EKĀDHIKENA PŪRVEŅA Apply the formula to find the squares of the numbers 15, 45, 85, 125, 175 and verify the answers. Algebraic proof: Apply Ekadhikena purvena to find the squares of the numbers 95, 225, 375, 635, 745, 915, 1105, 2545. ii) Vulgar fractions whose denominators are numbers ending in NINE : Find the recurring decimal form of the fractions 1 / 29, 1 / 59, |
NIKHILAM NAVATAS’CARAMAM DASATAH The formula simply means : “all from 9 and the last from 10” The formula can be very effectively applied in multiplication of numbers, which are nearer to bases like 10, 100, 1000 i.e., to the powers of 10 . The procedure of multiplication using the Nikhilam involves minimum number of steps, space, time saving and only mental calculation. The numbers taken can be either less or more than the base considered. The difference between the number and the base is termed as deviation. Deviation may be positive or negative. Positive deviation is written without the positive sign and the negative deviation, is written using Rekhank (a bar on the number). Now observe the following table.
Some rules of the method (near to the base) in Multiplication Eg : 94. Now deviation can be obtained by ‘all from 9 and the last from 10’ sutra i.e., the last digit 4 is from 10 and remaining digit 9 from 9 gives 06. b) The two numbers under consideration are written one below the other. The deviations are written on the right hand side. Eg : Multiply 7 by 8. Now the base is 10. Since it is near to both the numbers, 7 we write the numbers one below the other. 8 ----- Take the deviations of both the numbers from the base and represent _ 7 3 _ Rekhank or the minus sign before the deviations 8 2 ------ ------ or 7 -3 8 -2 ------- ------- or remainders 3 and 2 implies that the numbers to be multiplied are both less than 10 c) The product or answer will have two parts, one on the left side and the other on the right. A vertical or a slant line i.e., a slash may be drawn for the demarcation of the two parts i.e.,   d) The R.H.S. of the answer is the product of the deviations of the numbers. It shall contain the number of digits equal to number of zeroes in the base. _ Since base is 10, 6 can be taken as it is.i.e., 7 3 _ 8 2 _____________ / (3x2) = 6 e) L.H.S of the answer is the sum of one number with the deviation of the other. It can be arrived at in any one of the four ways. i) Cross-subtract deviation 2 on the second row from the original number 7 in the first row i.e., 7-2 = 5. ii) Cross–subtract deviation 3 on the first row from the original number8 in the second row (converse way of (i)) i.e., 8 - 3 = 5 iii) Subtract the base 10 from the sum of the given numbers. i.e., (7 + 8) – 10 = 5 i.e., 10 – ( 3 + 2) = 5 Hence 5 is left hand side of the answer. _ f) If R.H.S. contains less number of digits than the number of zeros in the base, the remaining digits are filled up by giving zero or zeroes on the left side of the R.H.S. If the number of digits are more than the number of zeroes in the base, the excess digit or digits are to be added to L.H.S of the answer.Thus 7 3 _ 8 2 ¯¯¯¯¯¯¯¯¯¯¯¯ 5 / Now (d) and (e) together give the solution _ 7 3 7 _ 8 2 i.e., X 8 ¯¯¯¯¯¯¯ ¯¯¯¯¯¯ 5 / 6 56 The general form of the multiplication under Nikhilam can be shown as follows : Let N1 and N2 be two numbers near to a given base in powers of 10, and D1 and D2 are their respective deviations from the base. Then N1 X N2 can be represented as  Case (i) : Both the numbers are lower than the base. We have already considered the example 7 x 8 , with base 10. Now let us solve some more examples by taking bases 100 and 1000 respectively. Ex. 1: Find 97 X 94. Here base is 100. Now following the rules, the working is as follows:  Ex. 2: 98 X 97 Base is 100.  