Shortcut method

Mathematics Arithmetic Calculation – Shortcut method, Prepared by
Dr.P.Ramasubramanian, P/CSE, SWCET
2017
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VEDIC MATHEMATICS
The word “Veda” means knowledge, which is derived from the root Vid, meaning to know
without limit. Vedic maths comes from the Vedic tradition of India. It is developed by Swami Bharati
Krishna Tirthaji in the early decades of the 20th
century. The Vedas are the most ancient record of
human experience and knowledge, passed down orally for generations and written down about 5,000
years ago. Medicine, architecture, astronomy and many other branches of knowledge, including
maths, are dealt with in the texts.Vedic Mathematics is based on 16 sutras dealing with mathematics
related to arithmetic, algebra, and geometry. The 16 basic Sutras or 'Word Formula' which can solve
all known mathematical problems by using an ancient yet simple system of MENTAL, ONE LINE
ARITHMETIC, used extensively by the Hindus more than 2,500 years ago.All Vedic Maths is based
on the understanding of Unity Consciousness which means they utilize processes or Number Bases
that correspond to:
0, 10, 100, 1000, 10000 etc all of which add to 1.
Advantages of Vedic Mathematics:
1. It is fun
2. It reduces time and mental strain
3. It offers a new and entirely different approach to the study of Mathematics based on pattern
recognition.
4. It allows for constant expression of a student's creativity, and is found to be easier to learn.
5. It improves one’s observation, imagination and visualization
6. It develops one’s mental agility, self confidence along with quick cross checking systems.
7. It converts a tedious subject into a playful and blissful one which students learn with smiles.
8. It offers a new and entirely different approach to the study of Mathematics based on pattern
recognition.
Mathematics is an interesting subject. If you don’t find it interesting, it simply means you
haven’t tried to understand it. Let you assume that, it is very simple and 100% logical. There is
nothing to be assumed and nothing to be confused about. So nothing to worry, if you come forward
with firm determination to learn mathematics.
We know that, the most basic things in mathematics are:
Addition, Subtraction, Multiplication and Division.
The Vedic Mathematics Sutras
This list of sutras is taken from the book Vedic Mathematics, which includes a full list of the
sixteen Sutras in Sanskrit, and the corresponding meaning was also given.

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Vedic Mathematics Sutras
S.No Sutra Meaning
1 Ekadhikena Purvena One more than the previous
2 Nikhilam Navatascharamam Dastah All from Nine and last from Ten
3 Urdhwa-tiryagbhyam Criss-cross
4 Paravartya Yojayet Transpose and Adjust
5 Sunyam Samyasamuchchaye When the samuchchaya is the same, the
samuchchaya is zero, i.e., it should be equated to
zero
6 (Anurupye) Sunyamanyat If one is in ratio, the other one is zero.
7 Sankalana-vyavkalanabhyam By addition and subtraction
8 Puranpuranabhyam By completion or non-completion
9 Chalana-Kalanabhyam Differential
10 Yavdunam Double
11 Vyastisamastih Use the average
12 Sesanyankena Charmena The remainders by the last digit
13 Sopantyadyaymantyam The ultimate and twice the penultimate
14 Ekanyunena Purvena One less than the previous
15 Gunitasamuchachayah The product of the sum of co-efficient in the factors
16 Gunaksamuchachayah When a quadratic expression is product of the
binomials then its first differential is sum of the two
factors.
10 – point circle/scale
The 10 – point circle illustrates the pairs of numbers whose sum is 10.
There are eight unique groups of three numbers that sum to 10.
Example: 1 + 2 + 7 = 10 1 + 3 + 6 = 10 1 + 4 + 5 = 10
0 + 5 + 5 = 10 0 + 1 + 9 = 10
2 + 3 + 5 = 10 0 + 2 + 8 = 10 0 + 1 + 9 = 10
0 + 3 + 7 = 10 0 + 4 + 6 = 10
The 10 – point scale is used for the following operation.
a. Using subtraction to simply addition
b. Using addition to simply subtraction
c. Simplifying addition by groups of 10

