# My Mathematical Magic Show: Part 4c

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.

For my third trick, I’ll present something that I first saw when pulling Christmas crackers with my family. I’ll give everyone a piece of paper with six cards printed. I’ll also have a large version of this paper shown at the front of the room (taken from http://diaryofagrumpyteacher.blogspot.com/2014/04/freebie-friday-magic-number-cards.html; see also this Google search if this link somehow goes down):

Here’s the patter:

Think of a number from 0 to 63. Then, on your piece of paper, circle the cards that contain your number. For example, if your number is 15, you’ll need to circle the card in the upper-left because 15 is on that card. You’d have to circle all the cards that contain 15.

(pause)

Is everyone done? (Points to someone) Which cards did you circle?

At this point, the audience member will say something like “Top left, top middle, and bottom right.” Then I will add the smallest numbers on each card (in this case, 1, 2, and 32) and answer in five seconds or less, “Your number was 35 (or whatever the sum is).” It turns out that the number is always the sum of the smallest numbers on the selected cards.

In yesterday’s post, I gave a similar but utterly unimpressive trick; the trick was unimpressive because it was obvious that the trick used our ordinary base-10 representation of whole numbers. The trick above is much more impressive because it uses binary (base-2) instead of base-10.

The cards above are carefully rigged using binary arithmetic, so that all numbers are written as sums of powers of 2. For example, on the card in the upper left, the first few numbers are

$1 = 1$

$3 = 2 + 1$

$5 = 4 + 1$

$7 = 4 + 2 + 1$

$9 = 8 + 1$

$11 = 8 + 2 + 1$,

and so on. On the right-hand side, I’ve written each number as the sum of powers of 2 (for the numbers at hand, that means 1, 2, 4, 8, 16, and 32). Notice that each expansion on the right hand side contains a 1. So, if the audience member tells me that her number is on the upper-left card, that tells me that there’s a 1 in the binary representation of her number.

Let’s now take a look at the first few number in the upper-middle card:

$2 = 2$

$3 = 2+1$

$6 = 4+2$

$7 = 4+2+1$

$10 = 8+2$

$11 = 8+2+1$,

and so on. Notice that each expansion on the right hand side contains a 2. So, if the audience member tells me that her number is on the upper-middle card, that tells me that there’s a 2 in the binary representation of her number.

Similarly, the upper-right card has numbers which contain 4 in its binary representation. The lower-left card has numbers containing 8. The lower-middle card has numbers containing 16. And the lower-right card has numbers containing 32. Happily for the magician, each of these numbers is also the smallest number on the card.

So, if the audience member will says “Top left, top middle, and bottom right,” then I know that the binary representation of her number contains 1, 2, and 32. Adding up those numbers, therefore, gives me the original number!

After explaining how the trick works, I’ll call up an audience member to play the magician and repeat the trick that I just performed. Then I’ll move on to the next magic trick in the routine.

# My Mathematical Magic Show: Part 4b

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.

For my third trick, I’ll present something that I first saw when pulling Christmas crackers with my family. I’ll give everyone a piece of paper with six cards printed. I’ll also have a large version of this paper shown at the front of the room (taken from http://diaryofagrumpyteacher.blogspot.com/2014/04/freebie-friday-magic-number-cards.html; see also this Google search if this link somehow goes down):

Here’s the patter:

Think of a number from 1 to 63. Then, on your piece of paper, circle the cards that contain your number. For example, if your number is 15, you’ll need to circle the card in the upper-left because 15 is on that card. You’d have to circle all the cards that contain 15.

(pause)

Is everyone done? (Points to someone) Which cards did you circle?

At this point, the audience member will say something like “Top left, top middle, and bottom right.” Then I will add the smallest numbers on each card (in this case, 1, 2, and 32) and answer in five seconds or less, “Your number was 35 (or whatever the sum is).” It turns out that the number is always the sum of the smallest number on the given cards.

To explain this trick to my audience, I’ll present the following conceptually similar trick using 20 cards. I’ll ask the audience to pick a number between 0 and 99 and then find the cards that contain that number.

Suppose that the audience member tells me that her number appears on these two cards:

The first card tells me that the number is in the 70s; the last card tells me that the ones digit is 2. So the answer must be 72. Stated another way, I can add the smallest number on each card (70 + 2) to get the answer.