Ex. 3: 75X95. Base is 100.  Ex. 4: 986 X 989. Base is 1000.  Ex. 5: 994X988. Base is 1000.  Ex. 6: 750X995.  Case ( ii) : Both the numbers are higher than the base. The method and rules follow as they are. The only difference is the positive deviation. Instead of cross – subtract, we follow cross – add. Ex. 7: 13X12. Base is 10  Ex. 8: 18X14. Base is 10.  Ex. 9: 104X102. Base is 100. 104 04 Ex. 10: 1275X1004. Base is 1000.102 02 ¯¯¯¯¯¯¯¯¯¯¯¯ 106 / 4x2 = 10608 ( rule - f ) ¯¯¯¯¯¯¯¯¯¯¯¯ 1275 275 Case ( iii ): One number is more and the other is less than the base.1004 004 ¯¯¯¯¯¯¯¯¯¯¯¯ 1279 / 275x4 = 1279 / 1100 ( rule - f ) ____________ = 1280100 In this situation one deviation is positive and the other is negative. So the product of deviations becomes negative. So the right hand side of the answer obtained will therefore have to be subtracted. To have a clear representation and understanding a vinculum is used. It proceeds into normalization. Ex.11: 13X7. Base is 10  Note : Conversion of common number into vinculum number and vice versa. Eg : __ 9 = 10 – 1 = 11 _ 98 = 100 – 2 = 102 _ 196 = 200 – 4 = 204 _ 32 = 30 – 2 = 28 _ 145 = 140 – 5 = 135 _ 322 = 300 – 22 = 278. etc The procedure can be explained in detail using Nikhilam Navatascaram Dasatah, Ekadhikena purvena, Ekanyunena purvena in the foregoing pages of this book.] Ex. 12: 108 X 94. Base is 100.  Ex. 13: 998 X 1025. Base is 1000.  Algebraic Proof: Case ( i ): Let the two numbers N1 and N2 be less than the selected base say x. N1 = (x-a), N2 = (x-b). Here a and b are the corresponding deviations of the numbers N1 and N2 from the base x. Observe that x is a multiple of 10. Now N1 X N2 = (x-a) (x-b) = x.x – x.b – a.x + ab = x (x – a – b ) + ab. [rule – e(iv), d ]= x [(x – a) – b] + ab = x (N1–b) + ab[rule–e(i),d] or = x [(x – b) – a] = x (N2 – a) + ab. [rule –e (ii),d] x (x – a – b) + ab can also be written as x[(x – a) + (x – b) – x] + ab = x[N1+N2 – x] + ab [rule – e(iii),d]. A difficult can be faced, if the vertical multiplication of the deficit digits or deviations i.e., a.b yields a product consisting of more than the required digits. Then rule-f will enable us to surmount the difficulty. Case ( ii ) : When both the numbers exceed the selected base, we have N1 = x + a, N2 = x + b, x being the base. Now the identity (x+a) (x+b) = x(x+a+b) + a.b holds good, of course with relevant details mentioned in case (i). Case ( iii ) : When one number is less and another is more than the base, we can use (x-a)(x+b) = x(x–a+ b)–ab. and the procedure is evident from the examples given. Find the following products by Nikhilam formula. 