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Table of Contents

ONESUBTRACTION BY NIKHILUM..........................................1

Deficit of Number...................................................................1Subtraction..............................................................................4

TWOMULTIPLICATION BY NIKHILUM...................................7

Product of numbers with deficit.............................................7Problems involving carry.......................................................9Multiplication of numbers with decimal fractions..............10Product of numbers with surplus..........................................13Vinculum..............................................................................15Product of a number with surplus and a number with deficit..............................................................................................15Tallying the answer..............................................................17

THREEFAST DESCENDANTS OF NIKHILUM.............................21

The First Corollary...............................................................21The Second Corollary...........................................................24Sub-corollary........................................................................25The Third Corollary..............................................................26

FOURMULTIPLICATION BY URDHWATIRYAK....................31

The Details of the Method....................................................31FIVE

DIVISION BY URDHWATIRYAK......................................39The Details of the Method:...................................................40Finding quotient in decimal format......................................43Three-digit Divisor...............................................................47

SIXSQUARE DRIVE....................................................................51

Duplex or Dwandwa-Yoga:..................................................51Squaring:...............................................................................53

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SEVENSQUARE ROOTS...................................................................57

Some tips about square roots................................................57Finding Square Root.............................................................58Square roots of non-exact squares........................................65

EIGHTQUADRATICS........................................................................67

Conventional method............................................................67Vaidik Method......................................................................68Quadratic Equations.............................................................74

NINEVAIDIK MNEMONICS.........................................................77

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ONESubtraction by Nikhilum

One of the most attractive techniques of Vaidik mathematics is multiplication by Nikhilum method. It is based on the following sutra-

|| fuf[kya uor% pjea nÓr% ||Nikhilum navataha charamam dashtaha

(Meaning: All from nine and the last from ten.)

In this chapter, we are going to learn the original application of this sutra, that is of finding out the deficit of a given number. We are also going to see, in this chapter, how to make the process of subtraction easier by application of this sutra. After mastering the technique of finding the deficit, we shall discuss, in the next chapter, the multiplication by Nikhilum.

Deficit of NumberThe above simple formula essentially gives us, deficit of a

number, also called as ‘Ten’s complement’. By deficit, we mean such a number that should be added to the original number to make it up to the next base of ten above. Here, we are defining base of a number as that power of 10, which is nearest to the number and larger than it. For example,Base for single digit numbers like 3, 7 or 8 is 10.Base for two digit numbers like 45, 69 or 83 is 100.Base for three digit numbers like 125, 667 or 859 is 1000.

The Base for number with fractional part will be decided by its integer portion. For a number with zero integer portion, the base will be 1 or some negative power of 10. For example,Base for 31.425 is 100, as base for 31 is 100.Base for 625.75 is 1000, as base for 625 is 1000.

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Applied Vedic Mathematics

Base for 0.125, 0.3 or 0.6705 is 1, i.e., 100.Base for 0.023, 0.0561 or 0.09 is 0.1, i.e., 10 –1.Base for 0.001, 0.003456 or 0.0092 is 0.01, i.e., 10-2.

The deficit of a number from its base is equal to the base minus the number. For example,Deficit of 3 is 10 – 3 = 7.Deficit of 8 is 10 – 8 = 2.Deficit of 47 is 100 – 47 = 53.Deficit of 885 is 1000 – 885 = 115.Deficit of 3.1415 is 10 – 3.1415 = 6.8585.Deficit of 0.0436 is 0.1 – 0.0436 = 0.0564, and so on.

Such deficits can be easily found by using the above sutra. One can tell verbally and instantaneously the answer. For finding the deficit of a number, start from its left most digit; subtract this digit from 9 and write down the answer. Then go on subtracting the subsequent digits up to the last but one digit from 9 and go on writing the answers. The last digit, according to the sutra, is to be subtracted from 10 and is the last digit of the answer. For example, to find deficit of 7853, we go as follows (of course, this is to be done mentally and only answer is to be written) –

9 9 9 10 - 7 8 5 3

2 1 4 7

Here, we subtracted the first three digits 7, 8 and 5 from 9 and the last digit 3 from 10 to get the answer as 2147. One advantage of this method over our conventional method is, we go here from left to right and thus can instantaneously write or tell the answer. In the conventional method, we have to start from the units place and work leftwards with carrying out as intermediate step. So, telling

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Chapter 1: Subtraction By Nikhilum

the deficit of a large number verbally, is difficult in our regular method; we have to use paper and pencil.

Exercise 1.1:(Level – Minimum middle school student)

Write down the deficits of the following numbers in the space provided. The answer is to be written directly, from left to right.1.53 2.67 3.119 4.294 5.617 6.32755 7.6994475 8.10026 9.3.1415 10.45.35 11.0.123 12.48000 13.0.0753 14.0.050 15.6.1008

You must be surprised by your own speed in solving the above exercises. But hold on! Did you get the answers to problem numbers 12 & 14 right? Most of the students make mistake in these two cases due to crave for speed! Usually, the answers given are 51999 and 0.949. You must be wondering, whether our sutra is wrong then! No, the sutra is not wrong; you have to remember the following sub-sutra or sub-rule:“When there is zero or there are zeroes at the end of a number, the first non-zero digit from the right end is considered as the last and is taken from ten. The ending zeroes remain zeroes in the answer

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also.” Thus, for such numbers the following procedure holds good-

9 10 - 4 8 0 0 0

5 2 0 0 0

Many students commit mistake in solving problem 13 also. The answers given are usually, 9.9247 or 0.9247. Here, we have to remember that the base for the given number 0.0753 is 0.1 and not 10 or 1. So, we start application of the sutra from the first non-zero digit from the left, i.e., 7 and retain the leading zeroes in the answer. Thus the answer is 0.0247. We can tally the answer by adding the original number and its deficit; it should be the base of the number. Here, 0.0753 + 0.0247 = 0.1000 or 0.1 proving that our answer is correct.

SubtractionMany people find calculating the sum of two numbers

easier than calculating the difference between two numbers. For such people, the Nikhilum method provides a way to represent the subtraction problem as an addition problem. Suppose, we want to find out, what is 8232 minus 6897. In our conventional method, we will start from the units place. As 7 can not be subtracted from 2, we borrow a ten from the tens place and subtract 7 from the resultant 12 giving 5. Then we move to tens place, carrying 1 to be added to 9. Then we subtract 10 from 13 and so on. The answer, we can calculate, is 1335. Most people would find difficulty in solving this problem orally.

In Nikhilum way, we first find out quickly the deficit of the second number (number to be subtracted), add this deficit to the first number and subtract the base of second number from the sum giving us the final answer. This process may seem to be longer, but

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Chapter 1: Subtraction By Nikhilum

its actual application is fast and easy. In the present case, we find sum of 8232 and 3103 (deficit of 6897), which is 11335. Subtracting the base of 6897, i.e., 10000, is nothing more than removing the leading 1 from the sum (11335) giving the final answer as 1335. Some more examples follow-

1. 73 - 49 73 + 51 = 124 Remove 1 in the hundred’s place, giving the answer as 24.

2. 234 - 67 234 + 33 = 267 Subtract 1 from the hundred’s place, giving the answer as 167.

3. 3453 - 2876 3453 + 7124 = 10577 Remove the leading 1, giving the answer as 577.

Why does it work ?Consider the problem of finding 73 - 49. We can represent this as 73 - 49 + 100 - 100. Regrouping, we get 73 + (100 - 49) - 100. But 100 - 49 is nothing but deficit of 49. So, the expression reduces to 73 + 51 – 100, i.e., 124 - 100. Removing the digit 1 from 124 is equivalent to subtracting 100 from the number and that gives us the final answer. Note that we are subtracting base of 49.

Exercise 1.2:(Level – Middle School)1.Manoj purchased a tooth-paste for Rs. 28=75 Ps. He gave a

hundred rupee note to the shop keeper. How much money should he get back?

2.In a Loksabha election, the winning candidate polled 1 lakh votes. If the second candidate got 67532 votes, what is the margin in votes of the winner over the second?

3.Boiling point of water is 100 degrees Celsius. If present temperature of water is 39.68 degrees C., what should be rise in temperature, so that it starts boiling?

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4.It takes 1 hour & 40 minutes to reach Ramtek from my house. If, I have already traveled for 53 minutes, how many more minutes are required to reach Ramtek?

5.A rim contains 10 meters of cloth. If 2.25 meters of cloth out of it, is given to a customer, how much will remain in the rim?

6.In a one-day Cricket match, Australia has set a target of 230 runs for India to win. If at the end of the 10 th over, India has made 53 runs, how many more runs are required to win?

7.Richa filled up petrol worth Rs. 121=40 Ps. in her scooter. If she gives 500 Rupee note, how much money should she get back?

8.A cooler requires 4 kg. of wood wool. If we already have 1.755 kg. of wood wool, how much additional wood wool be purchased?

9.Raju had a balance of Rs. 6342/- in his bank account. He paid MSEB bill of Rs. 1675/- by cheque, out of this account. How much balance would remain in his account after this?

10.There were 3000 tickets for a football match. If 1963 of them were sold, how many were not sold?

11.Cost price of a grade of steel is Rs. 5173/-. If the selling price is Rs. 8000/-, what is the profit?

12.The passing percentage for a medical entrance exam was 60%. If Prasad scored 38.79%, how much more percentage did he require to pass?

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TWOMultiplication by Nikhilum

Using the Nikhilum sutra, we can solve certain type of multiplication problems at very high speed compared to our conventional method. This speed advantage is available in the class of problems where the multiplicand and the multiplier have same base and both (or at least one of them is) are very near to the base. If the difference from the base is single digit, then the answer can be given verbally. Even if the bases are different, the problem can be modified and solved.

Product of numbers with deficitTo understand the method, let us find out the product of 9

and 7. Now, you may say, “what’s big deal about it? I know multiplication tables for both 9 & 7 and I can tell the answer immediately”. But, please have patience! We are starting from an easy problem, just to learn the method. Afterwards, we shall move to more difficult problems. At this point, however, note that both 9 & 7 are having deficit; that is, both are less than their common base 10. The steps in the method go as follows-

1)For the calculations, take base as that power of 10 which is nearest to the numbers to be multiplied. (The base need not be larger than the numbers, though in present case it is). Here, both 9 and 7 are nearer to 10, so take 10 as base.

2)Write the two numbers 9 & 7, one above and other below on the left hand side of colons, as shown in the adjoining box.

3)Write down their deficits or surpluses from the base (-1 and –3 respectively, both deficits in the present case) on the right hand

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9 : -1 x 7 : -3 6 / 3


side of colons as shown. The minus sign here, indicates that the number is less than the base, that is, it has deficit.

4)Draw a horizontal line below these. Write the product below this line in two parts, one on the left hand side and one on the right. To separate these two parts, draw a slant line under the two colons.

5)The left hand side part of the answer can be arrived at in any one of the following four ways –a)Sum of the two numbers minus base. 9 + 7 – 10 = 6.b)Base minus sum of the two deficiencies. 10 – (1 + 3) = 6.c)Cross subtract deficiency 3 on the second row from the

multiplicand 9 on the first row. 9 – 3 = 6.d)Cross subtract deficiency 1 on the first row from the

multiplier 7 on the second row. 7 – 1 = 6.6)Calculate the right hand part of the answer by multiplying the

two deficit figures (-1 and –3). The product is +3 or simply 3.Thus, 9 x 7 = 63.