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Example:
49 is close to 50 and is 1 short
So, 59 + 4 = 60 + 4 – 1 = 63
38 + 24 = 40 + 24 – 2 = 64 – 2 = 62
Subtracting Near a Base
When subtracting a number close to a multiple of 10, just subtract from the multiple of 10 and
correct the answer accordingly.
Example: Subtract 29 from 53.
Solution:
29 is close to 30, just 1 lower. So, subtract 30 from 53 making 23, then add 1 to make 24.
That is, 53 – 29 = 53 – (30 – 1)
= 53 – 30 + 1
= 24.
This process can be done mentally.
ADDITION AND SUBTRACTION
ADDITION:
In the convention process we perform the addition operation is as follows.

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234 + 403 + 564 + 721 written as
234
403
564
721
Step 1: 4 + 3 + 4 + 1 = 12 2 retained and 1 is carried over to left.
Step 2: 3 + 0 + 6 + 2 +1 (previous carry) = 12; 2 is retained and 1 is carried over to left
Step 3: 2 + 4 + 5 + 7 + 1(previous carry) = 19
As the addition process ends, the same 19 is retained in the left had most part of the answer.
Thus,
234
403
564
+721
_____
1922 is the answer
Vedic Addition
Step 1: Carry out the addition column by column in the usual fashion, moving from bottom to top or
top to bottom.
1 + 4 = 5; 5 + 3 = 8; 8 + 4 = 12;
The final result is more than 9. The tenth place ‘1’ is dropped once number in the unit place
i.e., 2 retained.
Now, you place a dot at the top 4. Thus column (1) of addition (right to left) as shown
below:
.
4
3
4
1
__
2
Step 2: Before coming to column (2) addition, the number of dots are to be counted. This shall be
added to the bottom number of column (2) and we proceed as above. Now, the Thus second
column becomes
.
3 dot=1, 1 + 2 = 3
0 3 + 6 = 9
6 9 + 0 = 9
2 9 + 3 = 12
__
2

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Here, 2 retained and ‘.’ is placed on top number 3
Step 3: Similarly, we can proceed as above for column (3)
2 i) dot = 1
4 ii) 1 + 7 = 8
. iii) 8 + 5 = 13
5 A dot is placed on 5 and proceed with retained
7 with retained unit place 3.
__
9 iv) 3 + 4 = 7, 7 + 2 = 9 Retain 9 in 3rd
digit i.e.,in 100th
place.
Step 4: Now the number of dots is counted. Here it is 1 only and the number is carried out left side
ie. 1000th
place
. .
Thus 234
403
.
564
+721
____
1922 is the answer.
Example 2:
Step 1: 6 + 4 = 10, 1 dot ; 0 + 8 = 8; 8 + 4 = 12;
Step 2: 2+2 (2 dots) = 4; 4+9 = 13: 1 dot and 3+0 = 3; 3+8 = 11;
1 dot and 1 answer under second column - total 2 dots.
Step 3: 3+2 (2 dots) = 5; 5+6 = 11:1 dot and 1+7 = 8; 8+7 = 15;
1 dot and 5 under third column as answer - total 2 dots.
Step 4: 4 + 2 ( 2 dots ) = 6; 6 + 5 =11:
1 dot and 1+3 = 4; 4+2 = 6. Total 1 dot in the fourth 6 column as answer.
1 dot in the 4th
column carried over to 5th
column (No digits in it) as 1.
Example 3:

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Example 4:
Example 5:
Exercises:
Vedic Addition – Checking
Example: 437 + 624 + 586 + 162 = 1809.
By using sum of digits method (beejank) method, the Beejanks are
437  4 + 3 + 7  14  1 + 4  5
624  6 + 2 + 4  12  1 + 2  3
586  5 + 8 + 6  19  1 + 9  10  1 + 0  1
162  1 + 6 + 2  9
Now
437 + 624 + 586 + 162  5 + 3 + 1 + 9  18  1 + 8  9

