Learn to multiply two numbers quickly (Part-5) - Vedic Maths tricks Video Lecture | Improve Your Calculations: Vedic Maths (English) - Class 6

Hi! I'm Samir.Let's extend our learnings on finding the product of
two numbers a bit further. Let us begin
by taking 2007 and 213
as an example.Here our working base
would be 2 x 1000 and
2x 100 and our differences
would be 7 and 13
respectively
Note here our multiplier is 2.We can now proceed to write down our answer
which would be in 3 parts.Let's begin
by writing the 3rd part
which would be simple product of 7 and 13
giving us 91. To find out the second part
we'll have to take the larger number and
choose as many digits from right as is the
difference in the number of the 0's
in our secondary bases. Which
in our case is 3-2,
1. So we will split 2007 as 200 and 7. 7
will come in the second part and 200 would
go in the first part along with the
difference obtained from the second number
which is 13, giving us 200 + 13 , 213. Now, since our
multiplier was 2 , we will have to make a minor adjustment by multiplying 213
x 2 , giving us 426 and the second part
would be14. Now since the
number of digits which could be accommodated in 2nd part was only one so one of the
14 will have to be carried to the first part giving us 426
+ 1, 427. Now we can club these parts and
get our final answer as 427491. Please
note that the 3rd part will always have number of digits equal to the number of
zeros in the base of the smaller number.
Let us now take 1993 x 197 as an
example here again our working bases would be 2x1000 and 2x100
respectively and our differences would be
-7 and -3.Please note the
multiplier again would be 2. Now we can again proceed to write down our answer which would be in
3 parts. Let's begin by 3rd part which would be the product of the differences
-3 x -7
Giving us + 21. To obtain the second part we will have to choose the larger number and choose
as many digits from right as is the
difference of the number of zeros in our
working base which in our case it 3-2, 1.
So 1993 will be split as 199 and 3. 3 will come to the second part and 199 will go to the
first part along with the difference obtained from the second number which is
-3, making the first part as 199 -3, 196. Since our multiplier was
2, we will now have to make a minor adjustment. We will multiply the 196 with 2 giving us 392. We will
have to do the same with the second part
which is 2x3, 6. Now we can
club these and get our final answer. Also note that the third part will have as many digits
as is the number of 0's in the base of the smaller number.So we
can now get our final answer as 392621.Let us try to multiply 2996 x
317
here our working bases would be 3x 1000 and 3 x100 respectively and the
differences would be -4 and +17. Now we can proceed to
write down our answer which would be in 3 parts. Let us begin by writing the 3rd part first
which would be -4 x 17 ,-68. To obtain the second part we will have to take the larger
number and choose as many digits from right as is the difference in the number of
0's in our working base which in our case would be 3-2, 1. So we will split
2996 as 299 and
6. 6 will come to the second part and 299 will
go to the first part along with the
difference obtained from the second
number which was +17
making the first part as 316. Since our multiplier is 3 we will have
to make a couple of adjustments. We will
multiply 316
with 3 giving us 948 and the second part
6 will also have to be multiplied by 3
giving us 18. Now before we can write
down the final answer, we will make some
changes
the second part can have only 1 digit, so the extra 1 will have to be
carried over to the first part making it
948 +1, 949 and the third part cannot be
left negative as per second rule so it will
borrow 1 from the second part which will
now become 8 -1, 7 and our final
part would become100 -68,
giving us 32, now we can join these
parts and write our final answer as
949732. This is it for now, lots of new stuff coming your way
stay tuned