More examples –

9 : -1 8 : -2 7 : -3x 9 : -1 x 7 : -3 x 9 : -1

8 / 1 5 / 6 6 / 3

The advantage of this method is not obvious for finding out product of single digit numbers shown in the above examples, since, anyway we know the multiplication tables for such numbers. In case of problems involving multiplication of two or more digit numbers, the advantage of ease and speed of the Nikhilum method becomes clear. This is because we are breaking down the problem in to subtraction and multiplication of single digits. The following examples illustrate this.

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Chapter 2: Multiplication By Nikhilum

97 : -03 96 : -04 786 : -214 896 : -104x 94 : -06 x 91 : -09 x 998 : -002 x 997 : -003

91 / 18 87 / 36 784 / 428 893 / 312

Problems involving carryUnder certain circumstances, some modifications are

required in the procedure mentioned above. Consider the problem of finding product of 7 and 4. According to the above method, the problem would be solved as follows-

7 : -3x 4 : -6

1 / 18We are getting 118 as answer, instead of the correct answer 28. The rule to be noted here is- We are allowed as many digits on the right hand side of the slant line as many zeroes are there in the base. Here, the base is 10; therefore, we are entitled only to one digit on the right hand side of slant line. So, what do we do with the extra digit? The problem is easily solved with the usual multiplication rule that the surplus portion on the right should always be “carried” over to the left of the slant line. Thus,

7 : -3x 4 : -6

1 / 18 = 28.

Similarly,

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88 : -12x 91 : -09

79 / 108 = 8008.

On the other hand, consider the following problem –

99 : -01x 97 : -03

96 / 3 = 963, which is again wrong!

Here, the vertical multiplication is giving a smaller number of digits than permissible. The solution is, as in case of decimal multiplication, filling up all such vacancies with zeroes on left. The correct procedure is as follows-

99 : -01x 97 : -03

96 / 03 = 9603.

Multiplication of numbers with decimal fractionsThe procedure for decimal multiplication is similar to the

one adopted in the conventional mathematics. First disregard the decimal points and find out the product of the resultant numbers; then, put the decimal point at the appropriate place. For example, suppose we want to find the product of 8.88 and 99.1; so, disregarding the decimal points, we get the bare numbers 888 and 991. We find their products with Nikhilum as follows-

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888 : -112x 991 : -009

879 / 1008 = 880008.

There are two places of decimals in 8.88 and one place in 99.1. So, there will be two plus one, that is three places of decimals in the answer. Thus, putting the decimal point after three digits from right, we get the final answer 880.008.

Likewise, the product of 9.99 and 2.35 would be calculated as- 999 : -001x 235 : -765

234 / 765 = 234765 23.4765

Here, one important tip is to always estimate the magnitude of the answer before placing the decimal point. This is done by rounding off, one or both the original numbers. For example, in the first problem above, of finding product of 8.88 and 99.1, round off the multiplier to 100. Then, 8.88 x 100 = 888 is the estimated answer. Our final answer comes in the vicinity of this number, and hence, our decimal point placement is correct. Similarly, in the second problem, round off 9.99 to 10. Then, 10 x 2.35 = 23.5 is the estimated answer, and our final answer comes in its vicinity.

Estimation of the magnitude of the answer is very important, as it avoids silly mistakes. For example, many of my students in hurry, tell that 0.98 x 0.8 is 7.840. Their working goes as follows. First, disregard the decimal points, giving bare numbers as 98 and 8. Then, to make the same base, consider 8 as 80. After this, calculate the product using Nikhilum method as-

98 : -02x 80 : -20

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78 / 40 = 7840.

So far, so good; the mistake occurs while putting the decimal point. In the original numbers, there are 3 places of decimals. So, put the decimal point in 7840 after 3 places from right, which gives 7.840 as answer. This answer is, however, wrong!

For estimating the magnitude of the answer, we round off 0.98 to 1. Then, 1 x 0.8 = 0.8. Our answer should come in the vicinity of 0.8. Thus, it should be 0.7840. What went wrong, in the above process? If we look carefully, to make the base same for both the numbers, we have really treated 0.8 as 0.80 and then disregarded the decimal point in it. In the final answer also, therefore, we should put decimal point after 4 places from right giving 0.7840 as the correct answer.


Calculate the following products. 1. 9 x 9 2. 9 x 5 3. 7 x 9 4. 9 x 8 5. 8 x 8 6. 94 x 98 7. 98 x 91 8. 92 x 97 9. 98 x 88 10. 78 x 97 11. 786 x 998 12. 896 x 997 13. 888 x 991 14. 113 x 998 15. 9998 x 9876 16. 99979 x 99999 17. 99.5 x 0.89 18. 0.895 x 98 19. 9.97 x 0.88 20. 99.98 x 998.2 21. 8.70 x 9

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Product of numbers with surplusIn all the examples, we considered up to now, the

multiplicand and the multiplier were just a little below their common base. But, does the Nikhilum method apply to product of numbers which are just above their base? And the answer is, Yes! The same procedure as in the previous case applies; only difference being, instead of cross-subtracting, we will have to cross-add here. The other rules, like how many digits are allowed to the right of slant line, remain same. A few examples of this type follow-

14 : +4 111 : +11 1005 : +005x 12 : +2 x 109 : +09 x 1009 : +009

16 / 8 120 / 99 1014 / 045


Calculate the following products. 1. 112 x 102 2. 112 x 109 3. 1.05 x 1.01 4. 10.15 x 101.0 5. 1.005 x 1001 6. 1.901 x 10.02 7. 1.18 x 10.2 8. 10009 x 10007 9. 10300 x 10005 10. 15001 x 1001011 1.13 x 1.09 12. 0.125 x 0.104

How does the method work?The Nikhilum method is based on representation of the

problem in a different way. It uses the following algebraic expansion to calculate the answer quickly-

(x – a) (x – b) = x {(x – a) – b} + ab.

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Here, x is the base and a & b are the deficits of the two numbers, respectively. For example, we may represent 95 x 97 as (100 – 5)(100 – 3). Here, x = 100, a = 5 and b = 3. Then, the right hand part of the answer is ab = 5x3 = 15; and left part is 100{(100 – 5) – 3} = 100{95 – 3} = 9200. The final answer is 9200 + 15 = 9215. Effectively, we are transforming the problem of product of double digit numbers to that of a subtraction and product of single digit numbers.

The product of numbers with surplus can similarly, be represented by the following algebraic expansion-

(x + a) (x + b) = x {(x + a) + b} + ab.


The following exercise have problems of both types discussed above, that is, product of numbers with deficit and product of numbers with surplus. Try to find the answer orally, without writing the intermediate steps. 1. 98 x 92 2. 99 x 97 3. 108 x 101 4. 95 x 95 5. 117 x 101 6. 132 x 102 7. 92 x 97 8. 107 x 103 9. 93 x 96 10. 105 x 107 11. 109 x 108 12. 93 x 92 13. 998 x 997 14. 996 x 994 15. 1004 x 1002 16. 1005 x 1008 17. 8675 x 9997 18. 9899 x 9996 19. 1115 x 1002 20. 97 x 88 21. 99998 x 99993 22. 635 x 999 23. 1035 x 1002 24. 99999 x 99999

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VinculumA very interesting and useful concept introduced by Swami

Bharati Krishna Tirthaji Maharaja in his famous book “Vedic Mathematics” is that of Vinculum. Vinculum, in simple words, means a negative digit. It may be embedded in a number. To understand the concept, let us recall our representation of a number in decimal number system. As we know, in the decimal system, each digit in a number has a place value; and the number is sum of these place values. Thus,

5342 is 5000 + 300 + 40 + 2.

What if some digit is negative? (Such negative number is represented by putting a bar above it.) The principle remains same! But, this time it will be algebraic sum.

For example,13 = 10 – 3 = 75342 = 5000 – 300 – 40 + 2 = 4662623 = 600 – 20 + 3 = 583, and so on.

One use of Vinculum, is to reduce the magnitude of a digit, in calculations. For example, if we want to multiply some number by 89, we will have to keep in our memory (for carry etc.) larger digits. But if the same number is represented as 111 (100 – 10 – 1), we have to deal with smaller digits and numbers. Another use is shown in the following class of problems.

Product of a number with surplus and a number with deficit

We have considered up to now, the examples where both the numbers were either below or above the base. But what should we do, in case, one number is above and the other is below the

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base? The answer is that we find the right hand portion as usual. But, as one of the numbers is positive (the surplus) and the other is negative (the deficit), the product will be negative. This negative part can be represented as Vinculum in the answer, from which, we can calculate the final answer. The following examples are given to illustrate the procedure.

12 : +2 19 : +9 108 : +08x 9 : -1 x 8 : -2 x 97 : -03

11 / -2 17 / -18 105 / -24 = 112 = 110 – 2 = 168 = 160 - 8 = 10524 = 108 = 152 = 10476

While finding the final answer also, we can make use of the Nikhilum sutra. Subtract one from the left hand portion and find the deficit of the right hand portion quickly, by the sutra. The final answer is just concatenation of these two.


Calculate the following products. 1. 109 x 98 2. 115 x 92 3. 103 x 99 4. 1027 x 998 5. 1081 x 995 6. 1150 x 989 7. 10073 x 9999 8. 10754 x 9998 9. 9975 x 12500 10. 1.2 x 811. 1.1 x 97 12. 1.14 x 9.2

How does the method work?

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Like in the previous two cases (both numbers positive and both numbers negative), here the algebraic expansion would be-

(x + a) (x – b) = x {(x + a) – b} - ab.Here, x is the base, a is the surplus of the first number and b is the deficit of the second numbers. For example, we may represent 104 x 97 as (100 + 4)(100 – 3). Here, x = 100, a = 4 and b = 3. Then, the right hand part of the answer is ab = 4x(-3) = -12; and left part is 100{(100 + 4) – 3} = 100{104 – 3} = 10100. The final answer is 10100 - 12 = 10088.

Tallying the answerIn many multiple-choice examinations, there is negative

marking. If our answer is wrong, we get penalty in the form negative marks. This is to avoid pure guess work by the student. Even if there is no negative marking, we strive for correct answer. A quick way to tally our answer and reduce the chances of silly mistake in the procedure, is to make use of the following sutra.

|| xqf.krleqPp;% leqPp;xqf.kr% ||Gunit Samuchchayaha Samuchchaya Gunitaha

(Meaning in the present context: The product of the digital roots of the numbers is equal to the digital root of their product.)