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Beejank of 1809  1 + 8 + 0 + 9  18  1 + 8  9 (verified)
Vedic Subtraction : - Checking
Example: 3274 – 1892 = 1382
3274  3 + 2 + 7 + 4  3 + 4  7 1892  1 + 8 + 9 + 2  2
3274 – 1892  7 – 2  5
1382  1+3+8+2  14  1 + 4  5
Hence verified.
Bar Numbers
29 is close to the number 30. So 29 can be rewritten as 13
13 means 30 – 1 = 29. 25 means 50 – 2 = 48
36 means 60 – 3 = 57 214 means 400 – 10 + 2 = 392
134 means 400 – 13 = 387 4 – 2 = 2
3 – 7 = 4 5 – 6 = 1
Negative numbers can be replaced with their bar number equivalent, so
4 3 5
- 2 7 6
---------
2 14 = 412 = (200 – 41 = 159)
--------
Arithmetic of Bar Numbers
The original problem can be rewritten in three different ways using bar numbers.
In each case, the result is 77.

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Example:
Doubling and Halving
Sutra 10: Yavdunam (Double)
Sub-Sutra 1: Anurupyena (Proportionately)
Lesson: Mentally multiplying and Dividing by 2
Mentally multiplying and Dividing by 4 and 8
Mentally multiplying and Dividing by 5, 50 and 25
Using number relationships to simplify a problem
Doubling
Adding a number to itself is called Doubling.
Example:
2 3
+ 2 3
------------
2 + 2  4 6  3 + 3
Mentally, we can double each column and then combine results.

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5 8  Double 50 and add double 8
+ 5 8
---------
1 1 6  100 + 16 = 116
Grouping columns may simply the problem
2 6 3  double 260 and add double 3
+ 2 6 3
-----------
5 2 6  520 + 6 = 526
Doubling can be used to multiply by 4, 8 ; just double the number twice.
Example: 35 x 4 = (35 x 2) x 2 = 70 x 2 = 140
163 x 4 = 326 x 2 = 652
35 x 8 = ((35 x 2) x 2) x 2 = 70 x 4 = 140 x 2 = 280
163 x 8 = 326 x 4 = 652 x 2 = 1304
Multiply by 5 by multiplying by 10 and halving the result.
Example:
26 x 5 = (26 x 10) / 2 = 260 / 2 = 130
Multiply by 50 by multiplying by 100 and halving the result.
Example:
43 x 50 = (43 x 100) / 2 = 4300/2 = 2150
Multiply by 25 by multiplying by 100 and halving the result twice.
Example:
68 x 15 = (68 x 100) / 4 = 6800 / 4 = 3400 / 2 = 1700
Divide by 5 by doubling and dividing the result by 10
Example:
320 / 5 = (2 x 320) / 10 = 640 / 10 = 64
Divide by 50 by doubling and dividing the result by 100
Example:
850 / 50 = (850 x 2) / 100 = 1700 / 100 = 17
Divide by 25 by doubling twice and dividing the result by 100
Example:

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325 / 25 = (325 x 4) / 100 = 1300 / 100 = 13
Digit Sums (Navshesh):
The digit sum is the sum of all the digits of a number and is found by adding all of the digits
of a number.
Note 1: If the sum of the digits is greater than 9, then sum the digits of the results again until the
result is less than 10.
2. When finding the Digit sum of a number, “9” and “group of numbers that sum to 9” , then
“9” can be “cast out”.
3. In case of a negative number add 9 to it. The single digit equivalent is 6.
4. This formula simply states that the Navshesh remains unchanged. In other words, Navshesh of
the digits before operation and after operation will remain unchanged.
Note: Keep finding the Digit Sum of the result until it’s less than 10; 0 and 9 are equivalent.
Sutra 15: Gunitasamuchachayah
(The product of the sum of co-efficient in the factors)
Sub-sutra 13: Samuchchyagunitah
(The sum of the coefficients in the product)
Example:
The digit sum of 35 is 3 + 5 = 8 and 142 is 1 + 4 + 2 = 7.
Example:
The digit sum of 57 is 5 + 7 = 12  1 + 2 = 3
The digit sum of 687 is = 6 + 8 + 7 = 21  2 + 1 = 3
Example:
The Digit sum of 94993 is 4 + 3 = 7 (cast out of 9’s)
The digit sum of 549673 is 7 (cast out 5+4, 9, and 6+3)
Example:
The digit sum of – 732 is = - (7 + 3 +2 ) = - 12 = 1( 1 + 2) = - 3 + 9 = 6
Example: 65 + 37 = 102
Navshesh of 65 is 2 and navshesh of 37 is 1 adding navshesh of both the numbers gives
3 and the answer side of Navshesh of 102 is 3.
Example: 873 – 439 = 434
Navshesh of 873 is 9 and navshesh of 439 is 7 Subtracting 7 from 9 gives 2 and the
answer side of Navshesh of 434 is 2.
Example: 106 × 108 = 11448
Navshesh of 106 is 7 and navshesh of 108 is 9 Multiplying 7 with 9 gives 63 whose

