# 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.