By digital root of a number, we mean the sum of the digits of the number. For example, if we have calculated the following-

10754 x 9998 = 107518492and, we want to make tally. Then, calculate the digital roots of the numbers involved as follows-10754 1 + 0 + 7 + 5 + 4 = 17 89998 9 + 9 + 9 + 8 = 35 8Expected digital root of the answer = 8 x 8 = 64 10 1

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Digital root of the answer 107518492 1 + 0 + 7 + 5 + 1 + 8 + 4 + 9 + 2 = 37 10 1As the digital root of the answer equals the expected digital root, we can say that our answer tallies. Please note that this is not the complete and fool-proof test. Just by rearranging the digits also, we get answer which is wrong but passes this test. That means, in the above example, if by mistake the answer came to 701518492, then also it will pass the test. However, we can be rest assured that if the answer fails this test, then it is definitely wrong.

We can accelerate the process of calculation of digital roots, by casting out nines. For example, you must have noticed that digital root of 9998 is 8. Just don’t bother about 9s! We do not have to calculate 9+9+9+8 = 35 = 8. Directly remove 9s and calculate digital root of the remaining number. Not only that, we can discard the digits that add to 9, and reduce the calculations. For example, in the following number, what remains is only 0 & 1.

7 0 1 5 1 8 4 9 2

By casting out 9 and the digits adding to 9, we quickly get the digital root as 1.

It is recommended to inculcate the habit of making this Gunit Samuchchaya test, as it will reduce the silly mistakes to a great extent.

Product of numbers with different basesIt is not necessary, that both the numbers should have the

same base. If the bases are different, we can transform the problem into the one having both numbers with the same base, solve the

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problem and adjust the result. For example, to multiply 94 by 9998, we transform the problem to 9400 x 9998 as

9400 : -06009988 : -0002

9398 / 1200

and then remove the borrowed last two zeroes, giving the answer 939812. This can be considered as a special case of multiplication of numbers with decimal fractions. Here, we consider the multiplicand 94 as 94.00 and then solve the problem.

Exercise 2.5:(Level – Middle school and above, preferably high school)

Solve the following problems.1.If cost of one Kg. of rice is Rs. 9=60 Ps., how much one has to

pay for 9.5 Kg. of rice?2.If one mango costs Rs. 9=30 Ps., what would be the cost of

dozen mangoes?3.If rate for construction of 1 cubic meter of brickwork is Rs. 992/-

, what would be cost of 1005 cubic meters of brickwork?4.What would be the simple interest for one year on amount of Rs.

1060/-, if the rate of interest is 10.05 percent? [Hint: Simple interest = Amount x Period x Rate of interest ]

5.A space ship is travelling at the speed of 9995 Kmph. How much distance would it travel in 10004 hours?

6.What is the area of square having side 1.6 m?7.What is the volume of water stored in a square tank of side 1.07

m. and depth 1 m.?8.Acceleration due to gravity is 9.8 meters per square second. If a

bomb is dropped from an airplane, how much velocity will it attain in 105 seconds? [Hint: Velocity = Acceleration x Time ]

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9.Giridhar seth has Rs. 21000/- with him. He purchased 88 pieces of toys, each piece costing Rs. 104/-. How much money would remain with him, after making the payment for the toys?

10.If a vendor purchases from Ahmedabad, 11 carpets for Rs. 10000/- and sells at Nagpur, each carpet for Rs. 990/-, then how much profit he would make?

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THREEFast Descendants of Nikhilum

We have already seen, how quickly we can solve multiplication problems using the Nikhilum sutra. But more dramatic and amazing are the up-sutras or sub-rules that are derived from the Nikhilum. These descendants or corollaries of Nikhilum enable us to solve certain types of problems at a speed that leaves one wonder-struck. In this chapter, we are going to learn these corollaries and their applications.

The First Corollary|| ;konwua rkonwuhÑR; ox± p ;¨t;sr~ ||

Yavadunam Tavadunikrutya Vargam Cha Yojayet(Meaning: Whatever the extent of its deficiency, lessen it still further to that very extent and also set up the square of that deficiency.)

This sub sutra is used to find out squares of numbers that are nearer to base, i.e., some power of 10. As we know, square of a number is multiplying the number by itself. The steps that we were carrying out for multiplication by Nikhilum get reduced still further because the multiplier and multiplicand are the same, in this case. The answer can be found out by mental working. The steps in the process are as follows-1.Take the nearest power of 10 as the base for the number, whose

square we are finding out. The base may be smaller or larger than the given number.

2.The square will be calculated in two parts. The left hand part is the given number minus the deficit (if the number is less than the base) or it is the given number plus the surplus (if the number is greater than the base).

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3.The right hand part of the answer is square of the deficit or square of the surplus, as the case may be. We are allowed as many digits on the right hand part, as many zeroes are there in the base of the number. The rule for carrying the extra digit to left or padding with zeroes is the same as explained in the previous chapter.

Let us take example of finding the square of 9. The mental steps would be carried out as follows-1.The base of 9 is 10.2.Left hand part of the answer = 9 – 1 = 8. Nine has deficit of 1

from its base 10. Therefore, we are lessening the number still further by that deficit to get the left hand part.

3.Right hand part of the answer is square of the deficit, i.e., 12 = 1.Thus, square of 9 is 81.

Similarly, square of 11 will involve the following steps-1.The base of 11 is 10, as it is the nearest power of 10.2.Left hand part of the answer = 11 + 1 = 12. Eleven has surplus of

1 from the base 10. Therefore, we are increasing that number still further by that surplus.

3.Right hand part of the answer is square of the surplus, i.e., 12 = 1.Thus, square of 11 is 121.

Some more examples will reinforce understanding of the method.82 = (8 – 2) / 22 = 6/4 = 6472 = (7 – 3) / 32 = 4/9 = 4962 = (6 – 4) / 42 = 2/16 = 3652 = (5 – 5) / 52 = 0/25 = 25

982 = (98 – 2) / 22 = 96/04 = 9604992 = (99 – 1) / 12 = 98/01 = 9801932 = (93 – 7) / 72 = 86/49 = 8649

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Chapter 3: Fast Descendants of Nikhilum

892 = (89 – 11) / 112 = 78/121 = 7921

122 = (12 + 2) / 22 = 14/4 = 144142 = (14 + 4) / 42 = 18/16 = 196152 = (15 + 5) / 52 = 20/25 = 225172 = (17 + 7) / 72 = 24/49 = 289

9992 = (999 – 1) / 12 = 998/001 = 9980011052 = (105 + 5) / 52 = 110/25 = 110251122 = (112 + 12) / 122 = 124/144 = 12544

While calculating the squares of numbers with decimal fraction, adopt the same procedure as explained in the last chapter. First, disregard the decimal point; calculate the square and then put the decimal point in the answer at appropriate place. So also, if there are zeroes at the end of the number, disregard those at first; calculate the square of the remaining number; then, put double the number of original zeroes at the end of this square to get the final answer.


Calculate squares of the following numbers 1. 14 2. 1.6 3. 0.17 4. 180 5. 98 6. 9.7 7. 8.9 8. 0.99 9. 930 10. 9.9311. 106 12. 10.713. 1.08 14. 109015. 111 16. 1.1217. 10.11 18. 11.319. 0.96 20. 99.8

23


21. 994 22. 9.9123. 1003 24. 1.00425. 10.09

The Second Corollary|| ,dkf/kdsu iwosZ.k ||

Ekadhiken Poorven(Meaning: By one more than the previous one)

This corollary is applicable to a special category of problems, viz., finding squares of numbers ending in 5. The steps involved in the procedure are as follows –1.Consider the left hand portion of the given number as the part

without the ending 5.2.Add one to this left hand portion. (Due to this, the method is

called 'Ekadhik'! ).3.The left hand part of the final answer is the product of numbers

obtained in steps 1 & 2 above.4.The right hand portion of the final answer is square of 5, i.e., 25.

Let us take example of finding square of 105. The mental steps involved are –1.Left hand part of the number is 10.2.Adding one to this number gives 11.3.Left hand part of the answer is 10x11 = 110.4.As the right hand part is 25, the final answer is 11025.

Some more examples will clarify the procedure.152 = (1x2)/25 = 225352 = (3x4)/25 = 1225452 = (4x5)/25 = 2025952 = (9x10)/25 = 90251152 = (11x12)/25 = 13225

24


1952 = (19x20)/25 = 38025

Sub-corollary|| vUR;;¨nZÓds·fi||

AntyaYorDashke Pee(Meaning: Even in case, where the end is ten.)

The previous rule is applicable not only to the squaring of number ending in 5, but also to the multiplication of two numbers whose last digits together total 10 and whose previous part is exactly the same.

For example, if we want to multiply 32 by 38; here, the sum of the last two digits (2 & 8) is 10, and their previous part is the same, viz., 3. So, the Ekadhiken Poorven rule will apply. The left-hand portion of the answer will be 3 multiplied by 4, the next higher number. Thus, we have 12 as our left hand part of the answer; and the right hand one is the product of the two end digits, 2 & 8, i.e., 16. So,

32 x 38 = 3x4 / 2x8 = 12/16 = 1216.

More examples – 73 x 77 = 7x8 / 3x7 = 56/21 = 5621103 x 107 = 10x11 / 3x7 = 110/21 = 110211032 x 1038 = 103x104 / 2x8 = 107/12/ 16 = 1071216

Note: If the end digits total to 10, then there must be two digits to the right of the slant line. If they total to 100, then there must be four digits in the right hand part. Examples –

41 x 49 = 4x5 / 1x9 = 20/09 = 2009197 x 103 = 1/97 x 1/03 = 1x2 / 97x3 = 2/0291 = 20291793 x 707 = 7x8 / 93x7 = 56/0651 = 560651

25


The rules for handling the decimal point and the ending zeroes, are the same as explained in the previous sections.


Calculate squares of the following numbers 1. 25 2. 4.5 3. 350 4. 0.85 5. 7.5 6. 1.05 7. 115 8. 1.85 9. 19.5 10. 0.125

Find the following products11. 54 x 56 12. 73 x 7713. 1.08 x 1.02 14. 9.3 x 0.9715. 1.12 x 11.8 16. 620 x 0.6817. 3.95 x 0.305 18. 1040 x 1060

The Third Corollary|| ,dU;wusu iwosZ.k ||

EkNyunen Poorven(Meaning: By one less than the previous one)

This sutra is applicable to multiplication problems where the multiplier digits consist entirely of nines, such as 9, 99, 999 etc. There are three categories of problems coming under this sutra.

a) The First Case: Multiplicand and the multiplier having same number of digits.

26


The product of such numbers consists of two parts. The left part is found simply by subtracting 1 from the multiplier and the right part is deficit of the multiplier. Suppose we have to multiply 7 by 9. Then subtract 1 from 7, to get the left hand part of the answer, i.e., 6. The right hand part of the answer is deficit of the multiplicand 7, i.e., 3. Thus the answer is 63. The steps can be represented as follows –

7 x 3 = (7-1) / (10-7) = 6/3 = 63

Similarly,3 x 9 = 2/7 = 274 x 9 = 3/6 = 369 x 9 = 8/1 = 8167 x 99 = (67-1) / (100-67) = 66/33 = 663391 x 99 = 90/09 = 9009864 x 999 = 863/137 = 863137979 x 999 = 978/021 = 97802187592165 x 99999999 = 87592164/12407835

= 8759216412407835Are you not amazed of your own speed of calculation of such products?

b) The Second Case: Multiplicand consists of smaller number of digits than the multiplier.