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Navshesh is 9 and the answer side of Navshesh of 11448 is 9.
Example: 512 ÷ 8 = 64
Navshesh of 512 is 8 and navshesh of 8 is 8 Dividing 8 with 8 gives 1 whose Navshesh
is 1 and the answer side of Navshesh of 64 is 1.
Checking with Digit sums of Addition and Multiplication Operation
Both addition and multiplication preserve digit sums.
Example:
If the sum of the digit sums does not equal, the digit sum of the sum, then there is a problem.
If the sum of the digit sums does not equal, the digit sum of the product , then there is a
problem.
Sutra 2: Nikhilam Navatascharamam Dastah
(All from NINE and Last from TEN)
When subtracting a number from a power of 10. Subtract all digits from 9 and last digits from 10.
Example:
If the number ends in zero, use the last non-zero number as the last number.

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If the number is less digits, then append zero’s to the start.
When subtracting from a multiple of a power of 10, just decrement the first digit by 1, then subtract
the remaining digits
 Subtract 8675 from 10000 Subtract 875 from 100000
10000 – 8675 = 1325 100000 – 875 = 99125
Subtract 564 from 1,000 and subtract from 357 from 1000.
Solution:
Each figure in 564 is subtracted from nine and the last figure is subtracted from 10, yielding 436.

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Example: Subtract 1049 from 10000
Number Splitting: Splitting number to simplify the problem
Quick mental calculations can be performed more easily if the numbers are “Split” into more
manageable parts.
Note: Please be careful to decide where to place the “split” line. It’s often best to avoid number
“carries” over the line.
The same can be done for subtraction also.

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The same can be done for multiplication also.
The same can be done for division also.
The split may also require more parts.
Sutra 3: Urdhwa-tiryagbhyam (Criss – Cross)
The multiplication of 2-digit numbers requires four single digit multiplies and series of
summations to combine the results.

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The vedic method requires
1. An addition (crosswise), a single digit multiplication (vertically) and possibly a carry. A carry
may be required when combining the crosswise and vertical results.
2. A subtraction (crosswise) and a single digit multiplication (vertically)
Example:

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Multiplying numbers just below base 100
Example:
Suppose you want to multiply 88 by 98.
Both 88 and 98 are close to 100. 88 is 12 below 100 and 98 is 2 below 100.
As before the 86 comes from subtracting crosswise:
88 – 2 = 86 (or 98 – 12 = 86). This is the Left Hand side Answer.
And vertically multiply 12 x 2 = 24. This is the right hand side answer.
So 88 x 98 = 8624
Multiplying numbers just over 100.
The multiplication of 3-digit numbers requires six single digit multiplies and series of

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Another Method
Add the excess number(“x”) to the other number "y" and then tag on the multiplication of
those two excesses: "x" and "y".
Example: 108 x 109 = 109 + 8 / 8 x 9
= 117 / 72
= 11,772
Example:
103 x 104 = 103 + 4 / 3 x 4 107 x 106 = 107 + 6 / 7 x 6
= 107 / 12 = 113 / 42
= 10712 = 11342
Example: Multiplying number close to 100 but on either side
The multiplication of larger numbers requires numerous single digit multiplies and series of

