Japanese Multiplication Method

The Japanese have a curious and easy Multiplication method which is really intriguing.

I call it 'The Chopstick Multiplication'.

Suppose you want to calculate

367 X 213=?

Here is what you do.

1.       Refer Fig-1 and draw drawings corresponding to each digit in 367 as shown from Bottom-left to Top-Right.  

Fig-1

2.       Next on top of the lines drawn in Fig-1, refer Fig-2 and draw lines corresponding to 213 from Left-Top to Bottom-Right.  

Fig-2

3.       These two sets of lines will intersect and form 5 Intersection sets. Intersection-1 to Intersection-5.

4.       Start from the bottom right and in the first Intersection set, we find that there are 21 points.  Take 1 from 21 and carryover 2 to add to the number of points in the next Intersection Set, Intersection Set-2.  (Refer Fig-3)  

Fig-3

5.       Now count the intersection points in Intersection Set-2 and add the carryover from previous Intersection Set-1 to this.  Again take the units place number and carryover the remaining part to the next Intersection Set and so on. (Refer Fig-4,5,6,7)

Fig-4

   

Fig-5

Fig-6

Fig-7

6.       Write down the digits we had taken at each intersection next to each other starting from Intersection-5 on extreme left to Intersection-1 on extreme right. And we get

** 78171 **

Which is the Answer

Chinese Method for Speedy Multiplication of any two integers

You must have read about Shakuntala Devi, the 'Human Calculator'. She use to crunch numbers real fast.

In 1977 in USA she competed with a computer to see who gives the cube root of 188138517 faster, she won. 

At the Southern Methodist University she was asked to give the 23rd root of a 201-digit number; she answered in 50 seconds. Her answer—546,372,891—was confirmed by calculations done at the U.S. Bureau of Standards by the Univac 1101 computer, for which a special program had to be written to perform such a large calculation.

But here's your chance at showing off your own Blazing Fast Multiplication prowess.

Suppose you need to find 2873 X 5964=?

That is an awful lot of multiplication right?

Wrong !! Ancient Chinese had  a dead simple method to do this, here is how:

1. Make a matrix which has n columns and m rows. Where n is the number of digits in number A and m is the number of digits in number B.

Here A=2873, so n=4

B=5964, so m=4 as well.

Let us write them as below in a table n X m (4X4) as in Fig-1 below.

Draw a diagonal line in each square.

 Fig-1

2. Now Start multiplying the number in each column with the number in every row and keep writing the numbers as below. Refer Fig-2

 Fig-2

Notice that 2X5=10, so 1 is written to the left of the diagonal in column 1, Row 1 and 0 is written to the right of diagonal in column 1, Row 1.

Likewise 2X4=8 is written as 0 written to the left of the diagonal in column 1, Row 4 and 8 is written to the right of diagonal in column 1, Row 4.

3. Starting from the bottom, mark the diagonal rows as shown in Fig-3

 Fig-3

So Diagonal D1 has the numbers

D1=2, similarily

D2=8,1,8

D3=2,2,2,1,7

D4=8,3,8,4,3,2,5

D5=0,2,4,2,6,5,1

D6=1,8,7,0,3

D7=1,0,4

And

D8=1

4.       Now sum the numbers in each Diagonal Row,  if the sum is less than 10 write down the digit else if greater than 10, then carryover the 10’s place to the next diagonal. As shown below:

D1=2

D2=8+1+8=17, so take 7 (and carryover 1 to D3)

D3=2+2+2+1+7+(1 from D2)=15, so take 5 (and carryover 1 to D4)

D4=8+3+8+4+3+2+5+(1 from D3)=34, so take 4 (and carryover 3 to D5)

D5=0+2+4+2+6+5+1+(3 from D4)=23, so take 3 (and carryover 2 to D6)

D6=1+8+7+0+3+(2 from D5)=21, so take 1 (and carryover 2 to D7)

D7=1+0+4+(2 from D6)=7 , so take 7

D8=1, so take 1

5   4.      Now simply write down the numbers taken at each of the diagonals starting from D1 on the extreme right to D8 on the extreme left

And we get the answer

** 17134572  **

Beauty is this method works for any size of numbers