In Vedic Mathematics, there are two types of techniques: specific techniques and general techniques. The specific techniques are those which are fast and effective but can be applied only to a particular combination of numbers. For example, the technique of squaring numbers ending with 5 is a specific technique because it can be used to square only those numbers that end with 5. It cannot be used to square any other type of number. On the other hand, the technique of multiplication as given by the Criss-Cross System is a general technique, as it can be used to multiply numbers of any possible combination of digits.

Thus, general techniques have a much wider scope of application than specific techniques because they deal with a wider range of numbers. In this book, we will give more emphasis to general techniques as they provide much wider utility. Chapter 1 deals with the specific techniques; from Chapter 2 onwards we will study general techniques.

We will discuss the following techniques in this chapter:

1. Squaring of numbers ending with 5
2. Squaring of numbers between 50 and 60
3. Multiplication of numbers with a series of 9’s
4. Multiplication of numbers with a series of 1’s
5. Multiplication of numbers with similar digits in the multiplier
6. Subtraction using the rule ‘All from 9 and the last from 10’

The first technique that we will discuss is how to instantly square numbers whose last digit is 5. Remember, squaring is multiplying a number by itself. When we multiply 6 by 6 we get the answer 36. This 36 is called the square of 6.

(a) Squaring of numbers ending with ‘5’

Squaring is multiplying a number by itself. Let us have a look at how to square numbers ending in 5.

(Q) Find the square of 65.

65X 65

——-

4225

• In 65, the number apart from 5 is 6.
• After 6 comes 7. So, we multiply 6 by 7 and write down the answer 42.
• Next, we multiply the last digits (5 X 5) and write

down 25 to the right of 42 and complete our multiplication.

• Our answer is 4225.

(Q) Find the square of 75. 7 5

X 7 5 = 5625

Apart from 5 the number is 7. The number that comes after 7 is 8. We multiply 7 with 8 and write the answer 56. Next, we multiply the last digits (5 X 5) and put 25 beside it and get our answer as 5625. Thus, (75 X 75) is 5625.

(Q) Find the square of 95.

Apart from 5 the digit is 9. After 9 comes 10. When 9 is multiplied by 10 the answer is 90. Finally, we vertically multiply the right-hand digits (5 X 5) and write the answer 25 beside it. Thus, the square of 95 is 9025.

(Q) Find the square of 105.

The previous examples that we solved were of two digits each. But the same technique can be extended to numbers of any length. In the current example, we will try to determine the square of a three-digit number: 105.

Apart from 5 the digits are 1 and 0, that is, 10. After 10

comes 11. We multiply 10 with 11 and write the answer as 110. We suffix 25 to it and write the final answer as 11025. The square of 105 is 11025.

So you can see how simple it is to square numbers ending with a 5! In fact, you can mentally calculate the square of a number ending with a 5. Just multiply the non-five numbers with the next number and then multiply the last digits (5 X 5) and add 25 after it.

A few more examples are given below:

152	=	225
252	=	625
352	=	1225
452	=	2025
552	=	3025
852	=	7225
1152	=	13225 (11 X 12 = 132)
2052	=	42025 (20 X 21 = 420)

 Thus we see that the technique holds true in all the examples.

The technique of squaring numbers ending with 5 is a very popular technique. Some educational boards have included it in their curriculum. In Vedic Mathematics, there is an extension to this principle which is not known to many people. This formula of Vedic Mathematics tells us that the above rule is applicable not only to the squaring of numbers ending in 5 but also to the multiplication of numbers whose last digits add to 10 and the remaining digits are the same.

Thus, there are two conditions necessary for this multiplication. The first condition is that the last digits should add to 10 and the second condition is that the remaining digits should be the same.

Let us have a look at a few examples:

66	107	91	51
X 64	X 103	X 99	X 59

In the above examples, it can be observed that the last digits in each case add up to 10 and the remaining digits are the same. Let us take the first example:

Here the last digits are 6 and 4 which add up to 10. Secondly, the remaining digits are the same, viz. ‘6’ and ‘6.’ Thus, we can find this is not the square, rather the product of these two numbers by the same principle which we used in squaring numbers ending with a 5.

• First, multiply the number 6 by the number that follows it. After 6 comes 7. Thus, (6 * 7) is 42.
• Next, we multiply the right-hand digits (6 * 4) and write the answer as 24. The complete answer is 4224.
• In the third example, we have to multiply 91 by 99. 107 * 103 =11021

• We multiply 9 by the number that follows it, 10, and write the answer as 90.
• We multiply the numbers (1 ´ 9) and write the answer as 09. The final answer is 9009.

