Posted January 11, 2013 By Presh Talwalkar. Read about me , or email me .

This is another fun math video on Youtube. The author explains an interesting algorithm to divide large numbers by 9.

I have to admit I was a bit puzzled at why the trick worked at first. Then I wrote out the math and got a deeper understanding.

Below is the trick and my explanation of why it works.

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The video: mental math multiplication

Video: Long Division trick

An example of the trick: dividing 221,013 by 9

Here is the algorithm outlined in the video.

Or in plain text:

1. Write down the leading digit of the number: in this case, 2.

2. Add that number to the next digit. Here we add 2 + 2 = 4. This is our next digit.

3. Repeat this process. That is, add the digit just obtained to the next digit of the number. So here we add 4 and 1 to get 5.

4. Repeat the process until the final digit. (We will have 24556 at this point.)

5. Now comes a slightly tricky part. We add the last digit obtained in the answer (6) to the last digit of the original number (3). We add those two together and divide by 9.

6. Add that to the final digit (in some cases, you will have a fraction, which means you have a remainder.). In our example, we add 6 and 3 to get 9. Dividing by 9 we get 1. So we add 1 to the final digit 6 to get 7.

7. Our answer is 24557.

Why does this work?

Here is the reason why the trick works. If you think about it carefully, here is what the trick really does.

The trick says if we start with a number abcdef, then dividing by 9 gives us the number:

a–(a+b)–(a+b+c)–(a+b+c+d)–(a+b+c+d+e + (a+b+c+d+e+f)/9)

(That is, we start by copying the first digit, and then we keep adding to the next digit of the number.)

The principle behind the trick is this. We will see what happens when we divide higher and higher powers of 10 by the number 9. Note that:

1/9 = 1/9

10/9 = 1 + 1/9

100/9 = 10 + 1 + 1/9

1000/9 = 100 + 10 + 1 + 1/9

10000/9 = 1000 + 100 + 10 + 1 + 1/9 …. Or more generally (can be proven by induction) (10)k/9 = (10)k-1 + (10)k-2…+ 1/9

In the base 10 system, dividing by 9 yields a “remainder” of 1 for each placeholder.

To figure out a number like 20,000 = 2(10)5, we more generally can deduce that:

2 (10)k/9 = 2[(10)k-1 + (10)k-2…+ 1/9]

That is, we end up with the number “2” in each placeholder position:

2 (10)k/9 = 2…22 + 2/9

So that’s the math behind the trick. Here is what happens when we divide a large number by 9:

abcdef / 9 = a00000 / 9 = aaaaa + a/9

b0000 / 9 = bbbb + b/9

c000 / 9 = ccc + c/9

d00 / 9 = dd + d/9

e0 / 9 = e + e/9

f / 9 = f/9

——————————————– When we add up each placeholder, we get: a–(a+b)–(a+b+c)–(a+b+c+d)–(a+b+c+d+e + (a+b+c+d+e+f)/9)

And this is exactly what we wished to prove.

For the example in the video we have:

It works like magic, but of course there is no real magic to it.

A warning about the trick

When you want to divide a number like 787 by 9, you will have some trouble. You will copy down the 7, but then you need to add 7 to 8 and you get 15. What do you do with that? You will need to carry over the 1 to make the leading digit an 8, and then leave the 6. But then you get another remainder when you need to add to the last digit…um, how is this supposed to work??

Let’s just say it gets overly complicated and it is MUCH better to divide the old-fashioned way.