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Example:
46 x 192 can be rewritten as 92 x 96 by doubling and halving.
46 x 192 = 92 x 96 = 8832 (vertically/ crosswise)
To multiply two numbers with a difference of 1
Procedure:
a. Square either number whichever is easier
b. If you square the larger number, subtract if from its square
c. If you square the smaller number, add it to its square.
Example: 34 x 33
a. Square the larger number 34 as it is easier using the method “Squaring a number ending in 4”
b. 342
= 352
– (34+35)
= 1225 – 69 = 1156
c. 1156 – 34 = 1122
Procedure:
a. Obtain the average of two numbers
b. Square the average
c. Subtract 1 from (b)
Example: 49 x 51
a. Average = 50
b. 502
= 2500
c. 2500 – 1 = 2499
a. Add 1 to the smaller number
b. Square the sum
c. Subtract 1 from the smaller number
d. Add (b) and (c) to get the answer

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Example: 27 x 24
a. Smaller number is 24. Therefore, 24 +1 = 25
b. 252
= 625
c. 24 – 1 = 23
d. 625 + 23 = 648
a. Obtain the average of the two numbers
Example: 77 x 73
a. Average = 75
b. 752
= 5625
c. 5625 – 4 = 5621
Procedure:
a. Obtain the average of the two numbers
Example: 44 x 38
a. Average = 41
b. 412
= 1681
c. 1681 – 9 = 1672
So same rule is applicable for case 1, but not for case 2.
We will take one more example of the same kind.
6 9
x 6 1
--------
42 09
--------
Here we can see that right digits sum is 10 i.e.(9+1) and left side digits are same. So we can now
apply the same method.
1. First, multiply the right side numbers(1 x 9) and the result is 09.
2. Second, multiply 6 by the number that follows it, i.e.7, so the result of (6 x 7) is 42.
3. And now the final output is 4209.

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Now, following are some of the problems for your practice.
1. 76 x 74
2. 33 x 37
3. 91 x 99
4. 85 x 85
5. 55 x 55
So this is all for today. Hope you have enjoyed the 'Vedic Math' tricks. We shall come up with
more tricks soon.
Multiplying Number near different bases
Squaring a number near a base (use the same technique)

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Sutra 1: Ekadhikena Purvena (One more than the previous)
Sutra 14: Ekanyunena Purvena (One less than the previous)
Sub-sutra 3: Adyamadyenantyamantyena (First by first and last by last)
Multiplying by 11
When the digit sum of the two digit number is less than 10
a. Add the numbers digit
b. Insert the sum between the digits
When the digit sum of the two digit number is more than 10
a. Obtain the digit sum
b. Increase the LHS of the two digit number by 1
c. Attach the unit digit of the digit sum
d. Attach the RHS of the number
Example 1: Multiply 45 by 11
Example 2: 57 x 11
Example 3: 243 x 11

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Example 4: 72 x 11
a. 7 + 2 = 9
b. 792
Example 5: 87 x 11
a. 8+7 = 15. It is more than 10.
b. LHS number is 8. Therefore, 8+1 = 9.
c. The unit digit of the digit sum is 5. Therefore, 95.
The RHS number is 7. Therefore, 957.
Example 6: 25 x 11 = 2 (2+5) 5
= 2 7 5
39 x 11 = 3 (3+9) 9
= 3 12 9 (nb, the "1" of the "12" gets carried over to the left)
= 429
When the 2-digit numbers start with the same digit and their last digits sum to 10, the product
of the two numbers can be easily computed.
Examples:
67 x 63 = 4221 48 x 42 = 2016 86 x 84 = 7224
71 x 79 = 5609 46 x 44 = 2024 47 x 43 = 2021
48 x 42 = 2016 49 x 41 = 2009
When the 2-digit numbers end with the same digit and their first digits sum to 10, the product
of the two numbers can be easily computed.