This magic trick looks utterly unimpressive because the trick is so obvious because base-10 arithmetic has been so utterly drilled into our heads since elementary school. So my audience is usually surprised to learn that the first magic trick, with the six cards with numbers from 1 to 63, is conceptually the same as this 0-99 trick. I’ll explain this in the next post.

# My Mathematical Magic Show: Part 4a

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.

For my third trick, I’ll present something that I first saw when pulling Christmas crackers with my family. I’ll give everyone a piece of paper with six cards printed. I’ll also have a large version of this paper shown at the front of the room (taken from http://diaryofagrumpyteacher.blogspot.com/2014/04/freebie-friday-magic-number-cards.html; see also this Google search if this link somehow goes down):

Here’s the patter:

Think of a number from 0 to 63. Then, on your piece of paper, circle the cards that contain your number. For example, if your number is 15, you’ll need to circle the card in the upper-left because 15 is on that card. You’d have to circle all the cards that contain 15.

(pause)

Is everyone done? (Points to someone) Which cards did you circle?

At this point, the audience member will say something like “Top left, top middle, and bottom right.” Then I will add the smallest numbers on each card (in this case, 1, 2, and 32) and answer in five seconds or less, “Your number was 35 (or whatever the sum is).”

Then I’ll ask someone else which cards they circled, and then tell him/her the number that he/she chose (by quietly and quickly adding the smallest numbers on the selected cards.) Then I’ll do this for a couple more people until everyone is convinced that I’m a genius.

Then I’ll start explaining why the trick works… which I’ll begin with tomorrow’s post.

# My Mathematical Magic Show: Part 3c

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.

In the last couple of posts, I discussed a trick for predicting the number of triangles that appear when a convex $x-$gon with $y$ points in the middle is tesselated. Though I probably wouldn’t do the following in a magic show (for the sake of time), this is a natural inquiry-based activity to do with pre-algebra students in a classroom setting (as opposed to an entertainment setting) to develop algebraic thinking. I’d begin by giving the students a sheet of paper like this:

Then I’ll ask them to start on the left box. I’ll tell them to draw a triangle in the box and place one point inside, and then subdivide into smaller triangles. Naturally, they all get 3 triangles.

Then I ask them to repeat if there are two points inside. Everyone will get 5 triangles.

Then I ask them to repeat until they can figure out a pattern. When they figure out the pattern, then they can make a prediction about what the rest of the chart will be.

Then I’ll ask them what the answer would be if there were 100 points inside of the triangle. This usually requires some thought. Eventually, the students will get the pattern $T = 2P+1$ for the number of triangles if the initial figure is a triangle.

Then I’ll repeat for a quadrilateral (with four sides instead of three). After some drawing and guessing, the students can usually guess the pattern $T=2P+2$.

Then I’ll repeat for a pentagon. After some drawing and guessing, the students can usually guess the pattern $T=2P+3$.

Then I’ll have them guess the pattern for the hexagon without drawing anything. They’ll usually predict the correct answer, $T = 2P+4$.

What about if the outside figure has 100 sides? They’ll usually predict the correct answer, $T = 2P+98$.

What if the outside figure has $N$ sides? By now, they should get the correct answer, $T = 2P + N - 2$.

This activity fosters algebraic thinking, developing intuition from simple cases to get a pretty complicated general expression. However, this activity is completely tractable since it only involves drawing a bunch of figures on a piece of paper.

# My Mathematical Magic Show: Part 3b

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.

This is a magic trick that my math teacher taught me when I was about 13 or 14. I’ve found that it’s a big hit when performed for grade-school children. Here’s the patter:

Magician: Tell me a number between 5 and 10.

Child: (gives a number, call it $x$)

Magician: On a piece of paper, draw a shape with $x$ corners.

Child: (draws a figure; an example for $x=6$ is shown)

Important Note: For this trick to work, the original shape has to be convex… something shaped like an L or M won’t work. Also, I chose a maximum of 10 mostly for ease of drawing and counting (and, for later, calculating).

Magician: Tell me another number between 5 and 10.

Child: (gives a number, call it $y$)

Magician: Now draw that many dots inside of your shape.

Child: (starts drawing $y$ dots inside the figure; an example for $y = 7$While the child does this, the Magician calculates $2y + x - 2$, writes the answer on a piece of paper, and turns the answer face down.