1) 7 X 4 2) 93 X 85 3) 875 X 994 4) 1234 X 1002 5) 1003 X 997 6) 11112 X 9998 7) 1234 X 1002 8) 118 X 105 Consider some two digit numbers (dividends) and same divisor 9. Observe the following example. i) 13 ÷ 9 The quotient (Q) is 1, Remainder (R) is 4. since 9 ) 13 ( 1 9 ____ 4 ii) 34 ÷ 9, Q is 3, R is 7. iii) 60 ÷ 9, Q is 6, R is 6. iv) 80 ÷ 9, Q is 8, R is 8. Now we have another type of representation for the above examples as given hereunder: i) Split each dividend into a left hand part for the Quotient and right - hand part for the remainder by a slant line or slash. Eg. 13 as 1 / 3, 34 as 3 / 4 , 80 as 8 / 0. ii) Leave some space below such representation, draw a horizontal line. Eg. 1 / 3 3 / 4 8 / 0 ______ , ______ , ______ iii) Put the first digit of the dividend as it is under the horizontal line. Put the same digit under the right hand part for the remainder, add the two and place the sum i.e., sum of the digits of the numbers as the remainder.Eg. 1 / 3 3 / 4 8 / 0 1 3 8 ______ , ______ , ______ 1 / 4 3 / 7 8 / 8 Now the problem is over. i.e., 13 ÷ 9 gives Q = 1, R = 4 34 ÷ 9 gives Q = 3, R = 7 80 ÷ 9 gives Q = 8, R = 8 Proceeding for some more of the two digit number division by 9, we get a) 21 ÷ 9 as The examples given so far convey that in the division of two digit numbers by 9, we can mechanically take the first digit down for the quotient – column and that, by adding the quotient to the second digit, we can get the remainder.9) 2 / 1 i.e Q=2, R=3 2 ¯¯¯¯¯¯ 2 / 3 b) 43 ÷ 9 as 9) 4 / 3 i.e Q = 4, R = 7. 4 ¯¯¯¯¯¯ 4 / 7 Now in the case of 3 digit numbers, let us proceed as follows. i) 9 ) 104 ( 11 9 ) 10 / 4 99 1 / 1 ¯¯¯¯¯¯ as ¯¯¯¯¯¯¯ 5 11 / 5 ii) 9 ) 212 ( 23 9 ) 21 / 2 207 2 / 3 ¯¯¯¯¯ as ¯¯¯¯¯¯¯ 5 23 / 5 iii) 9 ) 401 ( 44 9 ) 40 / 1 396 4 / 4 ¯¯¯¯¯ as ¯¯¯¯¯¯¯¯ 5 44 / 5 Note that the remainder is the sum of the digits of the dividend. The first digit of the dividend from left is added mechanically to the second digit of the dividend to obtain the second digit of the quotient. This digit added to the third digit sets the remainder. The first digit of the dividend remains as the first digit of the quotient. Consider 511 ÷ 9 Add the first digit Thus 9 ) 51 / 1 5 / 6 ¯¯¯¯¯¯¯ 56 / 7 Q is 56, R is 7. Extending the same principle even to bigger numbers of still more digits, we can get the results. Eg : 1204 ÷ 9 i) Add first digit 1 to the second digit 2. 1 + 2 = 3 In symbolic form 9 ) 120 / 4ii) Add the second digit of quotient 13. i.e., 3 to third digit ‘0’ and obtain the Quotient. 3 + 0 = 3, 133 iii) Add the third digit of Quotient 133 i.e.,3 to last digit ‘4’ of the dividend and write the final Quotient and Remainder. R = 3 + 4 = 7, Q = 133 13 / 3 ¯¯¯¯¯¯¯¯ 133 / 7 Another example. 9 ) 13210 / 1 132101 ÷ 9 gives 1467 / 7 Q = 14677, R = 8 ¯¯¯¯¯¯¯¯¯¯ 14677 / 8 In all the cases mentioned above, the remainder is less than the divisor. What about the case when the remainder is equal or greater than the divisor? Eg. 9 ) 3 / 6 9) 24 / 6 3 2 / 6 ¯¯¯¯¯¯ or ¯¯¯¯¯¯¯¯ 3 / 9 (equal) 26 / 12 (greater). We proceed by re-dividing the remainder by 9, carrying over this Quotient to the quotient side and retaining the final remainder in the remainder side. 9 ) 3 / 6 9 ) 24 / 6 / 3 2 / 6 ¯¯¯¯¯¯¯ ¯¯¯¯¯¯¯¯ 3 / 9 26 / 12 ¯¯¯¯¯¯¯ ¯¯¯¯¯¯¯¯ 4 / 0 27 / 3 Q = 4, R = 0 Q = 27, R = 3. When the remainder is greater than divisor, it can also be represented as 9 ) 24 / 6 2 / 6 ¯¯¯¯¯¯¯¯ 26 /1 / 2 / 1 ¯¯¯¯¯¯¯¯ 1 / 3 ¯¯¯¯¯¯¯¯ 27 / 3 Now consider the divisors of two or more