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When multiplying by a number with all digits equal to “9”, the product of the two numbers
can be easily computed in two parts.
When multiplying numbers, the average can sometimes be used to determine their product.
For example, 31 x 29, the average is 30.
Square this and subtract 1 to determine the product.

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To multiply a number by 21
Procedure:
a. Double the number
b. Multiply (a) by 10
c. Add the given number to (b)
Example: 67 x 21
a. 67 x 2 = 134
b. 134 x 10 = 1340
c. 1340+67 = 1407
Procedure:
a. Multiply the number by 3
Example: 74 x 31
a. 74 x 3 = 222
b. 222 x 10 = 2220
c. 2220 + 74 = 2294
a. Multiply the number by 3
Example: 28 x 41
a. 28 x 4 = 112
b. 112 x 10 = 1120
c. 1120 + 28 = 1148
To multiply a number by 51, 61, 71, 81, 91
a. Multiply the number of the 1st
digit of the multiplier
c. Add the given number to (b).
Example:

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43 x 91
a. 43 x 9 = 387
b. 387 x 10 = 3870
c. 3870 + 43 = 3913
To multiply 15, 25, 34, 45 etc
a. Half the multiplicand
b. Double the multiplier (ending in zero)
c. Multiply (a) and (b)
Example:
28 x 15 42 x 45 17 x 25
a. 14
2
28
 21
2
42
 a.
2
1
8
2
17

b. 15 x 2 = 30 45 x 2 = 90 b. 25 x 2 = 50
c. 14 x 30= 420 21 x 90 = 1890 c. 42550
2
1
8 
To multiply one number by another number when they both end in 5 and the sum
of their digit is even
Example: 35 x 75
Both number end in 5. The sum of the other digits 3+7 = 10 is even
Procedure
a. Multiply the 1st
digits
b. Obtain the average of the first digits
c. Add (a) and (b)
d. Attach 25 to (c)
Example: 35 x 75
a. 3 x 7 = 21
b. 5
2
10
2
73


c. 21+5 = 26
d. Answer = 2625

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To multiply one number by another number when they both end in 5 and the sum
of their digit is odd
Example: 35 x 65
a. Both number end in 5
The sum of the other digits 3+6 = 9 is odd
Procedure
a. Multiply the 1st
digits
b. Obtain the average of the first digits
c. Drop the
2
1
d. Add (a) and (c)
e. Attach 75 to (d)
Example: 85 x 35
a. 8 x 3 = 24
b.
2
1
5
2
11
2
38


c. 5
d. 24+5 = 29
e. Answer = 2925
To multiply a number by 19, 29, 39 etc
Procedure:
a. Add 1 to the multiplier (19, 29, 39 etc) to make it end in zero
b. Multiply the number by (a)
c. Subtract the number from (b)
Example:
24 x 19 36 x 59
a. 19 + 1 = 20 59 + 1 = 60
b. 24 x 20 = 480 36 x 60 = 2160
c. 480 – 24=456 2160 – 36 = 2124
To multiply any two numbers between 11 and 19
Procedure:

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a. Add either number to the units digit of the other
b. Multiply by 10
c. Multiply the unit digit
d. Add (b) and (c)
Example: 14 x 18
a. 14 + 8 = 22 OR 18 + 4 = 22
b. 22 x 10 = 220
c. 4 x 8 = 32
d. 220 + 32 = 252.
Nikhilam Method
• The formula is : “all from 9 and the last from 10” . This formula can be very effectively applied in
multiplication of numbers, which are nearer to bases like 10, 100, 1000 i.e., to the powers of 10 .
• The procedure of multiplication using the Nikhilam involves minimum number of steps, space,
time saving and only mental calculation. The numbers taken can be either less or more than the base
considered.
• The difference between the number and the base is termed as deviation. Deviation may be
positive or negative. Positive deviation is written without the positive sign and the negative deviation,
is written using Rekhank (a bar on the number).
Example:
MULTIPLICATION OF TWO DOUBLE DIGIT NUMBER CLOSE TO THE BASE
NUMBER
CASE 1: WHEN BOTH THE NUMBERS ARE LESS THAN THE BASE NUMBER
EXAMPLE: 98 X 97
Procedure:
1. Obtain how much each number is less than 100 and write them below each number with
negative sign.
2. Subtract these numbers cross-wise. This number is the first two digits of the answer (Left
Hand Side Part of the answer)
3. Multiply the difference. This number should be written in double digit. This number is the last
two digits of the answer (Right Hand Side Part of the answer)

