I dislike gimmicks — especially math gimmicks. I do everything in my power and ability to avoid teaching a gimmick. But! Every now and again, I like to have a little fun. In this particular case that I will show, I like to “wow” my students by multiplying two numbers without working through the traditional steps. When I do this in my Algebra, Calculus, or Statistics classes it is primarily as light entertainment and to remind my students that they are actually curious about mathematics.

Here’s the gimmick. First, I start with a simple multiplication problem like

$$\begin{array}{rr}
&83\\
\times&37\\
&\hline
\end{array}$$

And in one step, from right to left, I put down

$$\begin{array}{rr}
&83\\
\times&37\\
&\hline\\
&3071
\end{array}$$

And immediately there’s a reaction of surprise and amazement from a few students. “Wait, how did you do that?” To which I reply, “Let’s try a larger problem.” And I ask them to give me two, three-digit numbers. I’ll get something involving 7s and 9s, perhaps like

$$\begin{array}{rr}
&973\\
\times&879\\
&\hline
\end{array}$$

And in one step I write down

$$\begin{array}{rr}
&973\\
\times&879\\
&\hline\\
&855267
\end{array}$$

And at this point, the class is often flabbergasted. And now, a mini-lesson in multiplication to students who know how to multiply has begun.

I ask them to explain how I could write this down in one step, without having to do something like

$$\begin{array}{rr}
&83\\
\times&37\\
&\hline\\
&581\\
&2490\\
&\hline\\
&3071
\end{array}$$

I usually get responses like “you have it memorized” or “you did the multiplication in your head”. Of course, it is true that I did do some things in my head, but I didn’t work out the standard pen and paper algorithm out in my head.

So what did I do? Here I begin a discussion about place value. In my previous post, I discussed a different way to work out subtraction that really forced us to think about both place and value. And that is exactly what I bring back to focus in this multiplication gimmick.

I ask my students about what the “ones” place is with multiplication — that is, if I wanted a result in the ones place, what place(s) should I consider? The answer, of course, is that a result in the ones place can only be obtained by multiplication of the ones places. Thus, in \(83 \times 37\), the ones place is the result of \(3 \times 7\), which equals \(21\). Thus, the ones place is \(1\) with a carry of \(2\) (ie, \(20\)).

Next, how do we obtain the tens place? The tens place is nothing more than \(10s \times 1s\) and \(1s \times 10s\) with whatever carry we had from the ones place. Thus, in \(83 \times 37\) to obtain the tens place we have \(8 \times 7 + 3 \times 3 + 2 \mbox{ (carry)} = 67\) and as such, the tens place is \(7\) with a carry of \(6\) into the hundreds place.

The hundreds place is no more than \(10s \times 10s\) and in \(83 \times 37\) we have \(8 \times 3 = 24\) and with the carry of \(6\) this gives a total of \(30\), yielding \(0\) in the hundreds place and \(3\) in the thousands place. Hence our answer, \(3071\) as desired.

The whole point here is that if we think about and understand place value, we can see that the tens place can only be constructed by multiplying the tens and ones places of the multiplier and multiplicand. The hundreds place, in two-digit multiplication is only had by multiplying the two tens places together.

What of three-digit multiplication?

• ones place: ones times ones
• tens place: tens times ones plus ones times tens plus carry from the ones place
• hundreds place: hundreds times ones plus tens times tens plus ones times hundreds plus carry from tens place
• thousands place: hundreds times tens plus tens times hundreds plus carry from hundreds
• ten thousands place: hundreds times hundreds plus carry from thousands
• hundred thousands place: carry from ten thousands place

The mental arithmetic does become more taxing as the number of digits increases for the multiplication, but the work can still all be done on one line, albeit a bit more slowly and more carefully.

The actual underlying mathematics is simple expansion:
$$83 \times 37 = (80 + 3)(30 + 7) = \underbrace{2400}_{\mbox{hundreds}} + \underbrace{90 + 560}_{\mbox{tens}} + \underbrace{21}_{\mbox{ones}}$$

I’m simply doing it in my head, from right to left.

Outside of the mild diversion, it’s an easy segue into (for my Algebra class) or an easy refresher about (for my Calculus or Stats classes) polynomial expansion. If we let \(10 = x\), then we have
$$83 \times 37 = (8x + 3)(3x + 7) = 24x^{2} + 9x + 56x + 21$$

The final point is when we come back to the traditional method

$$\begin{array}{rr}
&83\\
\times&37\\
&\hline\\
&581\\
&2490\\
&\hline\\
&3071
\end{array}$$

and compare how the addition / grouping actually works. The first line, \(581\), is really the \(56x + 21\) part and the second line, \(2490\) is really the \(24x^{2} + 9x\) part.

If you’re teaching mathematics, practice this for yourself and bust it out to your students. I’ve had reasonably good success with (re-)engaging students and reminding them that when they said, “Whoa, how’d you do that?” that in that moment they were curious about mathematics and that’s a type of wonder they should try to maintain as they continue to learn more, mathematics or otherwise. This also works reasonably well with the younger crowd who are just getting the hang of multi-digit multiplication. Once they seem to have had a handle on the traditional method, show them the “cool” way. It’s a great way for them to get a sneak peek at some pre-algebra (so this is pre-pre-algebra).

Bonus challenge: can you think about how this applies to division?