Here, pad the multiplicand with zeroes on its left, so that it consists of same number of digits as that in the multiplier. The remaining procedure is as in the previous case. For example,

7 x 99 = 07 x 99 = 06/93 = 69379 x 999 = 079 x 999 = 078/921 = 78921798 x 99999 = 00798 x 99999 = 00797/99202 = 79799202

It may be noted here that while calculating the right hand side deficit of the multiplicand, the zeroes are also considered to be taking part in the process.

27


c) The Third Case: Multiplicand has more number of digits than the multiplier.

In this case, multiply the multiplicand by the base of the multiplier. Then subtract the original multiplicand from this product to get the final answer. The following examples will illustrate the procedure.

43 x 9 = (43x10)-43 = 430-43 = 387112 x 99 = 11200-112 = 1108811119 x 99 = 1111900-11119 = 1100781

Exercise 3.3:(Level – Middle school student)

Calculate the following products. 1. 3169 x 999 2. 4.103 x 9.9 3. 467 x 0.99 4. 78.326 x 999 5. 0.7567 x 99 6. 99.99 x 0.99 7. 12 x 0.09 8. 1.853 x 0.9 9. 19.5 x 0.099 10. 0.125 x 90

The following exercise consists of word problems based on all the three corollaries described above. You have to decide which technique suits best for the given problem and then solve it.

Exercise 3.4:(Level – High school student)1.What is the area of a square, having side 1.7 meters?

2.What is the area of rectangle having length 5.6 cms. And breadth

5.4 cms.?

3.If cost of a book is Rs. 9=90 Ps., how much amount would be

needed to purchase 54 copies of the book?

28


4.What is the area of triangle, having base 9 cms. And height 4.5

cms.?

5.A pair of shoes costs Rs. 999/-. A school wants to purchase these

shoes for the 22 players of its cricket teams A & B. How much

money will have to be paid?

6.What will be the compound interest for period of two years on

principal amount of Rs. 10000/-, if the rate of interest is 5% per

annum? [Hint: compound interest accounts for interest on

interest of previous year also.]

7.What will be the total number of soldiers on parade, if they are

standing in a square formation of 65 rows and 65 columns?

8.If average score of a batsman is 33 runs per match, then how

many runs he would have taken in 37 matches?

9.What is the volume of a cube having side 9 cms.?

10. If one ‘Hapoos’ mango costs Rs. 18/-, what will be the cost of

one & half dozen of these mangoes?

29

FOURMultiplication by UrdhwaTiryak

We have already seen the amazing power of the Nikhilum method in multiplication process. However, one serious limitation of the Nikhilum method is that it applies to a specific class of problems; the problems, where at least, one of the two numbers is very near to the base. Its corollaries also have the same draw back of being applicable to specific class of problems. Vaidik mathematics provides another method of multiplication, which is very general (applicable to any multiplication), shorter than the conventional method, although longer than the Nikhilum method. Its advantage over the conventional method is that it is suitable for mental working. It is based on the following sutra-

|| Å/oZfr;ZXH;ke~ ||UrdhwaTiryagabhyam

(Meaning: Vertically and cross-wise.)

In this method, we do the calculations from left to right, unlike in the conventional method, where we start from right, that is, units place. This enables us to calculate significant digits of the answer prior to the non-significant ones.

The Details of the MethodTo understand the application of this sutra, take an example

of finding product of 12 and 13. The following steps are involved.a)We multiply the left-most digit 1 of the multiplicand

vertically with the left-most digit 1 of the multiplier, giving product 1. This result becomes the left-most or the first part of the answer.

30

Chapter 4: Multiplication by UrdhwaTiryak

b)We then multiply 1 (of the multiplicand) by 3 (of the multiplier), and 1 (of the multiplier) by 2 (of the multiplicand) cross-wise, add the two results giving 5 as the sum. This sum becomes the middle or the second part of

the final answer.

c)At last, we multiply 2 (of the multiplicand) by 3 (of the multiplier) vertically and get 6 as their

product. This becomes the right-most or the third part of the answer.

d)The final answer is, then, 1/5/6 = 156.

The digits involved, in finding the three parts of the answer, can be understood by the following representative diagram.

vertical cross-wise vertical

Some more examples should make the process clear. 16 21 23 41 37x 11 x 14 x 21 x 41 x 33

1/7/6 2/9/4 4/8/3 16/8/1 9/30/21=176 =294 =483 =1681 =9/32/1

=1221

31

12x 13

(1 x 1) / (1 x 3) + (1 x 2) / (2 x 3)= 1 / 3 + 2 / 6= 156


The first three examples here, went exactly as per the procedure described above. In the fourth example, the first part of the answer is two digit number 16. But that does not matter; the answer is still, the concatenation of the three parts. In the fifth example, however, the second and the third part have two digits each. In such cases, the right-most digit of the part is put down there itself, and the remaining left digit or digits are carried over to the immediate left part. Thus, from the third part 21, we carried 2 to its left 30, making the middle part equal to 32. Then, from this 32, we carried 3 to its immediate left, i.e., the first part making it 9+3=12. The final answer is, therefore, 1221.

For the three digit numbers, the product consists of five components. The digits taking part in these five components are shown in the following representative diagram.

Note here that the middle or the third part of the answer consists of the sum of three products; the product of the first digit of the multiplicand & the last digit of the multiplier, the product of the middle digits of the multiplier & the multiplicand and the product of the first digit of the multiplier & the last digit of the multiplicand.

The following few examples should illustrate the procedure.

116 582 785 321x 114 x 231 x 362 x 052

1/2/11/10/24 10/31/33/14/2 21/66/77/46/10 0/15/16/9/2

32


= 13224 = 134442 = 284170 = 16692

The last example shows that the numbers involved in the multiplication need not have same number of digits. We can make them of the same length by left padding the smaller length number by appropriate number of zeroes.

The same technique can be extended to find product of two 4 digit numbers or two 5 digit numbers. In 4 digit numbers, the product will consist of 7 parts, while in 5 digit numbers the product will consist of 9 parts. To clarify further, study the procedure in the following two examples.

1021 6471x 2103 x 6212

2/1/4/5/1/6/3 36/36/56/36/17/15/2 = 2145163 = 40197852


Find the following products by UrdhwaTiryak method 1. 24 x 62 2. 45 x 58 3. 31 x 49 4. 33 x 34 5. 88 x 92 6. 78 x 87 7. 115 x 231 8. 242 x 568 9. 348 x 591 10. 719 x 88811. 5.4 x 5.7 12. 7.6 x 0.7713. 1.08 x 1.09 14. 0.79 x 0.8815. 1.12 x 11.9 16. 620 x 0.6717. 3.96 x 0.302 18. 1043 x 106219. 9.81 x 1.111 20. 1.231 x 5.13421. 7500 x 81

33


How does the method work?The UrdhwaTiryak method makes use of the fundamental

fact in the decimal number system, that each digit has a place value. The place value is, the given digit multiplied by some power of 10. Thus, number 23 is actually 20+3 or 2x10+3. The number 429 is actually 400+20+9 or 4x102 + 2x10 + 9. Algebraically, we can represent a two digit number in the form ax+b, where x is 10 and a & b are the two digits of the number. Similarly, a three digit number can be represented in the form ax2+bx+c, where x is 10 and a, b & c are the three digits of the number. Once this is understood, the product of two 2-digit numbers can be represented algebraically as follows-

a.x + bx c.x + d

a.c.x2 + (a.d + b.c)x + b.d

The answer is in three parts. The first part, which automatically gets place value of hundreds due to the x2 factor, is vertical product of first digits a and c of the original numbers. The second part, having tens place due to x factor, is the crosswise product of the digits, as explained earlier. And the third part bd in units place, is vertical multiplication of the second digits of the original numbers.

Similarly, the product of three digits numbers can be represented as follows- ax2 + bx + c

x dx2 + ex + f

a.d.x4 + (a.e +b.d)x3 + (a.f + b.e + c.d)x2 + (b.f + c.e)x + c.f

34


On the same lines, the working of this UrdhwaTiryak method for product of four or higher digit numbers can be explained.

Exercise 4.2:(Level – Preferably high school student or above)

Solve the following word problems by UrdhwaTiryak method.1.A dealer purchased 125 color TV sets. If the cost of each set is

Rs. 9820/-, determine the cost of all the sets together.

2.The monthly fee charged, from a student of class VI in a school

is Rs. 145/-. If there are, in all, 235 students in class VI of that

school, find the total monthly collection.

3.Veena deposited Rs. 7350/-, in a finance company which pays

16.5% interest per year. Find the simple interest she is expected

to get after four years.

4.An aeroplane flies at speed of 755 km. per hour. Find the

distance it will cover in 3 hours & 15 minutes.

5.In a school, the ratio of boys to girls is 1.25. If there are 384 girls

in the school, find the number of boys.

6.Rajesh works as an accountant in a bank. His gross salary is Rs.

8550/- per month. If he gets an increment of 14% in his salary,

what will be his new salary?

7.Calculate the number of hours in the month of May.

35


8.What is the volume of water in a rectangular tank of length 2.75

m., width 2.25 m., and depth 1 m.?

9.Ritu has taken a long term loan for purchasing a flat. She has to

pay Rs. 2235/- per month to the finance company as

installment towards the loan. How much total money, she will

pay over 20 years, to the company?

10.A group of 83 students and 7 teachers of a school, goes to

science exhibition. If the entrance fee is 75 paise per student

and one rupee fifty paise per teacher, how much total amount

in terms of rupees, the group has to give?

11.There are 127 workers in a factory. If, their average salary per

month is Rs. 1880/-, what is the total amount, that the factory

has to disburse each month?

12.A trader purchased a chair at a cost price of Rs. 687/-. He

obtained a profit of 34% by selling it. What was the sell price?

13.A train travels from station X to Y at an average speed 55 kmph

and from station Y to Z, at an average speed of 45 kmph. If, at

these speeds, the train takes 45 minutes from X to Y and 1 hour

15 minutes from Y to Z, what is the distance in Kilometers

from station X to Z?

14.Sale price of a ‘Saree’ is Rs. 630/-. If a discount of 14% is

given on it, what would be its price?

36


15.Coefficient of expansion for copper is 0.0000175 per degree

Celsius. What would be change in length in millimeters of a 10

meter long copper tube, if its temperature rises by 27.2 degrees

Celsius? [Hint: Change in length = Coefficient of expansion X

length X Change in temperature]

37

FIVEDivision by UrdhwaTiryak

We will see now the Vaidik Mathematics process of straight division. It is simple application of Urdhwa-tiryak Sutra. Swamiji, in his famous book 'Vedic Mathematics', has given several methods based on different sutras for performing division. However, he describes the Urdhwa-tiryak method and the sutra as the "crowning gem of all the sutras". The other methods have limitation that they are applicable to a particular class of problems. On the other hand, Urdhwa-Tiryak is very general method and quicker compared to our conventional method. Because of this reason and to avoid confusion, in the present book, only this method of division is covered.

