My method for multiplying numbers together - "the clop method"

[clop]

My method for multiplying numbers together - "the clop method".

One day at school, many years ago, the teacher was telling us about algebra and I was bored and my mind was wandering. Instead of listening to the teacher I came up with a novel technique to multiply large numbers together quickly without showing any working. I've never seen this method described by anyone else in more than 20 years, so I think I'm safe in calling it "the clop method of multiplication". If anyone knows this technique from elsewhere I would be very interested to see a link.

OK so here it is. Sorry about the lines of ...'s. I couldn't format the columns properly using spaces.

I'll start with the two-digit version.

If we multiply two numbers together, say 76 x 43, using the normal procedure, we do "3 times 6 is 18, (write 8, carry 1), 3 times 7 is 21 plus the carry 1 is 22 (write 22), next line so put a 0 on the right, then 4 times 6 is 24 (write 4, carry 2), 4 times 7 is 28 plus the carry 2 is 30 (write 30), and then do an addition from the right, 8 plus 0 is 8 (write 8), 2 plus 4 is 6 (write 6), 2 plus 0 is 2 (write 2), nothing plus 3 is 3 (write 3)" and then we have the answer 3268.

Let's do the same multiplication again but this time, instead of 76 x 43, we will use ab x cd where a, b, c & d are arbitrary digits.

d times b is bd (write bd), d times a is ad (write ad), next line so put a 0 on the right, then c times b is bc (write bc), c times a is ac (write ac), and then do an addition, which looks likes this

....................ab x
....................cd

............ad.....bd +
ac.........bc......0

ac..(ad+bc)..(bd)

so the three digits of the final answer are ac, (ad+bc) and bd.

We can save time by writing those digits out directly, from right to left, carrying any tens over into the next digit.

So for the original product, 76 x 43, where a=7, b=6, c=4, d=3 the digits of the answer are

(7x4) (7x3 + 6x4) (6x3)

You have to work from right to left, so it would go 6 times 3 is 18, write 8 carry 1, 21 plus 24 is 45 plus the carry is 46, write 6 carry 4, 7 times 4 is 28 plus the carry is 32, write 32, and the answer is 3268.

It looks complicated but there is an innate beauty to it - the right hand digit is the product of the two right hand digits (bd) mod 10, the second digit is the sum of the product of the diagonals (ad+bc) plus the carry mod 10, the third digit is the product of the two left hand digits (ac) plus the carry.

So for 23 x 14 it is 2 (carry 1), 8 plus 3 plus 1 is 12 is 2 (carry 1), 2 plus 1 is 3, answer 322. Very quick and easy.

Here is the three-digit version.

..............................................abc x
..............................................def

..............................af..........bf..cf +
............ae..............be..........ce...0
ad.........bd..............cd...........0...0

ad..(ae+bd)..(af+be+cd)..(bf+ce)..cf

It looks complicated again, but draw the lines between the digits and see the beautiful pattern. The product of the two digits on the right, then the sum of the products of the diagonals on the right, then the sum of the products of all three diagonals, then the sum of the products of the diagonals on the left, then the product of the two digits on the left. Work from right to left, carrying the tens into the next product as you go.

For 132 x 114 we get

8, 12+2 is 14 is 4 (carry 1), 4+3+2 plus the carry is 10 is 0 carry 1, 1+3 plus the carry is 5, 1 so the answer is 15048.

Once you get the hang of it you can easily extend it to doing 4 and 5 digit multiplications without any working.

What do you all think?

clop

[Zephir]

My method for multiplying numbers together

This is very well known method of ancient Chinese astronomers.
http://superstruny.aspweb.cz/images/...rojective2.gif

[01101001]

Zephir, welcome to BAUT Forum.

This is very well known method of ancient Chinese astronomers.
http://superstruny.aspweb.cz/images/fyzika/aether/projective2.gif

That very large (circa half-megabyte) animated GIF I think I understand, but could you possibly supply some text explaining what's going on? The animation's area is small and it's hard to read. It looks interesting.

By the way, that's a pretty big file and hard on members with low bandwidth or high bandwidth charges. I believe words would work better. Also, if you haven't, check the Rules For Posting To This Board, rule 8 about the permitted sources of hotlinked images; maybe the image is public domain or belongs to you. I hope it conforms. [Edit: it now looks like it's been removed from being in-line and turned into a clickable link.]

I don't mean to sound like a nag. Thanks for the post and, again, welcome to BAUT.

Edit, later: Maybe I can explain it. Once I knew what the answer was supposed to be (39 483) it made sense. I couldn't read it in the example. The digits of one factor (321) are represented as lines on one axis of a diamond grid. The digits of the other factor (123) are similarly represented as lines on the other axis of a diamond. Then the 5 vertical columns of intersecting points are added (with carries) from right to left. 3-8-14 (4, carry the 1)-9-3, and read out left-to-right 39483. (Of course, the handedness all depends on one's personal method of number notation.)

[antoniseb]

This is very well known method of ancient Chinese astronomers.

Hi Zephir welcome to the BAUT forum.

Personally I love the history of astronomy, and of calculation techniques. However, I have just edited your post so that the GIF file is a hyperlink rather than be hot-linked (embedded) into your message. Many of our members have slow connections to the internet, and this GIF was very rough on them.

[clop]

This is very well known method of ancient Chinese astronomers.

Thank you Zephir! I don't think it is very well known though. I teach it to a lot of people. I haven't yet found anyone who has seen it before (except you!).

clop

[01101001]

I haven't yet found anyone who has seen it before (except you!).

I didn't get that Zephir's cited (Chinese astronomer) method is the same as clop's.

In clop's method, you need to know the multiplication table for pairs of digits, and how to add.

With Zephir's you need to know how to count and how to add.

Edit: I retract. After I tried an example with larger digits -- that sample animation of 123 x 321 wasn't so realistic -- I see that the counting is sped up considerably if you know the multiplication table. When you have, say, 8 lines intersecting 9 lines, you could count 72 intersections, but it's better to just know.