Magician: Now connect the dots with lines until you get all triangles. Just be sure that no two lines cross each other.

Child: (connects the dots until the shape is divided into triangles; an example is shown)

Magician: Now count the number of triangles.

Child: (counts the triangles)

The reason this magic trick works so well is that it’s so counter-intuitive. No matter what convex $x-$gon is drawn, no matter where the $y$ points are located, and no matter how lines are drawn to create triangles, there will always be $2y + x - 2$ triangles. For the example above, $2y+x-2 = 2\times 7 + 6 - 2 = 18$, and there are indeed $18$ triangles in the figure.

This trick works by counting the measures of all the angles in two different ways.

Method #1: If there are $T$ triangles created, then the sum of the measures of the angles in each triangle is $180$ degrees. So the sum of the measures of all of the angles must be $180 T$ degrees.

Method #2: The sum of the measures of the angles around each interior point is $360$ degrees. Since there are $y$ interior points, the sum of these angles is $360y$ degrees.

The measures of the remaining angles add up to the sum of the measures of the interior angles of a convex polygon with $x$ sides. So the sum of these measures is $180(x-2)$ degrees.

These two different ways of adding the angles must be the same. In other words, it must be the case that

$180T = 360y + 180(x-2)$,

or

$T = 2y + x - 2$.

I’m often asked why it was important to choose a number between 5 and 10. The answer is, it’s not important. The trick will work for any numbers as long as there are at least three sides of the polygon. However, in a practical sense, it’s a good idea to make sure that the number of sides and the number of points aren’t too large so that the number of triangles can be counted reasonably quickly.

After explaining how the trick works, I’ll again ask a child to stand up and play the magician, repeating the trick that I just did, before I move on to the next trick.

# My Mathematical Magic Show: Part 3a

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.

For my second trick, I’ll show something that my math teacher taught me when I was about 13 or 14. Everyone in the audience has a piece of paper and a pen or pencil. Here’s the patter:

Magician: Tell me a number between 5 and 10.

Child #1: (gives a number, call it $x$)

Magician: On a piece of paper, draw a shape with $x$ corners. Don’t draw something really, really tiny… make sure it’s big enough to see well.

Audience: (draws a figure; an example for $x=6$ is shown) The Magician also draws this figure on the board.

Important Note: For this trick to work, the original shape has to be convex… something shaped like an L or M won’t work. Also, I chose a maximum of 10 mostly for ease of drawing and counting (and, for later, calculating).

Magician: Tell me another number between 5 and 10.

Child #2: (gives a number, call it $y$)

Magician: Now draw that many dots inside of your shape.The Magician also draws $y$ dots inside the figure on the board, an example for $y = 7$ is shown.

Audience: (starts drawing $y$ dots inside the figure) The Magician also calculates $2y + x - 2$ and says, “Now while you’re doing that, I’m going to write a secret number on the board,” discreetly writes the answer on the board, and then covers up the answer with a piece of paper and some adhesive tape.

Magician: Now connect the dots with lines until you get all triangles. Just be sure that no two lines cross each other. For example, your figure could look like this:

Audience: (quietly connects the dots until the shape is divided into triangles)

Magician: Now count the number of triangles.

Audience: (counts the triangles)

The reason this magic trick works so well is that it’s so counter-intuitive. No matter what convex $x-$gon is drawn, no matter where the $y$ points are located, and no matter how lines are drawn to create triangles, there will always be $2y + x - 2$ triangles. For the example above, $2y+x-2 = 2\times 7 + 6 - 2 = 18$, and there are indeed $18$ triangles in the figure.

In tomorrow’s post, I’ll explain why this trick works.

# My Mathematical Magic Show: Part 2c

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.

For my first trick, I chose the most boring of the routine. Everyone in the audience had a piece of paper and many had calculators. Here is the patter for the first trick:

To begin this trick, write down any three-digit number on your piece of paper. Just make sure that the first digit and the last digit are different.

(pause)

Now, reverse the digits and write down a new number. For example, if your number was 321, the new number will be 123.

(pause)

Now, subtract the small number from the big number. If your second number is larger, then put that number on top so that you can subtract the two numbers.