First, let us recapitulate various terms that we use in the division process. Let us take example of 38982 divided by 73. Our conventional procedure would go as follows-

5 3 4 <== quotient divisor ==> 73 ) 3 8 9 8 3 <== dividend

- 3 6 5 2 4 8 - 2 1 9 2 9 3 - 2 9 2

1 <== remainder

As one can see, our conventional method is some what long and cumbersome due to the large products (multiplication table of the divisor) involved. The Vaidik Maths method goes as follows-

38

Chapter 5: Division by Urdhwa Tiryak

The Details of the Method:

5 3 4 7 / 3 ) 3 8 9 8 / 3

Gross dividend => 39 38 / 13 Actual dividend => 38 24 29 / 1

1.Write the divisor in two parts, separating the parts by a slant line. Here, the divisor 73 is written as 7 / 3. The right hand part is called as Dhwajanka or the flag digit.

2.Write the dividend, in front of the divisor as in the conventional method, but keep the digits well spaced, so that we may write the next two lines under them in a more legible way. Divide this number also into two parts by a slant line, so that the number of digits on the right hand part are same in, both, the divisor and the dividend. Thus, we have divided 38983 as 3898 / 3.

3.Set up the next two lines for writing the intermediate gross dividend and intermediate actual dividend. Set the space above the dividend for writing the answer (the quotient) as usual.

4.Now, write the starting two digits of the divisor i.e. 38 on the 'actual dividend' line.

5.Divide this number 38, by the first digit of the divisor i.e. 7. The quotient would be 5 and the remainder would be 3 (38 - 7x5 = 38 - 35 = 3). Write the quotient at its usual place, at the top. To be systematic, write it above 8 of 38. Write the remainder 3, in the 'Gross dividend' line, between 8 and 9 of the main dividend.

6.Take down the next digit of the main dividend i.e. 9 and write it to the right of the remainder 3. These two digits together make the next gross dividend 39.

39


7.The actual dividend, however, would be this gross dividend minus the product of the last quotient and the flag digit. It would be, in the present case, 39 - 5x3 = 39 - 15 = 24. Write it down on the 'actual dividend' line, exactly below the gross dividend 39.

8.Divide this 24 again by 7 of the divisor; write the quotient 3 on the top line and the remainder 3 on the 'gross dividend' line. Pull the next digit of the main dividend to the right of remainder 3, making gross dividend 38. The actual dividend would be 38 - 3x3 = 29, which is written at the appropriate place.

9.Continue this procedure until all the digits of the main dividend are exhausted. The number obtained in the top most line is the quotient (here, 534) and in the actual dividend line after the slant line is the remainder.

To check the correctness of the solution, we make use of ‘Gunit Samuchchaya…’ sutra. In the present case, we have to ensure that digital root of the dividend minus the digital root of the remainder is equal to the product of the digital root of divisor and that of the quotient. In cases where the remainder is zero, the product of digital root of divisor and that of the quotient must equal the digital root of the dividend. Here,

Digital root of dividend = 3+8+9+8+3 = 31 = 4Digital root of remainder = 1 Digital root of divisor = 7+3 = 10 = 1Digital root of quotient = 5+3+4 = 12 = 3As 4-1 = 1x3 = 3, therefore, our calculation is correct.

We will consider now some additional examples to illustrate the process.

40


a)Divide 529 by 23. 2 32 / 3 ) 5 2 / 9 12/ 09 5 6 / 0

Here, Quotient = 23 and Remainder = 0. D.R. of 23 = 2+3 = 5. D.R. of 529 = 5+2 =7. For tally, D.R. of quotient X D.R. of divisor = D.R. of dividend So, 5 x 5 = 25 => 7. Thus, tallies.

b)Divide 3200 by 25.1 2 8

2 / 5 ) 3 2 0 / 0 12 30/ 403 7 20/ 0

Here, Q = 128 and R = 0. Tally is left for the reader.

c)Divide 56378 by 49.

1 1 5 04 / 9 ) 5 6 3 7 / 8

16 33 47 28 5 7 24 2 /28

Thus, Q=1150 and R=28. To check, we calculate digital roots as follows-

Dividend: 5+6+3+7+8 = 29 = 11 = 2 Remainder: 2+8 = 10 = 1 Divisor: 4+9 = 13 = 4

41


Quotient: 1+1+5+0 = 7As 2-1 is equal to 4x7 = 28 => 10 => 1, therefore our solution is correct.

d)Divide 7632 by 94. 0 8 19 / 4) 7 6 3 / 2 76 43 /22 7 76 11 /18

Here, Q = 81 and R = 18.

Finding quotient in decimal formatIn the examples discussed above, we were dealing with

integers. If the dividend was not completely divisible by the divisor, we were getting an integer as remainder. It is possible, however, to calculate the quotient in decimal fraction format using this method. The following example and its explanation illustrate the procedure.Divide 220 by 52 up to 3 places of decimals.

0 4 . 2 3 0 85 / 2) 2 2 / 0 0 0 0

22 / 20 20 10 4022 / 12 16 04 40

1.The division process up to finding the integer portion of the quotient is the same as explained earlier. We write divisor and the dividend in two parts as before. The two parts are separated by slant lines.

42


2.We, then, create space for writing gross dividend, actual net dividend and quotient as before.

3.We divide 22 by the first digit of the divisor, i.e. by 5 giving the quotient 4, which we write on top of 22 in place meant for quotient. The remainder 2, we write in gross dividend line below the zero of 220. We then pull down that zero, write it to the right of this 2, giving the gross dividend 20.

4.From this gross dividend, we subtract the product of quotient digit 4 and Dhwajanka 2, i.e. 8 giving 12, which we write in the net dividend line, below 20. If we wanted the answer in the quotient-remainder form, then our procedure would have been over here. But, we want the answer in decimal fraction form; so, we continue the process of division further.

5.We give decimal point after 4 in the quotient, placing it above the slant line of the divisor. Then, we put three additional zeroes to the right of the zero of 220. (Three zeroes, because, we want the answer up to three places of decimals.) Actually, we are dividing 220.000 (which is numerically same as 220) by 52.

6.The net dividend obtained in step 4, is divided by 5. This gives quotient 2, which is written after the decimal point in the quotient line and above the first zero after the slant line. The remainder, in this case, is also 2 and is written in the gross dividend line as usual. We then pull down the second zero, to the right of this 2, making the gross dividend to be 20. The net dividend is obtained as usual, by subtracting product of the quotient 2 and Dhwajanka 2, i.e. 4 from gross dividend 20. This is 16, which is written in the net dividend line.

7.The process is continued until all the zeroes to the right of slant line are exhausted. This gives us answer as 4.2308. As we want the answer upto 3 places of decimals, we round this off to 4.231.

43


To further clarify the procedure, two more examples are given below.1.Divide 7453 by 79 upto 3 places of decimals.

0 9 4 . 3 4 2 7 / 9) 7 4 5 / 3 0 0 0

74 115 / 63 60 50 074 34 / 27 33 14

So, the quotient is 94.342.

2.Divide 710.014 by 39 up to 3 places of decimals. Here, for marking slant line in the quotient, we consider only its integer part 710 and mark it to the left of zero in the units place. After drawing the slant line, we disregard the decimal point and write fractional part after the integer. The rest of the procedure is as in the previous two examples.

1 8 . 2 0 5 63 / 9) 7 1 / 0 0 1 4 41/ 80 20 21 64 7 32/ 8 2 21 19

Rounding off up to the third place of decimal, we get the answer as 18.206.

How does the method work?As discussed in the last chapter, the UrdhwaTiryak method

makes use of the fundamental fact in the decimal number system, that each digit has a place value. The place value is, the given digit multiplied by some power of 10. Thus, number 23 is actually 20+3 or 2x10+3. The number 429 is actually 400+20+9 or 4x102 + 2x10 + 9. Algebraically, we can represent a two digit number in the form

44


ax+b, where x is 10 and a & b are the two digits of the number. Similarly, a three digit number can be represented in the form ax2+bx+c, where x is 10 and a, b & c are the three digits of the number. This process can be extended further to represent larger numbers. To understand the logic behind the division by UrdhwaTiryag method, consider the problem – 695457 divided by 57. Representing these numbers algebraically (assuming x = 10) we carry out the division process as follows-

1x4 + 2x3 + 2x2 + 0x + 1 = 122015x + 7 ) 6x5 + 9x4 + 5x3 + 4x2 + 5x + 7 -(5x5 + 7x4) x5 + 2x4

= 10x4 + 2x4 = 12 = 12x4 + 5x3

-( 10x4 +14x3) 2x4 – 9x3

= 20x3 – 9x3 = 11 = 11x3 + 4x2

-( 10x3 +14x2) x3 – 10x2

= 10x2 – 10x2 = 0 = 0x2 + 5x -( 0x2 + 0x) 5x = 5 = 5x + 7 -(5x + 7) 0 = 0Compare the figures in red above, with the figures in the actual dividend line, in the division process shown below; and the logic behind the UrdhwaTiryag division will be clear.

45


1 2 2 0 15 / 7) 6 9 5 4 5 / 7 19 25 14 05/ 07 6 12 11 0 5/ 0

In fact, the vaidik method is mechanical replica of the algebraic process that happens behind the scene as described above. If you feel comfortable with the algebraic process, then initially you may use it.

Three-digit DivisorWe shall now consider problems involving divisor with 3

digits. The overall procedure remains same as described before; only intricacy increases. The difference occurs in the calculation of actual dividend from the gross dividend. Let us see the steps in division of 186116 by 578-

0 3 2 2 5 / 78) 1 8 6 1 / 1 6

18 36 51/31 16 1 18 15 13/ 1 0

1.Here, we put the slant line in the divisor after two digits from the right. Of course, we could have put the slant line after one digit from right; but, in that case we will have to know the multiplication table for 57, a two digit number. This would defeat the basic nature of vaidik mathematics methods, viz., ease of calculations. So, there are two digits in the Dhwajanka.

2.As we have put the slant line after two digits from right in the divisor, we have to put the slant line in the dividend also, after two digits from right. The slant line divides the dividend in left

46


part (1861 here) and right part (16 here). As before, the left part procedure gives us the quotient, while the right one gives the remainder.

3.We put the first digit of the dividend (1) on the actual dividend line, and divide it by 5 of the divisor. The quotient 0, is written in the quotient line, as usual. The remainder 1 is written in gross dividend line, 8 from the dividend is pulled down and written to the right of it. This makes the gross dividend at this stage 18 and the procedure is usual.

4.To calculate the actual dividend, we subtract from gross dividend, the product of earlier quotient digit 0 and first digit of the Dhwajanka 7. The product is zero, so actual dividend is 18.

5.By dividing this 18 by 5 and with usual procedure, we get quotient 3 and gross dividend 36. From this 36, we subtract the sum of products of last quotient digit 3 by first digit of Dhwajanka 7 and last-but-one quotient digit 0 by second digit of Dhwajanka 8. This cross product is explained by following figure.

0 3 7 8 21 + 0 = 21 (Because of this cross-product, the word Tiryak comes in the

name of the method.) Thus, the actual dividend is 36 – 21 = 15.6.The remaining procedure is similar. The calculations of actual

dividends are explained below.36 – (3x7 + 0x8) = 36 – 21 = 1551 – (2x7 + 3x8) = 51 – 38 = 1331 – (2x7 + 2x8) = 31 – 30 = 116 – (0x7 + 2x8) = 16 – 16 = 0 (The remainder)

7.Please note that, while calculating the remainder, we assumed zero in the quotient line.