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CASE 2: WHEN BOTH THE NUMBERS ARE MORE THAN THE BASE NUMBER
EXAMPLE: 104 X 108
Procedure:
1. Obtain how much each number is more than 100 and write them above each number with
positive sign.
2. Add these numbers cross-wise. This number is the first three digits of the answer (Left Hand
Side Part of the answer)
3. Multiply the difference. This number should be written in double digit. This number is the last
two digits of the answer (Right Hand Side Part of the answer)
CASE 3: WHEN ONE NUMBER IS MORE THAN THE BASE 100 AND THE OTHER
NUMBER IS LESS THAN THE BASE NUMBER 100
EXAMPLE: 96 X 109
Procedure:
1. Obtain how much each number is less or more than 100 and write them with sign negative or
positive.
2. Subtract cross-wise, if the sign is negative OR Add cross-wise if the sign is positive.
3. Multiply step 2 answer by 100
4. Multiply the difference.
5. Subtract step 4 from step 3.
6. This number should be written in double digit. This number is the last two digits of the answer
(Right Hand Side Part of the answer)

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Note:
The same method can be applied when both the numbers are close to 10, 100, 10000 etc. And also
with the slight modification when both the numbers are close to 20, 30, 40 etc, 200, 300, 400 etc,
2000, 3000, 4000 etc can be mentally calculated in just few seconds.
The conventional approach is-
35
x 35
-------
175
105
--------
1225
--------
In above problem, we followed the following steps:
1. In first step, we multiply 5 by 35, get 175 and wrote it below the line.
2. In second step, we multiply 3 by 35, get 105, wrote it below the first step and leave one space
from right.
3. In last, we add results from both the steps and get 1225 as answer.
Now here is the magical trick or quicker way to do this calculation using Vedic Math (to square any
number with a 5 on the end). Let us have a look on the same example once again, following 'Vedic
Math' steps to solve it.
1. In 35, the last digit is 5 and other number is 3.
2. Add 1 to the top left digit 3 to make it 4 (i.e. 3+1=4) (See the image below).
3. Then multiply original number '3' with increased number i.e. '4'. Like 3 x 4, and we get 12.
4. Now you can see that this is the left hand side of the answer.
5. Next, we multiply the last digits, i.e 5 x 5 and write down 25 to the right of 12.
6. And here we come up with a desired answer, 1225

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MULTIPLICATION
Example 1 :
2 Digit by 2 Digit Multiplication : Format  Multiplicand x Multiplier
Procedure:
1. Multiply the right hand digit of the multiplicand (unit digit) and multiplier (unit digit). Write the
sum and left out the carry.
2. Perform cross multiplication and add the two products with the previous carry in the first step.
Write the sum and left out the carry.
3. In last step, multiply the left hand digit of both multiplicand and multiplier and the carry in the
second step. Write the sum.
Example 1: Multiply 43 x 28
Example 2 :
Procedure:
2. Perform cross multiplication of units and ten position and add the two products with the previous
carry in the first step. Write the sum and left out the carry.
3. Perform cross multiplication of units and 100th
position plus multiplication of 10th
position and
add the three products with the previous carry in the 2nd
step. Write the sum and left out the carry.
4. Perform cross multiplication of 10th
and 100th
position and add the two products with the
previous carry in the 3rd
5. In last step, multiply the left hand digit of both multiplicand and multiplier and add the carry in
the 4th
step. Write the sum.