(pause)

Your difference is probably a three-digit number. However, if you ended up with a two-digit number, you can make it a three-digit number by putting a 0 in the hundreds place.

Next, I want you to reverse the digits of the difference to make a new three-digit number. Write this new number under the difference.

(pause)

Finally, add the last two three-digit numbers that you wrote down.

If everyone follows the instructions and does the arithmetic correctly, everyone will get a final answer of 1,089.

The next part of my mathematical magic show is showing everyone why the trick works. Yesterday, I gave an explanation suitable for upper elementary students. Today, I’ll give a more abstract explanation using algebra.

The succinct explanation can be found on Wikipedia:

The spectator’s 3-digit number can be written as 100 × A + 10 × B + 1 × C, and its reversal as 100 × C + 10 × B + 1 × A, where 1 ≤ A ≤ 9, 0 ≤ B ≤ 9 and 1 ≤ C ≤ 9. (For convenience, we assume A > C; if A < C, we first swap A and C.) Their difference is 99 × (AC). Note that if AC is 0 or 1, the difference is 0 or 99, respectively, and we do not get a 3-digit number for the next step.

99 × (A − C) can also be written as 99 × [(A − C) − 1] + 99 = 100 × [(A − C) − 1] − 1 × [(A − C) − 1] + 90 + 9 = 100 × [(A − C) − 1] + 90 + 9 − (A − C) + 1 = 100 × [(A − C) − 1] + 10 × 9 + 1 × [10 − (A − C)]. (The first digit is (A − C) − 1, the second is 9 and the third is 10 − (A − C). As 2 ≤ A − C ≤ 9, both the first and third digits are guaranteed to be single digits.)

Its reversal is 100 × [10 − (A − C)] + 10 × 9 + 1 × [(A − C) − 1]. The sum is thus 101 × [(A − C) − 1] + 20 × 9 + 101 × [10 − (A − C)] = 101 × [(A − C) − 1 + 10 − (A − C)] + 20 × 9 = 101 × [−1 + 10] + 180 = 1089.

However, I don’t particularly care for the succinct explanation, and so I’d prefer to give my audience the following explanation. Let’s write our original three-digit number as $ABC$, which of course stands for $100 \times A + 10 \times B + C$. Then, when I reverse the digits, the new three-digit number will be $CBA$, or $100 \times C + 10 \times B + A$.

Of course, because the first number is bigger than the second number, this means that the first hundreds digit is bigger than the second hundreds digit. This means that the first ones digit has to be less than the second ones digit. In other words, when we subtract, we have to borrow from the tens place. However, the tens digits are the same for both numbers. That means that I have to borrow from the hundreds place also.

I’ll illustrate this for both subtraction problems:

Now I’ll subtract. The hundreds digit will be $A - 1 - C$. The tens digit will be $9 + B - B$, or simply $9$. Finally, the ones digit will be $10 + C - A$. This is a little hard to write on a board, so I’ll add some dotted lines to separate the hundreds digits from the tens digit from the ones digit:

The next step is to reverse the digits and add:

I’ll begin with the ones digit:

$(10 + C - A) + (A - 1 - C) = 10 - 1 = 9$.

No matter what, the ones digit is a 9.

Continuing with the tens digits, I get $9 + 9 = 18$. I’ll write down $8$ and carry the $1$ to the next column.

Finally, adding the hundreds digits (and the extra $1$), I get

$1 + (A - 1 + C) + (10 + C - A) = 1 - 1 + 10$.

Therefore, no matter the values of $A$, $B$, and $C$, the end result must be $1089$.

To complete the routine, I’ll ask a volunteer (usually a young child) to play the magician and repeat the trick for the audience. I consider this an important pedagogical step — the child enjoys being the magician on stage, while the audience lets the routine sink in one more time before I move on to the next magic trick.

# My Mathematical Magic Show: Part 2b

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.

For my first trick, I chose the most boring of the routine. Everyone in the audience had a piece of paper and many had calculators. Here is the patter for the first trick:

To begin this trick, write down any three-digit number on your piece of paper. Just make sure that the first digit and the last digit are different.

(pause)

Now, reverse the digits and write down a new number. For example, if your number was 321, the new number will be 123.

(pause)

Now, subtract the small number from the big number. If your second number is larger, then put that number on top so that you can subtract the two numbers.

