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.