digits whose last digit is 9,when divisor is 89. We Know 113 = 1 X 89 + 24, Q =1, R = 24 10015 = 112 X 89 + 47, Q = 112, R = 47. Representing in the previous form of procedure, we have 89 ) 1 / 13 89 ) 100 / 15 / 11 12 / 32 ¯¯¯¯¯¯¯ ¯¯¯¯¯¯¯¯¯¯ 1 / 24 112 / 47 But how to get these? What is the procedure? Now Nikhilam rule comes to rescue us. The nikhilam states “all from 9 and the last from 10”. Now if you want to find 113 ÷ 89, 10015 ÷ 89, you have to apply nikhilam formula on 89 and get the complement 11.Further while carrying the added numbers to the place below the next digit, we have to multiply by this 11. 89 ) 1 / 13 89 ) 100 / 15 ¯¯ / 11 11 11 / first digit 1 x 11 ¯¯¯¯¯¯¯¯ 1 / 24 1 / 1 total second is 1x11 22 total of 3rd digit is 2 x 11 ¯¯¯¯¯¯¯¯¯¯ 112 / 47 What is 10015 ÷ 98 ? Apply Nikhilam and get 100 – 98 = 02. Set off the 2 digits from the right as the remainder consists of 2 digits. While carrying the added numbers to the place below the next digit, multiply by 02. Thus 98 ) 100 / 15 ¯¯ 02 02 / i.e., 10015 ÷ 98 gives 0 / 0 Q = 102, R = 19 / 04 ¯¯¯¯¯¯¯¯¯¯ 102 / 19 In the same way 897 ) 11 / 422 ¯¯¯ 103 1 / 03 / 206 ¯¯¯¯¯¯¯¯¯ 12 / 658 gives 11,422 ÷ 897, Q = 12, R=658. In this way we have to multiply the quotient by 2 in the case of 8, by 3 in the case of 7, by 4 in the case of 6 and so on. i.e., multiply the Quotient digit by the divisors complement from 10. In case of more digited numbers we apply Nikhilam and proceed. Any how, this method is highly useful and effective for division when the numbers are near to bases of 10. * Obtain the Quotient and Remainder for the following problems. 1) 311 ÷ 9 2) 120012 ÷ 9 3) 1135 ÷ 97 4) 2342 ÷ 98 5) 113401 ÷ 997 6) 11199171 ÷ 99979 | ||||||||||||||||||||||||||||||||||||||||||||||||||||||||
ŨRDHVA TIRYAGBHYĀM Urdhva – tiryagbhyam is the general formula applicable to all cases of multiplication and also in the division of a large number by another large number. It means “Vertically and cross wise.”(a) Multiplication of two 2 digit numbers. Ex.1: Find the product 14 X 12 i) The right hand most digit of the multiplicand, the first number (14) i.e., 4 is multiplied by the right hand most digit of the multiplier, the second number (12) i.e., 2. The product 4 X 2 = 8 forms the right hand most part of the answer. ii) Now, diagonally multiply the first digit of the multiplicand (14) i.e., 4 and second digit of the multiplier (12) i.e., 1 (answer 4 X 1=4); then multiply the second digit of the multiplicand i.e., 1 and first digit of the multiplier i.e., 2 (answer 1 X 2 = 2); add these two i.e., 4 + 2 = 6. It gives the next, i.e., second digit of the answer. Hence second digit of the answer is 6. iii) Now, multiply the second digit of the multiplicand i.e., 1 and second digit of the multiplier i.e., 1 vertically, i.e., 1 X 1 = 1. It gives the left hand most part of the answer. Thus the answer is 16 8. Symbolically we can represent the process as follows :  The symbols are operated from right to left . Step i) :  Step ii) :  Step iii) :  Now in the same process, answer can be written as 23 13 ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ 2 : 6 + 3 : 9 = 299 (Recall the 3 steps) Ex.3 41 X 41 ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ 16 : 4 + 4 : 1 = 1681. What happens when one