(Lack of) Math Videos
I just saw a math video that showed a technique for multiplying two-digit numbers. I’ll share this trick with you in a minute. But I am dismayed that this video — along with so many other math videos I’ve seen on the web — are willing to show these tricks but not explain them. So I’m going to provide the trick, which I thought was pretty cool, along with an explanation as to why it works.
Before I do, though, I have to vent. Math videos on the web are a real problem. The ones I’ve seen generally do very little to promote conceptual understanding. Instead, the videos present skills without explanation, and the implicit message is that you don’t need to understand why it works — just do exactly as the videos say, and you’ll get the right answer. The issue is that students then see mathematics as a series of disconnected rules they need to memorize, instead of seeing it as the beautifully interconnected discipline that it is.
Okay, I’ll now step down from my soapbox and share the trick with you…
Take two numbers, both between 10 and 99, that meet the following criteria:
- Their tens digits are the same.
- The sum of their units digits is 10.
Here are some problems that meet these criteria:
- 22 × 28
- 74 × 76
- 33 × 37
- 55 × 55
I will now state the obvious — there are only 81 multiplication problems to which this trick applies. Still, I think it’s fun to explore the math underlying the trick.
To find the product, then, follow this process:
- Take the tens digit, and multiply it by one more than itself.
- Multiply the units digits.
- Finally, concatenate (put together) these two results to make a three- or four-digit number. (As noted in the comments below, if the result of Step 2 is a one-digit number, you’ll also need to put a 0 in front of it.)
Let’s take the first problem above, 22 × 28, to see how this works in practice.
- The tens digit is 2, so multiply 2 by one more than itself: 2 × 3 = 6.
- The units digits are 2 and 8, so multiply them: 2 × 8 = 16.
- Consequently, 22 × 28 = 616.
More generally, the trick could be explained with symbols. If the tens digit of each number is n, and the units digits are p and q, then we’re trying to find the product of (10n + p)(10n + q). Using the FOIL method from algebra gives
(10n + p)(10n + q) = 100n2 + 10n(p) + 10n(q) + pq = 100n2 + 10n(p + q) + pq
But since the criteria implies that p + q = 10, this becomes
100n2 + 10n(p + q) + pq = 100n2 + 100n + pq
This can be further refined to
100(n2 + n) + pq
and the important point, then, is realizing that n2 + n = n(n + 1). Which is just another way of saying that we’re going to take the tens digit, n, and multiply it by one more than itself.
And that’s it. The first part of the expression above, 100(n2 + n), gives the hundreds (and perhaps the thousands) digit, and the second part, pq, gives the tens and units digits.