(Note: The right-hand part should always be filled in with a two-digit number. Thus, we have to convert the number 9 to 09.) In the last example, we have to multiply 51 by 59. 51 * 59 =3009

• We multiply 5 with the next number 6 and write the answer as 30.
• Next, we multiply (1 * 9) and write the answer as 09. The final answer is 3009

This formula of Vedic Mathematics works for any such numbers whose last digits add up to ten and the remaining digits are the same. The same formula works while squaring numbers ending with 5 because when you square two numbers ending with 5, then the right-hand digits add to 10 (5 + 5) and the remaining digits are the same (since we are squaring them).

Let us look at a few other examples where the right-hand digits add to 10 and the remaining digits are the same.

72 X 78 = 5616

84 X 86 = 7224

23 X 27 = 621

89 X 81 = 7209

106 X 104 = 11024

1003 X 1007 = 1010021

This was the first specific technique that we studied. The next technique that we will discuss is also related to squaring. It is used to square numbers that lie between 50 and 60.

(b) Squaring of numbers between 50 and 60

We have taken four different examples above. We will be squaring the numbers 57, 56, 52 and 53 respectively. We can find the answer to the questions by taking two simple steps as given below:

• Add 25 to the digit in the units place and put it as the left-hand part of the answer.
• Square the digits in the units place and put it as the right- hand part of the answer. (If it is a single digit then convert it to two digits.)
• Find the square of 57. 5 7

X 57=3249

• In the first example we have to square 57. In this case we add 25 to the digit in the units place, viz. 7. The answer is 32 which is the LHS (left-hand side) of our answer. (Answer at this stage is 32 )
• Next, we square the digit in the units place ‘7’ and get the answer as 49. This 49 we put as the right-hand part of our answer. The complete answer is 3249.

(Q) Find the square of 56. 5 6

X56 = 3136

In the second example, we add 25 to 6 and get the LHS as

31. Next, we square 6 and get the answer 36 which we put on the RHS. The complete answer is 3136.

(Q) Find the square of 52. 5 2

 X52 = 2704

In the third example, we add 2 to 25 and get the LHS as 27. Next, we square 2 and get the answer 4 which we will put on the RHS. However, the RHS should be a two-digit number. Hence, we convert 4 to a two-digit number and represent it as

04. The complete answer is 2704.

(Q) Find the square of 53. 5 3

 X 53 = 2809

In the last example, we add 3 to 25 and get the answer as

28. Next, we square 3 and get the answer as 9. As mentioned in rule B, the answer on the RHS should be converted to two digits. Thus, we represent the digit 9 as 09. The complete answer is 2809.

On similar lines we have:

51² = 2601

52² = 2704

54² = 2916

55² = 3025

58² = 3364

(c) Multiplication of numbers with a series of 9’s

In my seminars, I often have an audience challenge round. In this round, the audience members ask me to perform various mental calculations and give them the correct answer. They generally ask me to multiply numbers which involve a lot of 9’s in them. The general perception is that the higher the number of 9’s the tougher it will be for me to calculate. However, the truth is exactly the opposite—the higher the number of 9’s in the question, the easier it is for me to calculate the correct answer. I use two methods for this. The first method is given below and the second method is explained in the chapter ‘Base Method of Multiplication.’

Using the method given below, we can multiply any given number with a series of nines. In other words, we can instantly multiply any number with 99, 999, 9999, 99999, etc.

The technique is divided into three cases. In the first case, we will be multiplying a given number with an equal number of nines. In the second case we will be multiplying a number with a higher number of nines. In the third case, we will be multiplying a number with a lower number of nines.

Case 1

(Multiplying a number with an equal number of nines.)

(Q) Multiply 654 by 999.

654 * 999 653 346

• We subtract 1 from 654 and write half the answer as 653. Answer at this stage is 653      .
• Now we will be dealing with 653. Subtract each of the digits 6, 5 and 3 from 9 and write them in the answer one by one.
• Y9
• YYY

6     5     3

3     4     6

• 9 minus 6 is 3. 9 minus 5 is 4. 9 minus 3 is 6.
• The answer already obtained was 653 and now we suffix to it the digits 3, 4 and 6. The complete answer is 653346.

(Q) Multiply 9994 by 9999.

9994 * 9999 = 99930006

We subtract one from 9994 and write it as 9993. This becomes our left half of the answer. Next, we subtract each of the digits of 9993 from 9 and write the answer as 0006. This becomes the right half of the answer. The complete answer is 99930006.

(Q) Multiply 456789 by 999999.