47


One more example should clarify the procedure further.

0 8 5 4 4 8 / 23) 7 0 3 1 9/ 8 5

70 63 71 59/48 285 7 70 47 37 36/28 273

So, 7031985 divided by 823 gives quotient 8544 and remainder 273.


Solve the following division problems by UrdhwaTiryak method. Find quotient and remainder in the first 10 problems and quotient up to 3 places of decimals in the remaining ones.

1. 24347 62 2. 1450 58 3. 6631 49 4. 3300 34 5. 88888 92 6. 785672 87 7. 115256 231 8. 242000 568 9. 3486 591 10. 71922 88811. 54 57 12. 76334 7713. 105.8 23 14. 7967.82 8815. 11255 119 16. 62032 6717. 3.96 3.1 18. 104.35 6.219. 29.81 11.1 20. 1231 5.1321. 7500 8.1


Solve the following word problems by UrdhwaTiryak method.

48


1.A student has obtained total 1836 marks out of 2700 in an exam.

What is her performance in terms of percentage of marks?

2.A car travels a distance of 48.75 km. in 45 minutes. What is its

speed expressed in Kmph?

3.A cricket batsman has scored 7567 runs in 63 matches. What is

his average run rate? (Find up to 2 places of decimals.)

4.A businessman purchased goods worth Rs. 88000/- and sold it

for Rs. 98650/-. What was his profit percentage? (Find up to 2

places of decimals.)

5.There are 53 officers of different grades working in a branch

office of a company. Their total salary per month is Rs.

419000/-. What is average salary per officer in a month? (Give

only the Rupee part, i.e. integer part.)

6.In an engineering college, there are 1367 boys and 78 girls. What

is the ratio of boys to girls? (Find up to 2 places of decimals,

i.e. XX.XX boys : 1 girl).

7.How many hours would be required for an aeroplane travelling at

a speed of 245 kmph. to cover a distance of 1082 km.? (Find

up to 3 places of decimals.)

49

SIXSquare Drive

We have already seen two methods of finding squares of numbers. One was "Yavadunam..." method and the other was "Ekadhiken Poorven" method. These methods were applicable only to a particular class of numbers and were not universally applicable. The first method was applicable to the numbers which were nearer to the base; the second method was applicable to the numbers which were ending with 5. As square of a number is the number multiplied by itself, we can also use the quick multiplication methods, that we learnt earlier, for finding out squares of numbers. But a more direct method is available in Vaidik Mathematics for finding squares. For using this method, we have to learn first, how to find "Dwandwa-Yoga" or Duplex of a number.

Duplex or Dwandwa-Yoga:The duplex of a single digit number is defined as its square.

If we denote duplex by D, (i.e. D7 means duplex of 7) thenD5 = 5 x 5 = 25D9 = 9 x 9 = 81D4 = 4 x 4 = 16 and so on.

The duplex of a double digit number is defined as twice the product of its first digit and the second digit. For example,D23 = 2 x (2 x 3) = 12D45 = 2 x (4 x 5) = 40D36 = 2 x ( 3 x 6) = 36D88 = 2 x (8 x 8) = 128D30 = 2 x (3 x 0) = 0

50

Chapter 6: Square Drive

D05 = 2 x (0 x 5) = 0Note that, duplex of 5 would be 25, if it is represented as

single digit number; however, if it is represented as 05, then its duplex is 0.

The duplex of a three digit number is defined as sum of the duplex of the middle digit and double the product of the first and the third digit. For example,D123 = D2 + D13 = (2 x 2) + 2 x (1 x 3) = 4 + 6 = 10D503 = D0 + D53 = (0 x 0) + 2 x (5 x 3) = 0 + 30 = 30D317 = D1 + D37 = (1 x 1) + 2 x (3 x 7) = 1 + 42 = 43D437 = D3 + D47 = (3 x 3) + 2 x (4 x 7) = 9 + 56 = 65D075 = D7 + D05 = (7 x 7) + 2 x (0 x 5) = 49 + 0 = 49

Again, note that duplex of 75 is 2 x 7 x 5 = 70; but if it is represented as 075, then its duplex becomes 49.

The duplex of a four digit number is twice the sum of the product of first & last digits and the product of middle two digits. For example,D6537 = 2 x (6 x 7) + 2 x (5 x 3) = 2 x {(6 x 7) + (5 x 3)} = 2 x (42 + 15) = 114D2035 = 2 x {(2 x 5) + (0 x 3)} = 2 x (10 + 0) = 20D1890 = 2 x {(1 x 0) + (8 x 9)} = 2 x (0 + 72) = 144

For finding out duplex of five or more digits numbers, the same procedure is extended. For example,D34532 = 2 x (3 x 2) + 2 x (4 x 3) + 52 = 12 + 24 + 25 = 61D50306 = 2 x (5 x 6) + 2 x (0 x 0) + 32 = 60 + 0 + 9 = 69D132645 = 2 x {(1 x 5) + (3 x 4) + (2 x 6)} = 2 x (5 + 12 + 12) = 58and so on.

51



Find the duplex of the following numbers. 1. 53 2. 48 3. 30 4. 07 5. 353 6. 405 7. 800 8. 080 9. 008 10. 123411. 6050 12. 700913. 9000 14. 078515. 11255 16. 6203217. 39600 18. 104354

Squaring:Once the process of finding the Dwandwa Yoga is clear,

finding square of a number is easy. The square of a single digit number is, obviously, its duplex.

The square of a double digit number consists of three parts, as follows-the first part is duplex of the left digit of the number;the second part is duplex of the double digit number itself; and, the third part is duplex of the right digit of the number.

The square of a three digit number consists of five parts, as follows-the first part is duplex of the left-most digit of the number;the second part is duplex of the first two digits of the number;the middle part is duplex of the number itself;the fourth part is duplex of the last two digits of the number;and, the fifth part is duplex of the last digit of the number.

52


The same procedure is extended for numbers with four or more digits. The following examples would make the idea clear.452 = D4/D45/D5 = 16/40/25 = 16/40/25 = 2025502 = D5/D50/D0 = 25/0/0 = 25003672 = D3/D36/D367/D67/D7 = 9/36/78/84/49 = 1346895432 = D5/D54/D543/D43/D3 = 25/40/46/24/9 = 2948498972 = D8/D89/D897/D97/D7 = 64/144/193/126/49

=78/23/15/10/9 = 8046093082 = D3/D30/D308/D08/D8 = 9/0/48/0/64 = 9486421342 = D2/D21/D213/D2134/D134/D34/D4 = 4/4/13/22/17/24/16

= 4553956and so on.

The squares of numbers involving decimal fraction can be found by the usual procedure- Disregard first the decimal point, calculate the square of the resultant number and then put the decimal point at appropriate place.

Why does the method work?To understand the working of this method, consider a three

digit number with digits a, b, and c. We can represent this number algebraically as ax2 + bx + c, where x = 10. Then, the expression for square of this number can be derived as-

ax2 + bx + c ax2 + bx + c

a2x4 + 2abx3 + (2ac + b2) x2 + 2bcx + c2 = D(a)x4 + D(ab)x3 + D(abc) x2 + D(bc)x + D(c)

But as x = 10, the above expression can be represented as D(a) + D(ab) + D(abc) + D(bc) + D(c)

53


This last expression is the one that we are using to quickly calculate square of the number.


Find the squares of the following numbers using The Dwandwa-Yoga method. The intermediate steps and the answer may be written directly in front of the number.

1. 392. 423. 5.84. 0.735. 0.0846. 1437. 8.058. 45.79. 5.4810. 0.35711. 0.078312. 437113. 12.5514. 70.0915. 530.816. 0.375217. 0.0902118. 3276719. 61.31120. 5.300221. 892.8222. 34256223. 23.001124. 17.100

54



Solve the following word problems using Dwandwa-Yoga method.

1.What will be the area of a 3.2 m by 3.2 m. room?

2.What is the area of a square with side 4.23 cm. ?

3.What is the ratio of area of 28 mm diameter bar to area of 10

mm diameter bar? [ Hint: Area of cross-section of a ‘d’

diameter bar is ? d2 / 4 .]

4.On a sunny and calm day, a ship leaves its port and starts

travelling with a constant acceleration of 20 kilometer per hour

per hour. How much distance will it travel in 2 hours & 45

minutes? [Hint: use the formula (s = ut + 0.5 at2 ) with usual

notations of physics. ]

5.What volume of water is stored in a circular tank of internal

diameter 3.78 m. and depth of water 1.4 m. ? [ Hint: Take

value of ? as 22/7. ]

55

SEVENSquare Roots

Having seen the process of finding square using the Dwandwa-Yoga method, we will discuss in this chapter the opposite process; that of finding square root of a given number. In the conventional method of finding the square root, the divisor goes on becoming larger in each step. This increases the calculation time as well as the complexity. In Vaidik mathematics, we use the Dwandwa-Yoga method, which is akin to the process of division by Urdhwa-Tiryag method. It also gives us the usual benefit of Vaidik mathematics, viz., providing simple, easy, and straight procedure.

Some tips about square roots1.The square root of a number with 'n' digits will contain n/2 or

(n+1)/2 digits.2.For finding out square root, we have to arrange the given number

in two-digits groups starting from right to left. If there are odd number of digits in the given number, then obviously the last (left-most) group will contain one digit only. The number of digits in the square root can also be found by counting these digit-groups.

3.The squares of single digit integers are 0, 1, 4, 9, 16, 25, 36, 49, 64 and 81. It follows that square of any integer must end with either 0, 1, 4, 5, 6, or 9. It can not end with 2, 3, 7, or 8.

4.As corollary, a number can not be an exact square, if it ends with 2, 3, 7, or 8.

5.Also, a number can not be an exact square, if it ends with an odd number of zeroes.

56

Chapter 7: Square Roots

Finding Square RootLet us now understand the process of finding out the square

root through examples. Take problem of finding out square root of 529. The steps involved are as follows-1.Divide the given number into

groups of two digits starting from right. Here, the right-most group consists of two digits 2 & 9 and the remaining group consists of single digit 5. (If somebody insists upon exactness of our definition of 'two-digit groups' then consider this last group as 05.) As two groups are there, we know that

there would be two digits in the integer part of the square root of 529.

2.Set up a table as shown in the figure above, similar to the one we used for division process.

3.Here, write 529 with digits well spaced apart in the dividend position. Draw a slant line after the first group (of single digit 5) from left, separating it from rest of the number.

4.We are now ready to find out the first digit of the answer. This is to be written above the first (left-most) group, in the quotient line. This number should be largest single-digit number whose square is less than or equal to the number in the first group of the dividend. In the present case the dividend number is 5. There are two single-digit numbers 1 and 2, whose squares are less than 5; We choose the largest among them and write it. So, we write 2 in the quotient line, above 5. We also draw a slant line after this digit in the quotient line.

57

2 / 3 <--Ans.

