The Law of Doubled Digit Transposition Over Multiplication

My son is learning long division at school. He was intimidated by it, so we started practicing for just a few minutes every day after school. The format is simple: I generate two numbers, a big dividend and a small divisor, and he divides them.

One day I wanted the quotient to be 99 because it’s a fun number—what with two 9s and all. I picked 8 as the divisor because 8 and 9 are friends. These choices led to a dividend of 792. So I told my son, “792 divided by 8.”

He solved the problem, and then we checked the result on a calculator. But I accidentally divided by 9 instead of 8. And this craziness appeared:

$$\frac{792}{9} = 88$$

88? What were the chances? 88 is a fun number, too, what with two 8s and all. I wondered if this was true:

$$77 \times 8 = 88 \times 7$$

It was true! I got a little bolder, wondering if this was true:

$$22 \times 9 = 99 \times 2$$

It was also true!

At this point, I needed a proof that this would work for any two digits. I hate to admit it, but I left my son behind and worked it out on my own. He’s not really doing algebra yet, and I needed an answer now.

Let’s call one of the digits a and the other b. A number xx is formed by $(x \times 10 + x)$. I set out to prove that the following would hold for any $a, b \in [0, 9]$:

$$(a \times 10 + a) \times b = (b \times 10 + b) \times a$$

I got this! This equation is just begging to be simplified.

$$(a \times 11) \times b = (b \times 11) \times a$$

Oh. Of course this will hold. It’s just the associative and commutative properties of multiplication. I was fooled into thinking there was more magic here than there really was.

At least now I know which number is the most fun: 11.


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