456789 * 999999456788 543211

We subtract 1 from 456789 and get the answer 456788. We write this down on the LHS. Next, we subtract each of the digits of 456788 (LHS) from 9 and get 543211 which becomes the RHS of our answer. The complete answer is 456788543211.

More examples:

7777	65432	447	90909
X 999	X 99999	X 99	X 99999
7776 2223	65431 34568	446 553	90908 09091

The simplicity of this method can be vouched by the examples given above. Now we move to Case 2. In this case, we will multiply a given number with a higher number of nines.

Case 2

(Multiplying a number with a higher number of nines.)

(Q) Multiply 45 with 999.

45* 045

045 * 999 = 044955

There   are   three   nines   in   the   multiplier.   However, the multiplicand 45 has only two digits. So we add a zero and convert 45 to 045 and make it a three-digit number. After having done so, we can carry on with the procedure explained in Case 1.

First we subtract 1 from 045 and write it down as 044. Next, we subtract each of the digits of 044 from 9 and write the answer as 955. The complete answer is 044955.

(Q) Multiply 888 with 9999.

We convert 888 to 0888 and make the digits equal to the number of nines in the multiplier. Next, we subtract 1 from 0888 and write the answer as 0887. Finally, we subtract each digit of 0887 from 9 and write the answer as 9112. The final answer is 08879112 which is 8879112.

(Q) Multiply 123 by 99999.

The multiplicand is a three-digit number and the multiplier is a five-digit number. Therefore, we add two zeros in the multiplicand so that the digits are equal in the multiplicand and the multiplier.

We now subtract 1 from 00123 and write the LHS of the answer as 00122. Next, we subtract each of the digits of the LHS of the answer from 9 and write it down as 99877 as the RHS of the answer. The complete answer is 12299877.

Other examples:

 X 9999 X 99999 X 999999    X 9999999

162                    5555                         363                         10101

0161 / 9838 05554 / 94445 000362 / 999637 0010100 / 9989899

We can see that this technique is not only simple and easy to follow but also enables one to calculate the answer in the mind itself. This is the uniqueness of these systems. As you read the chapters of this book, you will realize how simple and easy it is to find the answer to virtually any problem of mathematics that one encounters in daily life and especially in the exams. And the approach is so different from the traditional methods of calculation that it makes the whole process enjoyable.

Case 3 of this technique deals with multiplying a number with

a lower number of nines. There is a separate technique for this in Vedic Mathematics and requires the knowledge of the Nikhilam Sutra (explained later in this book). However, at this point of time, we can solve such problems using our normal practices of instant multiplication.

(Q) Multiply 654 by 99.

In this case the number of digits is more than the number of nines in the multiplier. Instead of multiplying the number 654 with 99 we will multiply it with (100 – 1). First we will multiply 654 with 100 and then we will subtract from it 654 multiplied by 1.

654 * 99 = 65400-654 = 64746

(Q) Multiply 80020 by 999.

We will multiply 80020 with (1000 – 1).

80020000 – 80020 = 79939980

This method is so obvious that it needs no further elaboration.

(d) Multiplication of numbers with a series of 1’s

In the previous technique we saw how to multiply numbers with a series of 9’s. In this technique we will see how to multiply numbers with a series of 1’s. Thus, the multiplier will have numbers like 1, 11, 111, etc.

Let us begin with the multiplier 11.

(Q) Multiply 32 by 11.

32*11 = 352

• First we write the right-hand  digit 2 as it is.

(Answer =   2).

• Next, we add 2 to the number on the left 3 and

write 5. (Answer =   52).

• Last, we write the left-hand digit 3 as it is.

Thus, the answer is 352.

(Q) Multiply 43 by 11.

43*11 = 473

Write the last digit 3 as it is. Next we add 3 to 4 and get 7. Finally we write 4 as it is. The complete answer is 473.

(Q) Multiply 64 by 11.

64*11 =704

In this example we write down the last digit 4 as it is. Next, we add 4 to 6 and get the answer 10. Since 10 is a two-digit answer, we write down the 0 and carry over 1. Finally, we add 1 to 6 and make it 7. The complete answer is 704.

(Q) Multiply 652 by 11. 6 5 2

*11 = 7172

The logic of two-digit numbers can be expanded to higher numbers. In the given example we have to multiply 652 by 11.

• We write down the last digit of the answer as 2. (Answer =   2).
• Next, we add 2 to 5 and make it 7. (Answer =    72).
• Next, we add 5 to 6 and make it 11. We write down 1 and carry over 1. (Answer is 172).
• Last, we take 6 and add the one carried over to make it

7. (Final answer is 7172).

(Q) Multiply 3102 by 11.