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Example: 118 x 486
Example 3 :
Procedure:
position and
and 100th
position plus cross multiplication of units and
1000th
position and add these four products with the previous carry in the 3rd
step. Write the sum and
left out the carry.
and 1000th
position and multiply 100th
position and add these
products with the previous carry in the 4th
and 1000th
position and the two products with the previous
carry in the 5th
the 6th

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Example 4 :
Procedure:
position and
and 100th
position plus cross multiplication of units and
1000th
position and add these four products with the previous carry in the 3rd
step. Write the sum and
left out the carry.
position plus cross multiplication of 10th
and
1000th
position and multiply 100th
position and add these products with the previous carry in the 4th
and 1000th
position and cross multiplication of 10000th
position and 10th
position and add these products with the previous carry in the 5th
step. Write the sum
and left out the carry.

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and 10000th
position and direct multiplication of 1000th
position and these two products with the previous carry in the 6th
step. Write the sum and left out the
carry.
8. Perform the cross multiplication of 1000th
and 10000th
position and add these products with the
previous carry in the 7th
the 8th
Note:
1. If you want to multiply a three digit number by a two digit number (say 481 x 39), then put a
zero in the left side of the double digit number (039) and then multiply applying 3 digit number by 3
digit number as it is 3 digit by 3 digit multiplication.
2. Similarly, one can do it for 4 digit x 2 digit, 4 digit x 3 digit multiplication etc.
Square of a number Ending in 1
Procedure
a. Square the next lower number (always end in zero)
b. Obtain the sum of the next lower number and the number being square.
c. Add (a) and (b) to get the answer.

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Example: 41 x 41 = 412
a. The next lower number is 40. Square of 40 = 1600
b. 40 + 41 = 81
c. 1600 + 81 = 1681
Procedure:
a. Square the next higher number (always end in 5)
b. Obtain the sum of the number being squared and the next higher number
c. Subtract (b) from (a) to get the answer
Example: 64 x 64 = 642
a. The next higher number is 65. Therefore, 652
= 4225
b. 64 + 65 = 129
c. 4225 – 129 = 4225 – 100 – 30 + 1 = 4096
Procedure:
a. Multiply the 1st
number by its consecutive next number.
b.Attach 25.
Example:
35 x 35 = 352
a. 1st
digit is 3 and next digit is 4. Therefore, 3 x 4 = 12.
b. Attach 25.
c. Answer = 1225
152
= 1 x 2 / 5 x 5 = 2 / 25 = 225
352
= 3 x 4 / 5 x 5 = 12 / 25 = 1,225
452
= 4 x 5 / 5 x 5 = 20 / 25 = 2,025
752
= 7 x 8 / 5 x 5 = 56 / 25 = 5625
852
= 8 x 9 / 5 x 5 = 72 / 25 = 7225
952
= 9 x 10 / 5 x 5 = 90 / 25 = 9, 025
1252
= 12 x 13 / 25 = 156 / 25 = 15625
Procedure

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a. Square the next lower number (always end in 5)
b. Obtain the sum of the number being squared and is next lower number
c. Add (a) and (b) to get the answer
Example: 56 x 56 = 562
a. The next lower number is 55. Therefore, 552
= 3025
b. 56 + 55 = 111
c. 3025 + 111 = 3025 + (100+10+1) = 3136
Procedure:
a. Square the next higher number (always end in zero)
b. Obtain the sum of the number being squared and the next higher number
c. Subtract (b) from (a) to get the answer
Example: 39 x 39 = 392
a. The next higher number is 40. Square of 40 = 1600
b. 39 + 40 = 79
c. 1600 – 79 = 1521.
Find out the Square of a number near 100 (close to 100 or lesser than 100)
a. Subtract the base difference from the given number
b. Add the square of the difference number.
Example: Find the square of 97.
1. 97 is 3 less than 100 (that is – 3).
2. 972
= 97 – 3/ 32
= 94 / 09 = 94 09
2. 962
= 96 – 4/ 42
= 92 / 16 = 92 16
2. 872
= 87 – 13/ 132
= 74 / 1,69 = 7569 (The carry 1 is added to the 4)

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