(pause)

Your difference is probably a three-digit number. However, if you ended up with a two-digit number, you can make it a three-digit number by putting a 0 in the hundreds place.

Next, I want you to reverse the digits of the difference to make a new three-digit number. Write this new number under the difference.

(pause)

Finally, add the last two three-digit numbers that you wrote down.

If everyone follows the instructions and does the arithmetic correctly, everyone will get a final answer of 1,089.

The next part of my mathematical magic show is showing everyone why the trick works. The explanation depends on the mathematical sophistication of the audience. Today, I’ll give an explanation suitable for upper elementary students. Tomorrow, I’ll give a different explanation using algebra.

For today’s explanation, I’ll give an example:

Of course, because the first number is bigger than the second number, this means that the first hundreds digit is bigger than the second hundreds digit. This means that the first ones digit has to be less than the second ones digit. In other words, when we subtract, we have to borrow from the tens place.

However, the tens digits are the same for both numbers. That means that I have to borrow from the hundreds place also.

So, when I subtract, I’m guaranteed that the middle number will be a $9$. Also, I’m guaranteed that the hundreds digit and the ones digit will add up to $9$. In this example, the sum of the hundreds digits and the ones digits is equal to

$2 - 1 + 11 - 3$.

We know that $11-1$ is equal to $10$, and adding $2$ and subtracting $3$ is like subtracting $1$. Therefore, the sum of the hundreds digit and the ones digit must be equal to $9$.

The next step was reversing the digits of the difference:

Finally, I asked you to add these last two numbers. Remember, I had rigged things so that the hundreds and ones digits add up to $9$. So the last digit of the sum must be $9$. Also, I rigged things so that the tens digit must be $9$. So, when I add, I get a sum of $18$, and I leave the $8$ and carry the $1$. Finally, the hundreds and ones digits add up to $9$ again. Adding the extra $1$, I write down a sum of $10$.

In tomorrow’s post, I’ll explain how the trick works using algebra.

# My Mathematical Magic Show: Part 2a

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.

For my first trick, I chose the most boring of the routine. Everyone in the audience had a piece of paper and many had calculators. I also had a small white board to write on at the front of the room. I began,

To begin this trick, write down any three-digit number on your piece of paper. Just make sure that the first digit and the last digit are different.

After waiting 10 seconds, I then said,

Now, reverse the digits and write down a new number. For example, if your number was 321, the new number will be 123.

And, to be sure my instructions are clear, I’ll write these numbers on my white board:

Next, I’ll say:

Now, subtract the small number from the big number. If your second number is larger, then put that number on top so that you can subtract the two numbers.

After waiting a minute or so, I’ll say,

Your difference is probably a three-digit number. However, if you ended up with a two-digit number, you can make it a three-digit number by putting a 0 in the hundreds place.

Next, I want you to reverse the digits of the difference to make a new three-digit number. Write this new number under the difference.

After everyone’s done, I’ll give my final instruction:

Finally, add the last two three-digit numbers that you wrote down.

After everyone’s done, I’ll point to someone and say, “Your final number was 1,089.” If he followed my instructions and did the arithmetic correctly, he’ll say, “You’re right.” Then I’ll point to someone else and say, “You also got 1,089.” She’ll also say, :”You’re right.” Then I’ll say, “Everyone got 1,089, right?”

Another (and more dramatic) way to end the routine is to hand a book of mine to someone, with the following instructions:

Your last number should have four digits. Cross out the last digit; you now will have a number with only three digits. Turn to that page number in this book. Then find the word on that page corresponding to the number you crossed out. For example, if you crossed out a one, point to the first word on the page. If you crossed out a two, point to the second word on the page. And so on.

Got it? The word you’re pointing to is XXXXXX.

And of course, I’ll get this right because, before starting the routine, I had already memorized the ninth word on page 108 of my book (the XXXXXX above). This looks really dramatic because it looks essentially random to the audience.

In tomorrow’s post, I’ll explain how the trick works.

# My Mathematical Magic Show: Part 1

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 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.

Before showing my own tricks, however, I have to pay homage to Arthur Benjamin, who is a Professor of Mathematics at Harvey Mudd College who’s also made a second career doing mathematical magic shows. See his Amazon page for the books that he’s published and his webpage at Harvey Mudd.