3102 * 11 =34122

• We write down 2 as it is. (Answer is  2)
• We add 2 to 0 and make it 2. (Answer is  22)
• We add 0 to 1 and make it 1. (Answer is  122)
• We add 1 to 3 and make it 4. (Answer is   4122)
• We write the first digit 3 as it is. (Final answer is 34122)

Similarly,

41	303	1309	2901265
X 11	X 11	X 11	X 11
451	3333	14399	31913915

When we multiply a number by 11 we write the last digit as it is. Then we move towards the left and continue to add two digits at a time till we reach the last digit which is written as it is.

Since the multiplier 11 has two 1’s we add maximum two digits at a time. When the multiplier is 111 we will add maximum three digits at a time because the multiplier 111 has three digits. When the multiplier is 1111 we will add maximum four digits at a time since the multiplier 1111 has four digits.

We have already seen how to multiply numbers by 11. Let us have a look at how to multiply numbers by 111.

(Q) Multiply 203 by 111.

203*111

• We write down the digit in the units place 3 as it is in the answer.
• We move to the left and add (3 + 0) = 3.
• We move to the left and add (2 + 0 + 3) = 5 (maximum three digits).
• We move to the left and add (0 + 2) = 2.
• We take the last digit 2 and write it down as it is.

(Q) Multiply 201432 by 111.

201432 * 11122358952

• The (2) in the units place of the multiplicand is written as the units digit of the answer.
• We move to the left and add (2 + 3) and write 5.
• We move to the left and add (2 + 3 + 4) and write 9.
• We move to the left and add (3 + 4 + 1) and write 8.
• We move to the left and add (4 + 1 + 0) and write 5.
• We move to the left and add (1 + 0 + 2) and write 3.
• We move to the left and add (0 + 2) and write 2.
• We move to the left and write the single digit (2) in the answer.

Thus, the complete answer obtained by each of the steps above is 22358952.

Similarly,

The simplicity of this method is evident from the examples. In most cases you will get the answer within a minute. In fact, the beauty of this technique is that it converts a process of multiplication to basic addition.

Using the same method, we can multiply any number by a series of 1’s.

If you want to multiply a number by 1111 you can use the method given above. The only difference will be that we will add maximum four numbers at a time (because there are four ones in 1111) and when the multiplier is 11111 we will be multiplying maximum five digits at a time. An example of the former type is given below:

(Q) Multiply 210432 by 1111.

210432 * 1111 233789952

• We write down the last digit 2 as it is                    – 2
• Add (2 + 3) = 5                                                       – 52
• Add (2 + 3 + 4) = 9                                              – 952
• Add (2 + 3 + 4 + 0) = 9                                      – 9952
• Add (3 + 4 + 0 + 1) = 8                                    – 89952
• Add (4 + 0 + 1 + 2) = 7                                  – 789952
• Add (0 + 1 + 2) = 3                                       – 3789952
• Add (1 + 2) = 3                                           – 33789952
• Add (2) = 2                                                – 233789952

Thus, the product of (210432 * 1111) = 233789952.

(e) Multiplication of numbers with a series of similar digits in multiplier

This technique is basically an extension of the previous technique. In technique (c) we saw how to multiply a number with a series of 9’s and in technique (d) we saw how to multiply a number with a series of 1’s.

A question may arise regarding how to multiply numbers with a series of 2’s, like 2222, or with a series of 3’s, like 333, and such other numbers.

Let us have a look at a few examples:

(Q) Multiply 333 by 222.

The question asks us to multiply 333 by 222. Now, carefully observe the logic that we apply in this case

333 * 222

=  333 * 2 * 111 (because 222 is 2 multiplied by 111)

=  666 * 111 (because 333 multiplied by 2 is 666)

Therefore, multiplication of 333 by 222 is the same as multiplication of 666 by 111. But, we have already studied the procedure of multiplying a number by 111 in the previous sub- topic. Our answer will be as follows:

666*111 = 73926

Therefore, 333 * 222 is 73926

(Q) Multiply 3021 by 333.

The multiplicand is a normal number and the multiplier is a series of 3’s. We do not know how to multiply a number with a series of 3’s but we know how to multiply a number with a series of 1’s. Thus, we will represent the expression in such a manner that the multiplier is 111.

3021 * 333

= 3021 * 3 * 111 (because 333 is same as 3 * 111) =

9063 * 111 (because 3021 * 3 is 9063)

We have already learnt how to multiply 9063 by 111. We can easily complete the multiplication.

