most magicians. So I will place the deck behind my back and select your card.

(I place the deck behind my back, wait about 10 seconds as if I’m trying to find the card, select a card, and then show it to the audience with complete and utter confidence that I’ve found the right card.)

At this point in the routine, there is one chance in 52 that I drew the correct card. If that happens to happen, then I would take a deep bow and end the show. If someone asks how I did it, I would explain that magicians have to keep some tricks a secret.

However, I’ve performed this trick dozens of times over the last 30 years, and I have yet to select the correct card even once. So, in the highly likely event that I pull the wrong card, the audience will say something like, “No, that’s not it.” I turn to the audience with a straight face, shrug my shoulders, and say,

Most magicians find your card.

And that’s the end of the show as the audience howls in laughter.

It’s really important to perform this “trick” after performing several real magic tricks. Indeed, after seeing a few highly impressive mathematical magic tricks, the audience is expecting me to pull out the right card, and the deliberately repetitive patter above builds the tension in the room as the audience tries to figure out how on earth I’m going to pull out the correct card from a thoroughly shuffled deck.

It’s also important to get in touch with my inner Bud Abbott or Super Dave Osborne or any of the legendary straight men of comedy, as I can’t so much as smile during the routine lest I give away the joke.

If anyone complains, I explain that this was a mathematical magic trick… after all, the probability of me pulling the correct card is $\displaystyle \frac{1}{52}$ .

Mean Green Math

Geometree

Lessons from teaching gifted elementary school students: Index (updated)

Another poorly written word problem (Part 3)

Careers in Industry for Math Majors

Teachers Who Are Funnier Than Their Students

My Mathematical Magic Show: Part 8c

My Mathematical Magic Show: Part 8b

My Mathematical Magic Show: Part 8a

My Mathematical Magic Show: Part 7

My Mathematical Magic Show: Part 6