Hey friends!, as i was doing calculations mainly
some huge multiplications, divisions and squares, i found that they are
consuming longer time than the whole problem has to be solved. So i tried to
browse for the better shortcuts of doing them. After sparing some time in
practicing those methods i gradually reduced the time of doing those
calculations. So here in this post i will write the methods of using them.

__NOTE:__

Before entering into the explanation, i want
the reader of this article has to remember one key point before starting
implementing. That is, almost all methods uses the technique of

**excess carried over.**As one single place will have only one digit, if the method gives the result in 2 digits for one place then the excess will be forwarded to the next place. Ex: If units place has 12 as a result, then 2 will be places in units place and 1 will be forwarded to the next cell i.e., tens place.

__Firstly, we will deal with__*MULTIPLICATION*:
To multiply two numbers (of two or
more digits), split each number into two parts. If the first number is a1 + b1
and the second number is a2 + b2, then the product of the two numbers is:

(a1 x a2) + (a1 x b2 + b1 x a2) + (b1 x b2)

The solution comprises three parts
(as shown by the boxes and arrows above): the head, the middle, and the tail.

- The digits on the right are multiplied vertically to get the tail part: b1 x b2 (excess carried over)
- All digits are multipled crosswise and added together to get the middle part: a1 x b2 + b1 x a2 (excess carried over)
- The digits on the left are multiplied vertically to get the head part: a1 x a2

Here is a simple example to
illustrate this technique.

23 x 41 = 943

23 x 41 = 943

The steps are:

- 3 x 1 = 3
- 2 x 1 + 3 x 4 = 14, put down 4 and carry over 1
- 2 x 4 = 8, plus the 1 carried over, is 9

The speed gain using this technique
(over the conventional method of multi-line long multiplication) becomes more
apparent when handling larger numbers. Here is another example involving excess
carryover at each stage.

108 x 64 = 6912

108 x 64 = 6912

The steps are:

- 8 x 4 = 32, put down 2 and carry over 3
- 10 x 4 + 8 x 6 = 88, plus the 3 carried over, is 91; put down 1 and carry over 9
- 10 x 6 = 60, plus the 9 carried over, is 69

Maybe this method looks some what
crazy and funny but it really works on every calculation.

__Now we will deal with__*DIVISIONS*:
It is very easy to understand with an
example rather than understanding with the theoretical method.

Take for example, 716769 ÷ 54. Yes,
you too CAN work it out manually -- and in one line -- without having to reach
for the calculator! "on top of the flag". The trick is to reduce
the divisor to a mentally manageable value by putting its other digits "on
top of the flag". In this example, the divisor will be reduced to 5
(instead of 54) by pushing the 4 up the flagpost, as shown below. Corresponding
to the number of digits flagged on top (in this case, one), the rightmost part
of the number to be divided is split to mark the placeholder of the decimal
point or the remainder portion.

Now observe carefully as we walk
through the steps of this example:

716769 ÷ 54 = 13273.5

- 7 ÷ 5 = 1 remainder 2. Put the quotient 1, the first digit of the solution, in the first box of the bottom row and carry over the remainder 2
- The
product of the flagged number (4) and the previous quotient (1) must be subtracted
from the next number (21) before the division can proceed. 21 - 4 x 1 = 17

17 ÷ 5 = 3 remainder 2. Put down the 3 and carry over the 2 - Again
subtract the product of the flagged number (4) and the previous quotient
(3), 26 - 4 x 3 = 14

14 ÷ 5 = 2 remainder 4. Put down the 2 and carry over the 4 - 47
- 4 x 2 = 39

39 ÷ 5 = 7 remainder 4. Put down the 7 and carry over the 4 - 46
- 4 x 7 = 18

18 ÷ 5 = 3 remainder 3. Put down the 3 and carry over the 3 - 39
- 4 x 3 = 27. Since the decimal point is reached here, 27 is the raw
remainder. If decimal places are required, the division can proceed as
before, filling the original number with zeros after the decimal point

27 ÷ 5 = 5 remainder 2. Put down the 5 (after the decimal point) and carry over the 2 - 20 - 4 x 5 = 0. There is nothing left to divide, so this cleanly completes the division.

This takes a little bit time to
understand. But its easy to solve rather than BALDING your head :P

As we are clear about the
Multiplications and Divisions, without any waste of time we will deal with

**:**

__SQUARES__
The HUGE knowledge needed to solve
the following method is (a + b)

^{2}= a^{2}+ 2ab + b^{2}
Yeah you heard right....... above formula is enough to solve the
following method. Te method is as follows.

First, a nifty shortcut! The square
of a number ending in 5 is almost a no-brainer.

If n is the number formed by the preceding digit/s (before the 5), get the product of n and n+1.

Then just append 25 (i.e. 5 x 5) to this product.

If n is the number formed by the preceding digit/s (before the 5), get the product of n and n+1.

Then just append 25 (i.e. 5 x 5) to this product.

For example, 75

7 x 8 = 56; therefore solution is 5625.

^{2}:7 x 8 = 56; therefore solution is 5625.

Another example, 115

11 x 12 = 132; therefore solution is 13225

^{2}:11 x 12 = 132; therefore solution is 13225

For other cases of squaring, the same
shortcut techniques used in multiplication may be utilised.

Especially the general-purpose Urdhva Tiryagbhyam (Vertically and Crosswise) formula.

To get the square of a number (of two or more digits), simplify by splitting it into at least two parts, a and b.

Especially the general-purpose Urdhva Tiryagbhyam (Vertically and Crosswise) formula.

To get the square of a number (of two or more digits), simplify by splitting it into at least two parts, a and b.

Thus (a + b)

^{2}= a^{2}+ 2ab + b^{2}
The solution comprises three parts,
neatly fitting the three boxes shown above. Just adjust for excess carry over.

- the
head: a
^{2} - the middle: crosswise multiplication and doubling a x b x 2
- the
tail: b
^{2}

Here is a simple example to
illustrate this technique.

23

23

^{2}= 529
The steps are:

- tail:
3
^{2}= 9, put it down in the rightmost box - middle: 2 x 3 x 2 = 12, put down the 2 in the middle box and carry over the 1
- head:
2
^{2}= 4, plus the 1 carried over, is 5 in the left box

Another example.

108

108

^{2}= 11664
The steps are:

- tail:
8
^{2}= 64, put down the 4 and carry over the 6 - middle: 10 x 8 x 2 = 160, plus the 6 carried over, is 166; put down the 6 and carry over the 16
- head: 10 x 10 = 100, plus the 16 carried over, is 116.

So i hope that the reader has
understood these crazy methods and makes his MATH simple and avoid ambiguity
and avoid wastage of time.

Thanks for reading my post..... :) :)

--- PaVaN KoUndiNyA

Very easy to understand. It sure does simplify things lol

ReplyDeleteLol ye very useful :D

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