9063 * 111 = 1005993

On the basis of the above examples it can be seen that there is no need to explain the procedure of multiplication. The procedure is the same as observed in the previous sub-topic. Basically, we have to convert a series of 2’s, 3’s, 4’s, etc. in the multiplier to a series of 1’s by dividing it by a certain number.

Next, we have to multiply the multiplicand by the same number

(a) Multiply 1202 by 44.

1202*44

4808*11 = 52888

In this case, we have divided the multiplier 44 by 4 to obtain a series of 1’s (11). Since we have divided the multiplier by 4 we will multiply the multiplicand 1202 by 4. Thus, we have the new multiplicand as 4808. When 4808 is multiplied by 11 the answer is 52888. This is also the answer to the original question of 1202 by 44.

(b)Multiply 2008 by 5555.

2008 5555

10040 * 1111 = 11154440

Ans: The product of 2008 and 5555 is 11154440

(C)Multiply 10503 by 888.

10503                   84024

X 888               X 111 9326664

Ans: The product of 10503 and 888 is 9326664.

(f) Subtraction using the rule ‘All from 9 and the last from 10’

Subtraction using the rule ‘All from 9 and the last from 10’ is one of the elementary techniques of Vedic Mathematics. Basically, it is used to subtract any number from a power of ten. The powers of ten include numbers like 10, 100, 1000, 10000, etc.

So if you want to learn a method by which you can quickly subtract a number from a power of ten, then this technique can come to your aid.

When we go to the market to buy something, we generally give a `100 note to the shopkeeper and calculate the change that we should get after deducting the total amount of groceries. In such a situation, this technique can come to our aid.

(Q) Subtract 54.36 from 100.

1 0 0.0 0 – 5 4.3 6

We are asked to subtract 54.36 from 100. In this case, we generally start from the right and subtract 6 from 0. But, we realize that it is not possible to subtract 6 from 0 and so we move to the number to the left and then borrow one and give it to zero and make it ten and so on.

This whole procedure is slightly cumbersome and there is a possibility of making a mistake too.

Vedic Mathematics provides a very simple alternative. The approach of Vedic Mathematics is explained by the rule ‘All

from 9 and the last from 10.’ It means that we have to subtract each digit from nine and subtract the last digit from 10. This will give us the answer.

The number to be subtracted is 54.36. We have to subtract all the digits from nine except for the last digit which will be subtracted from ten. Thus,

9 – 5 = 4

9 – 4 = 5

9 – 3 = 6

10 – 6 = 4

The final answer is 45.64.

(Q) Subtract 3478.2281 from 10000.

10000.0000 – 3478.2281

In this case, we will subtract the digits 3, 4, 7, 8, 2, 2 and 8 from 9 and the last digit 1 from 10. The respective answers will be 6, 5, 2, 1, 7, 7, 1 and 9. Thus, the final answer is 6521.7719

More examples:

(a)1000.000	(b) 10000.00	(c)10.000000
-363.633	-9191.09	-7.142857
636.367	808.91	2. 857143

d) 10000                10000 – 23 – 0023 = 9977

The examples prove the simplicity and efficiency of this system. In examples (d) and (e) we added zeros in the number below so that we get the accurate answer.

In this chapter we have seen six simple yet quick techniques of Vedic Mathematics. I wanted to begin the book with these easy techniques so that we can prepare ourselves for the comprehensive techniques that will follow in the forthcoming chapters.

You must have observed that the techniques that we have employed in this chapter work with a totally different approach. We have found answers to our questions using a completely different approach. In all the chapters of this book you will discover that the method used in solving the problems is far more efficient than the normal systems that we have been using, thus enabling us to produce outstanding results.

EXERCISE

Q. (1) Find the product of the following numbers whose last digits add to ten.

(a) 45 * 45

(b) 95 * 95

(c) 111 * 119

(d) 107 * 103

Q. (2) Find the squares of the following numbers between 50 and 60.

• 56
• 51
• 53

Q. (3) Find the product of the following numbers which are multiplied by a series of nines.

(a) 567 * 999

(b) 23249 * 99999

(c) 66 * 9999

(d) 302 * 99999

Q. (4) Find the product of the following numbers which are multiplied by a series of ones.

(a) 32221 * 11

(b) 64609 * 11

(c) 12021 * 111

(d) 80041 * 111

Q. (5) Find the product of the following numbers which are multiplied by a series of same numbers.

(a) 7005 * 77

(b) 1234 * 22

(c) 2222 * 222

(d) 1203 * 333

Q. (6) Subtract the following numbers from a given power of ten.

(a) 1000 minus 675.43

(b) 10000 minus 7609.98

(c) 10000 minus 666 (d) 1000 minus 2.653