4 ) 5 / 2 9

Gross 12 09

Net 12 0


5.From the first digit of the square root (answer) calculated in the previous step, we get the divisor for the rest of the process. It is double of the first answer digit. In the present case, double of 2, that is 4 would act as divisor for rest of the process. Write this 4 in the position of the divisor.

6.Subtract square 4 of the first quotient digit from the first group number 5 and write the remainder 1, in the gross dividend line. This is to be written below the first digit of the dividend number to the right of slant line. Now take down the first digit 2 of the right part of the dividend and write it in front of the remainder 1 on the gross line. The gross dividend now is 12. For the first digit of the right part of the dividend, the net dividend is same as the gross dividend. So, simply write down this 12 in the net or actual dividend line also.

7.Divide this net dividend 12 by the divisor 4. Write down the quotient 3 after the slant line in the quotient line. Write the remainder 0 on the gross dividend line below the second digit 9 of the right part of the dividend. Take down this 9 and write it in front of the 0 on the gross dividend line making the gross dividend as 9.

8.The net dividend is gross dividend minus the duplex of the quotient part to the right of slant line. Here, the quotient to the right of slant line is 3. Duplex of 3 is 9. Thus, net dividend is 9 minus 9, that is 0. We write this in the net dividend line. As there are no further digits in the dividend and 0 in the net dividend, our process is over.

The square root is the number in the quotient line, that is 23.

Two more examples of finding square root of a three digit number are given below. They should consolidate further the understanding of the above procedure.

Square root of 784 2/ 8

58


4) 7/ 8 4 38 64Ans: 28 38 0Square root of 729 2/ 7 4) 7/ 2 9 32 49Ans: 27 32 0

Now let us consider a four digit number 1225 and find out its square root. Here, the two-digits groups would be 12 and 25. So, the answer would consist of, again, two digits. The only difference from the above three examples would be the position of slant line in the dividend; it would be after first two digits (first group) of the given number. The solution is as follows-

Square root of 1225 3/ 5 6) 12/ 2 5 32 25Ans: 35 32 0

Study the following two more such examples and clarify the method further-



59


Let us now understand to extend the above procedure to longer numbers. Assume that we want to find square root of 16384. The procedure is shown in the following chart and explained after it.

1 / 2 8 . 0 <-- Quotient (Answer) 2 )1 / 6 3 8 4Gross dividend --> 06 23 38 64Net dividend--> 6 19 6 0

1.Divide the given number into two-digit groups starting form right. Here, the groups would be 1, 63 and 84. This tells us that there would be three digits in the integer portion of the square root. In the present problem, this knowledge is significant.

2.Set up a chart similar to the division by Urdhwa-Tiryag process and write the given number with digits spaced well apart in the dividend line. Draw a slant line after the first group, which in the present case consists of single digit 1.

3.Write the first digit of the answer (square root) on the quotient line. This would be largest such single digit number that its square is less than or equal to the first group of the dividend. In the present case, such number is 1 and we write it in the quotient line above the 1 of the dividend. We also draw a slant line in the quotient line above the one in the dividend line.

4.The divisor for the rest of the process would be double of this quotient digit; in the present problem, it is 2 and is written in the usual place for divisor.

5.Subtract square of the quotient digit 1 from the first group of the dividend. It is 1 minus 1, that is 0. Write this remainder 0 in the gross dividend line below the first digit 6 of the right part of the dividend. Pull down that 6 and write it on the gross dividend line, in front of the 0. The gross dividend is now 06, i.e., 6. For the first digit of the dividend to the right of slant

60


line, the gross and net dividend are same. So, write this 6 on the net dividend line also, as it is.

6.Divide the net dividend 6 by the divisor 2, write the quotient 2 after the slant line in quotient and write the remainder 2 on the gross dividend line below the second digit 3 of the part of the dividend after the slant line. (Hold your surprise and doubt about the fact that, here, the quotient should be 3 and remainder 0, until the reading of next point! After that the doubt will vanish!)

7.Pull down the 3 in the dividend line and write it in front of the remainder 2 in the gross dividend line, making the gross dividend as 23. The net dividend is gross dividend minus the duplex of the quotient to the right of slant line. Here, the number to the right of slant line in the quotient is 2. Its duplex is 4. So, the net dividend is 23 minus 4, that is 19. This 19 is written on the net dividend line below 23 of the gross dividend line. Note that if we had taken the quotient to be 3 and the remainder 0 in the last step, the gross dividend would have been 3 only. Subtracting 9 (duplex of quotient 3) from this gross dividend would give us negative net dividend. This would complicate the procedure terribly. So, we avoid it by taking the wise decision in the last step. (You must have observed the same phenomenon in the examples given earlier, of calculation of square root of 784 and 729).

8.Divide the net dividend 19 by the divisor 2, write the quotient 8 on the quotient line and the remainder 3 on the gross dividend line below the third digit 8 of the right part of the dividend.

9.Pull down this 8 and write it in front of the remainder 3 on the gross dividend line, making the gross dividend 38. The net dividend is gross dividend minus the duplex of the number to the right of slant line in the quotient. This number is 28 right now and its duplex is 2 x 2 x 8 = 32. Subtracting 32 from the gross dividend 38, we get the net dividend as 6.

61


10.Divide this net dividend 6 by the divisor 2, write quotient 0 on the quotient line and the remainder 6 on the gross dividend line below the last digit 4 of the dividend. Pull down this 4 and write it on the gross dividend line in front of the remainder 6, making the gross dividend as 64. To get the net dividend, subtract from this 64, the duplex of 280 (the number to the right of slant line in the quotient) that is 82 + 2 x 2 x 0 = 64. The net dividend is 0 and the division process stops.

11.You may be tempted to assume that square root of 16384 is 1280; but, we have estimated the number of digits in the integer portion of the square root to be 3 (see step 1). So, put decimal point after three digits in the answer making it 128.0 . As 128.0 is nothing but 128, we say that the answer is 128.

We must always check back the answer. Here also we can use our earlier "Gunit Samuchaya...." sutra, but the direct method of calculation of square of the answer is more advisable. It would be the complete check.1282 = D1/D12/D128/D28/D8 = 1/4/20/32/64 = 16384, the original number. Hence, our answer is checked.

Two more examples are given below to further consolidate the understanding of the procedure.

Square root of 552049: 7 / 4 3 . 0 14 ) 55 / 2 0 4 9 62 60 24 9 62 44 0 0 Ans: 743

Square root of 4553956: 2 / 1 3 4 . 0 0 4 ) 4 / 5 5 3 9 5 6 05 15 23 19 25 16

62


5 14 17 2 1 0 Ans: 2134

How does the method work?Finding out square root is reverse process of finding square

by "Dwandwa Yoga" method. Consider the process of finding out square root of 552049 shown above. We know that the square root would be a three digit number. Let us denote these digits by p, q, and r respectively. Then, by our "Dwandwa Yoga" method-

552049 = D(p) / D(pq) / D(pqr) / D(qr) / D(r)We find out p to be 7. We can now subtract D(7), i.e. 49, whose place value is 490000 from the original number and reduce the problem to-

552049 – 490000 = 62049 = D(pq) / D(pqr) / D(qr) / D(r)The part 62 out of the number 62049 is coming as the first dividend in the tabular form shown earlier. In the above expression D(pq) is 2 x p x q. As p is found to be 7, D(pq) is 14 x q. Therefore, we divide 62049 (actually, 62000) by 14, find q to be 4 (written in quotient line in the table) and D(pq) to be 56. The problem then reduces to-

62049 – 56000 = 6049 = D(pqr) / D(qr) / D(r)But, D(pqr) = 2 x p x r + q2 . As q is 4, so q2 is 16, with place value 1600. So, the problem now is-

6049 – 1600 = 4449 = 14 x r / D(qr) / D(r)We divide 4449 by 14 and find r to be 3. Finally, we check whether

4449 – 4200 = 249 = D(qr) / D(r)to ensure that we have not gone overboard in the division process. These steps have been simplified, shortened, and systematized by the tabular form method described in the previous sections.

Exercise 7.1:

63


(Level – High school students and above)Tell how many digits would be there in the square root of

the following numbers and then calculate their square root.

1. 1024 2. 5329 3. 3364 4. 7921 5. 2209 6. 1936 7. 4225 8. 441 9. 5929 10. 136911. 13689 12. 2822413. 118336 14. 70728115. 21025 16. 60372917. 56644 18. 61152419. 290521 20. 29160021. 1172889 22. 7402881623. 1532644 24. 498628925. 9412624

Square roots of non-exact squaresSquare roots of numbers that are not exact squares of an

integer will involve fractional portion. We may not get the exact answer in such cases. Depending upon the acceptable error in the answer, we calculate the square root up to a certain number of decimal places.

The procedure for finding square roots in such cases is same as above. Only thing to be done more carefully is placement of decimal point, which was somewhat obvious in the earlier cases. If we want to find the answer up to more number of decimal places, we have to add extra zeroes at the end of the given number. Please note that these zeroes should not be considered while calculating the position of the decimal point.

64


The calculation of square root of 4554516 which is not an exact square, is shown in the chart below. We are stopping after three places of decimals. So, there would be error involved in the answer.

2 / 1 3 4 . 1 3 1 4 ) 4 / 5 5 4 5 1 6 5 15 24 25 41 36 5 14 18 8 15 8

Let us know the amount of error involved due to stoppage at three places of decimals, by calculating back the square of the answer.

2134.1312 = D2/D21/D213/D2134/D21341/D213413/D2134131/ D134131/D34131/D4131/D131/D31/D1

= 4/4/13/22/21/38/32/28/31/14/10/6/4 = 4554515.125064

So, we are making error in the decimal portion only.

Exercise 7.2:(Level – High school students and above)

Tell how many digits would be there in the integer part of square root of the following numbers and then calculate their square root up to three places of decimals.

1. 73 2. 100 3. 755 4. 258 5. 2300 6. 3000 7. 4222 8. 7352 9. 4008 10. 713811. 13789 12. 2800413. 118426 14. 707000

65


15. 46.24 16. 66.585617. 2331.9241 18. 15227.5619. 72.795024 20. 1491.504421. 53.24 22. 163.523. 228.81 24. 79.7925. 865.2

66

EIGHTQuadratics

One of the important topics of high school mathematics is dealing with quadratics. As the quadratics occur quite frequently in many problems of science and engineering, the mastery over their solution becomes unavoidable. As we know, the quadratic expressions are of the form-

ax2 + bx + c

where a, b and c are the known constants. One of the problems given to students is the factorization of the quadratic expressions into their binomial factors.

Conventional methodIn our present day conventional method of factorization of

a quadratic expression, we split the middle term bx in such a fashion, that the sum of the split coefficients equals the middle coefficient b and their product equals the product of the first term a and the independent term c. The split equation takes the form-

ax2 + b1x + b2x + csuch that b1+b2 equals b and (b1 . b2) equals a.c. After that we take out ax common from the first two terms and b2 common from the last two terms. In the next step, we take out (x + b1/a) common from the resultant two terms and get the final answer. The procedure is illustrated in factorization of the following two expressions-

x2 + 7x + 12 = x2 + 4x + 3x + 12 = x (x + 4) + 3 (x + 4) = (x + 3) (x + 4) -Ans.

