The internet is great. It allows any mild-mannered mathematician to rant to the world about the inconsistency of punctuation or provide advice about what to do with those unending extra credit requests. Every action has an equal and opposite reaction says one of those laws of Newtown physics. In that way, these mild-mannered mathematicians are also then responsible for those videos showing off “fast ways to multiply!”. For if they hadn’t felt empowered to blurt out to the world what they thought was meaningful and relevant then maybe some of these other folks wouldn’t have felt the need to make such videos / tutorials about how to multiply.

A simple query for “multiplication trick” will garner well over a dozen hits on popular video sharing websites. This author hasn’t had the patience to sit through and watch each video. However, just watching a few of these videos, there are two conclusions:

• These are all gimmicks.
• The explanations are terrible or just incomplete.

Now with that said, it doesn’t look like the video makers are actually doing anything mathematically incorrect. It’s just that the advertisement that “here’s a cool, fun, fast, awesome, spectacular way to multiply that you’ve never seen before because those educators of yours don’t know how to teach” isn’t quite correct. Students are, in fact, taught these other methods, but the “traditional” way tends to be safer.

Interested in knowing what’s out there in terms of multiplication algorithms? Go here.

Here is how the traditional way works (hopefully there are no alignment issues, this was tested on a few browsers and looks fine). Consider multiplying 28 by 45.
\[\begin{align}
&&28\\
&\times&45\\
\hline
&&140\\
&&1120\\
\hline\\
&&1260\\
\end{align}\]

Here is how it works in one of the popular “alternative” methods.
\[\begin{align}
& & 28\\
& \times & 45\\
\hline\\
&&1260\\
\end{align}\]

See! It’s great! All in one line! No steps!

Oy … here’s what’s going on in the alternative way.

Explanation 1

1. Resolve the ones / units place. The only way to get a numeral in the units place is to multiply the ones places together. Since \(8\times\ 5 = 40\), the units place is 0 with a carry of 4 into the tens place.
2. Resolve the tens place. The tens place is a result of tens times ones or ones times tens. Thus, \(2\times\ 5 = 10\) and \(4\times\ 8 = 32\). This means that the tens place in the original problem is the ones place of \(10 + 32 + 4 = 46\) (remember there is a carry of 4 from the ones place multiplication). Thus, the tens place in the original problem is 6 and there is another carry of 4 into the hundreds place.
3. Resolve the hundreds place. The hundreds place is a result of the tens times tens. Thus, \(2\times 4 = 8\) and adding the 4 from the tens place carry, the result is 12. In other words, the hundreds place is 2 with a carry of 1 into the thousands place.
4. Therefore, the result is 1260.

Of course, this explanation is terrible since one has to read it (scroll to the bottom for a “good” explanation). But in a video demonstration it is, admittedly, actually quite appealing. The demonstrations are flawless and it almost feels like magic that the numbers just all happen to resolve correctly. But here’s what’s really going on (in excruciating detail) and why all students of algebra have already seen this.

Explanation 2

\[\begin{eqnarray}
28 \times 45 & = & (8 + 20) \times (5 + 40)\\
& = & (40) + (100 + 320) + (800)\\
& = & (4\times 10 + 0) + ((10 + 32)\times 10) + (8\times 100)\\
& = & (0) + ((10 + 32 + 4)\times 10) + (8\times 100)\\
& = & (0) + (46\times 10) + (8\times 100)\\
& = & (0) + ((40\times 10 + 6)\times 10) + (8\times 100)\\
& = & (0) + (6\times 10 + 40\times 10\times 10) + (8\times 100)\\
& = & (0) + (6\times 10) + ((8 + 4)\times 100)\\
& = & (0) + (6\times 10) + (12\times 100)\\
& = & (0) + (60) + (1200)\\
& = & 1260
\end{eqnarray}\]

Yes, this is a very long progression of the arithmetic. And in the video demonstrations a lot of the steps are combined into one large step as given in Explanation 1. In any case, the point here is that this is nothing more than expanding \((a+b)\dot\ (c+d)\). There is nothing fancy about it. It’s just a gimmick. Things have been wrapped up a little differently.

There is a substantial downside though. The alternative method relies on a reasonable amount of mental math. There is no “error-checking”. If there were an error in the calculation, the calculator (in the “old” days, a calculator was a person whose job was to, well, calculate) would have to go through and redo all the calculations. Whilst, in the standard method, all the steps are written out and it is far easier to spot check and debug if need be.

Incomplete Explanations

So, there is nothing actually wrong with demonstrating another way to do something, even if it is a little gimmicky. But with arithmetic and mathematics in general, the problem is exactly in how things are explained. The explanations given are really just procedural. Do this. Then that. Then that. Then this. And now it’s done. This is one of the biggest problems with mathematics education. It’s not just about a set of steps to be mindlessly carried out! Understand why they are being done that way and ask, “Can this be generalized?”!

If one is going to go through the trouble of demonstrating another way, explain it conceptually at least and show how it can be generalized!!

Here is a more complete explanation of the mechanics of what’s going on with the alternative way. For simplicity and without loss of generality, this explanation will consider only the case of multiplying two positive integers of arbitrary size. The full generalization is left as an exercise for the reader.

1. The ones place is always achieved by multiplying the ones places together of both integers with any carry going into the tens place.
2. The tens place is achieved by multiplying the tens place of the first number with the ones place of the second number (think: tens times ones gives tens) and adding it to the result of multiplying the ones place with the tens place. Add any carry from the ones. If the final result is greater than nine then there is carry that goes to the hundreds place.
3. The hundreds place is achieved in three ways: hundreds times ones, ones times hundreds, and tens times tens. Add carry. If the final result is greater than nine then there is carry that goes to the thousands place.
4. The thousands place is achieved as follows: thousands times ones, ones times thousands, hundreds times tens, tens times hundreds. Resolve carry.
5. The ten thousands place is achieved as follows: ten thousands times ones, ones times ten thousands, thousands times tens, tens times thousands, hundreds times hundreds. Resolve carry.
6. Continue until all leading digits are zero.

Here it is in action.

An example

\(123\times 456 = \)?

1. Ones: \(3\times 6 = 18\). So ones place is 8, with a carry of 1.
2. Tens: \(2\times 6 + 5\times 3 + 1 = 28\). So tens place is 8, with a carry of 2.
3. Hundreds: \(1\times 6 + 4\times 3 + 2\times 5 + 2 = 30\). So hundreds place is 0, with a carry of 3.
4. Thousands: \(1\times 5 + 4\times 2 + 3 = 16\). So thousands place is 6, with a carry of 1.
5. Ten Thousands: \(1\times 4 + 1 = 5\). So ten thousands is 5, with no carry.
6. \(123\times 456 = 56088\)

I hope this makes sense on a more general level. If this is fully understood, then show that \(12.3\times 45.6 = 560.88\) by first generating the hundredths place.