67


2x2 + 9x + 9 = 2x2 + 6x + 3x + 9 = 2x (x + 3) + 3 (x + 3) = (x + 3) (2x + 3) -Ans.

In this method, the student has to write the intermediate steps. In the Vaidik method, a student can directly write the final answer.

Vaidik MethodFor factorization of quadratic expressions, we use the

following two sub-sutras in Vaidik mathematics.

a. || vkuq#I;s.k ||

Anuroopyen(Meaning: Proportionately.)

b. || vk|e~ vk|su vUR;e~ vUR;su ||

Adyam Adyen Antyam Antyen(Meaning: The first by the first and the last by the last.)

The procedure for the factorization is as follows-1.For splitting the middle term we use 'Anuroopyen' sutra. Split

the middle coefficient into two such parts that the ratio of the first coefficient to the first split part is the same as the ratio of the second split part to the last coefficient; i.e. a/b1 = b2/c . This ratio gives one of the two factors.

2.Obtain the second factor by dividing the first coefficient of the quadratic by the first coefficient of the first factor already found and the last coefficient of the quadratic by the last

68

Chapter 8: Quadratics

coefficient of that (first) factor. Here, we are using 'Adyam Adyen Antyam Antyen' sutra.

For example, consider the quadratic 2x2 + 7x + 6.1.The middle term 7 is split into two such parts 4 and 3, that the

ratio of the first coefficient to the first part of the middle coefficient, i.e. 2:4 and the ratio of the second part to the last coefficient, i.e. 3:6 are the same. Now, this ratio 1:2, i.e. x+2 is one factor.

2.And, the second factor is obtained by dividing 2x2 + 6 (the first and last term of the original quadratic) by the first and last coefficient of the factor obtained in the earlier step.

So, we say 2x2 + 7x + 6 = (x + 2) (2x + 3)

For checking the correctness of the answer, we may multiply the factors so found and check whether they are giving our original equation back. But a quicker way is to use our familiar “Gunit Samuchchayaha, Samuchchya Gunitaha” sutra here. In the present context it means “The sum of coefficients of the quadratic expression is equal to the product of sum of the coefficients of the factors”. For the example taken above, we note that- 2x2 + 7x + 6 = (x + 2) (2x + 3) 2 + 7 + 6 = (1 + 2) (2 + 3) 15 = 3 x 5Besides checking the correctness of the answer, this sutra can be some times intuitively used to split the middle term.

For our convenience, we can classify the quadratic expressions given for factorization, into four categories. The process of factorization, then becomes quite mechanical. This

69


classification also makes computerization of the method very easy. Students should use this for practice only. Afterwards, they would get the answers intuitively. The categories are as follows-

Type 1: All three terms positive.

Example: 6 x2 + 19 x + 15

While splitting the middle term in such cases, start from 1x and step by 1x up to (b/2 + 1)x, to find appropriate proportion. In the present case, we try the following splits-

Split First ratio Second ratio1x+18x 6:1 18:15 No2x+17x 6:2 17:15 No3x+16x 6:3 16:15 No4x+15x 6:4 15:15 No5x+14x 6:5 14:15 No6x+13x 6:6 13:15 No7x+12x 6:7 12:15 No8x+11x 6:8 11:15 No9x+10x 6:9 10:15 Yes!

So, the first ratio is 6:9 i.e. 2:3 and the first factor is (2x+3). The second factor is easily found by "Adyam Adyen..." sutra as (3x+5). For obtaining this, we have divided the first coefficient 6 by the first coefficient 2 of the first factor and the last coefficient 15 by the last coefficient 3 of the first factor.

Note that all the coefficients in the two factors would be positive in the present case.

Type 2: Second coefficient b positive, last one c negative.

Example: 12x2 + 13 x - 4

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While splitting the middle term in such cases, start from (b+1)x and step by 1x up to (b - c/2 + 1)x, to find appropriate proportion. In the present case, we try the following splits-

Split First ratio Second ratio14x-1x 12:14 -1:-4 No15x-2x 12:15 -2:-4 No16x-3x 12:16 -3:-4 Yes!

The ratio is found to be 12:16 i.e. 3:4. Therefore, the first factor is (3x+4). And the second factor would be (4x-1). The answer as usual can be checked by the "Gunit Samuchchaya...." sutra as follows-

12 + 13 – 4 = (3 + 4) x (4 – 1) = 7 x 3 = 21. So, checked!

Notice that one of the factors will have negative term in such cases.

Type 3: Second coefficient b negative, last one c positive.

Example: 12x2 - 31 xy + 20y

While splitting the middle term in these cases, start from -1x and go up to (b/2)x, in steps of -1x to find the appropriate ratio. In the present case, we try the following splits-

Split First ratio Second ratio-1xy-30xy 12:-1 -30:20 No-2xy-29xy 12:-2 -29:20 No-3xy-28xy 12:-3 -28:20 No-4xy-27xy 12:-4 -27:20 No-5xy-26xy 12:-5 -26:20 No-6xy-25xy 12:-6 -25:20 No-7xy-24xy 12:-7 -24:20 No

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-8xy-23xy 12:-8 -23:20 No-9xy-22xy 12:-9 -22:20 No-10xy-21xy 12:-10 -21:20 No-11xy-20xy 12:-11 -20:20 No-12xy-19xy 12:-12 -19:20 No-13xy-18xy 12:-13 -18:20 No-14xy-17xy 12:-14 -17:20 No-15xy-16xy 12:-15 -16:20 Yes!

The ratio is found to be 12:-15 i.e. 4:-5. Therefore, the first factor is (4x-5y). And the second factor would be (3x-4y). The answer can be checked by the "Gunit Samuchchaya...." sutra as follows-

12 - 31 + 20 = (4 - 5) x (3 - 4) = (-1) x (-1) = 1. So, checked!

Notice that both the factors will have negative term in such cases.

Type 4: Second coefficient b and the last term c negative.

Example: 6x2 - 13 x - 19

While splitting the middle term in these cases, start from 1x and go up to (|b| + |c|/2)x, in steps of 1x to find the appropriate ratio. In the present case, we try the following splits-

Split First ratio Second ratio1x-14x 6:1 -14:-19 No2x-15x 6:2 -15:-19 No3x-16x 6:3 -16:-19 No4x-17x 6:4 -17:-19 No5x-18x 6:5 -18:-19 No6x-19x 6:6 -19:-19 Yes!

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The ratio is found to be 1:1. Therefore, the first factor is (x+1). And the second factor would be (6x-19). The answer as usual can be checked by the "Gunit Samuchchaya...." sutra as follows-

6 - 13 - 19 = (1 + 1) x (6 – 19) = 2 x (-13) = -26. So, checked!

Notice that the second factor will have negative term in such cases.


Factorize the following quadratic expressions. You may write the answer on the same line.

1.6x2 + 13x + 52.9x2 + 18x + 83.4x2 + 12x + 54.2x2 + 11x + 155.3x2 + 10x – 256.2x2 + 5x – 37.12x2 + 7x – 128.6x2 + 5x – 69.12x2 - 23xy + 10y2

10.8x2 - 26x + 1511.9x2 - 18x + 812.3x2 - 10x + 813.15x2 - 11xy - 12y2

14.7x2 - 6x – 115.6x2 - 5x – 3416.2x2 - 9x – 5

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Quadratic EquationsThe quadratic equation is of the form-

ax2 + bx + c = 0

We have already seen the method of factorization of quadratic expressions. For example,

2x2 + 7x + 5 = (x + 1) (2x + 5)If this expression is equated to zero, we readily get the solution in the form of two equations-(x + 1) = 0 x = -1 and(2x + 5) = 0 x = -2.5

However, for some equations, there may not be such integer factors. For example-

6x2 + 5x – 3 = 0

For solving such quadratic equations quickly, we use the following Vaidik mathematics method-

Rule: The first differential of a quadratic expression is equal to the square root of its discriminant. Representing this symbolically,

D1 = discriminant = b2 – 4ac

Thus,

D1 = d ( 6x2 + 5x – 3 ) = 12x + 5 = 25 + 72 = 97 dx

So, the first root is given by 12x + 5 = 97

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and the second root is given by 12x + 5 = - 97


Find roots of the following quadratic equations. You may write the answer on the same line.

1.4x2 - 4x + 1 = 02.7x2 - 5x - 2 = 03.x2 - 11x + 10 = 04.7x2 - 9x - 1 = 05.9x2 - 13x - 2 = 06.4x2 + 13x + 3 = 07.14x2 - 15x - 9 = 08.12x2 - 19x + 5 = 09.7x2 - 22x + 16 = 010.81x2 - 90x + 25 = 011.16x2 - 8x - 3 = 012.4x2 - 9 = 0

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NINEVaidik Mnemonics

It is very interesting to note that there existed a numerical code to convert the numbers from digits to alphabets of the Sanskrit Devanagari script and further create Verses or Shlokas out of these, so as to facilitate memorizing these numbers. The conversion code can be summarized by the sutra- dkfn UkOk] Vkfn UkOk] Ikkfn IkaPkd] ¸kkfn v"Vd] {k% 'kwU¸kEk~ The following table shows in detail the meaning of this sutra in the form of code using Devanagari alphabets to represent the ten digits-

1 2 3 4 5 6 7 8 9 0dka

[kkha

Xkga

?kgha

³gna

pca

Ncha

Tkja

>jha

Vta

Btha

MÃa

<dha

.kña

Rkta

Fktha

nda

/kdha

Ikpa

Qpha

Ckba

Hkbha

Ekma

;ya

jra

Ykla

Okva

'kÏa

"kÌa

Lksa

gha

{kkÌa

The vowels and matras did not have any code attached and thus made no difference. In conjunct consonants, only the last consonant was considered. As most of the digits have more than one alphabet representation and matras & vowels did not count, we can form multiple words to represent the same number. For example, 'bat' and 'gap' both represent the number 31. This facility

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gave the poet of the mnemonic Shloka, more freedom to use different words to represent a number.

An interesting example given by Bharati Krishna Tirthaji in his famous book "Vedic Mathematics" is the following Shloka-

XkksIkhHkkX¸kEk/kqOkzkRk J`fX³'kksnf/kLkfU/kXk ||

[kYkTkhfOkRk[kkRkkOk XkYkgkYkkjLka/kj ||It has two literal meanings; one it is in praise of Lord Krishna and other it is in praise of Lord Shiva. Its third meaning is important to mathematicians because it codes an evaluation of / 10 up to 32 places of decimals. The coding as can be found from the chart given earlier is as follows-

Xkks Ikh Hkk X¸k Ek /kq Okzk Rk 3 1 4 1 5 9 2 6

J` fX³ 'kks n f/k Lk fU/k Xk 5 3 5 8 9 7 9 3

[k Yk Tkh fOk Rk [kk Rkk Ok 2 3 8 4 6 2 6 4

Xk Yk gk Ykk j Lka /k j 3 3 8 3 2 7 9 2

As known = 3.1415926535897932384626433832792....


This coding can be used, for remembering intermediate as well as final results of calculations in various numerical computations, besides remembering important numbers such as .

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