of the results i.e., either in the last digit or in the middle digit of the result, contains more than 1 digit ? Answer is simple. The right – hand – most digit there of is to be put down there and the preceding, i.e., left –hand –side digit or digits should be carried over to the left and placed under the previous digit or digits of the upper row. The digits carried over may be written as in Ex. 4. Ex.4: 32 X 24 Step (i) : 2 X 4 = 8 Step (ii) : 3 X 4 = 12; 2 X 2 = 4; 12 + 4 = 16. Here 6 is to be retained. 1 is to be carried out to left side. Step (iii) : 3 X 2 = 6. Now the carried over digit 1 of 16 is to be added. i.e., 6 + 1 = 7. Thus 32 X 24 = 768 We can write it as follows 32 24 ¯¯¯¯ 668 1 ¯¯¯¯ 768. Note that the carried over digit from the result (3X4) + (2X2) = 12+4 = 16 i.e., 1 is placed under the previous digit 3 X 2 = 6 and added. After sufficient practice, you feel no necessity of writing in this way and simply operate or perform mentally. Ex.5 28 X 35. Step (i) : 8 X 5 = 40. 0 is retained as the first digit of the answer and 4 is carried over. Ex.6Step (ii) : 2 X 5 = 10; 8 X 3 = 24; 10 + 24 = 34; add the carried over 4 to 34. Now the result is 34 + 4 = 38. Now 8 is retained as the second digit of the answer and 3 is carried over. Step (iii) : 2 X 3 = 6; add the carried over Thus 28 X 35 = 980. 48 47 ¯¯¯¯¯¯ 1606 65 ¯¯¯¯¯¯¯ 2256 Step (i): 8 X 7 = 56; 5, the carried over digit is placed below the second digit. Algebraic proof :Step (ii): ( 4 X 7) + (8 X 4) = 28 + 32 = 60; 6, the carried over digit is placed below the third digit. Step (iii): Respective digits are added. a) Let the two 2 digit numbers be (ax+b) and (cx+d). Note that x = 10. Now consider the product (ax + b) (cx + d) = ac.x2 + adx + bcx + b.d = ac.x2 + (ad + bc)x + b.d Observe that i) The first term i.e., the coefficient of x2 (i.e., 100, hence the digit in the 100th place) is obtained by vertical multiplication of a and c i.e., the digits in 10th place (coefficient of x) of both the numbers; b) Consider the multiplication of two 3 digit numbers.ii) The middle term, i.e., the coefficient of x (i.e., digit in the 10th place) is obtained by cross wise multiplication of a and d; and of b and c; and the addition of the two products; iii) The last (independent of x) term is obtained by vertical multiplication of the independent terms b and d. Let the two numbers be (ax2 + bx + c) and (dx2 + ex + f). Note that x=10 Now the product is ax2 + bx + c x dx2 + ex + f ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ad.x4+bd.x3+cd.x2+ae.x3+be.x2+ce.x+af.x2+bf.x+cf = ad.x4 + (bd + ae). x3 + (cd + be + af).x2 + (ce + bf)x + cf Note the following points : i) The coefficient of x4 , i.e., ad is obtained by the vertical multiplication of the first coefficient from the left side :  ii)The coefficient of x3 , i.e., (ae + bd) is obtained by the cross –wise multiplication of the first two coefficients and by the addition of the two products;  iii) The coefficient of x2 is obtained by the multiplication of the first coefficient of the multiplicand (ax2+bx +c) i.e., a; by the last coefficient of the multiplier (dx2 +ex +f) i.e.,f ; of the middle one i.e., b of the multiplicand by the middle one i.e., e of the multiplier and of the last one i.e., c of the multiplicand by the first one i.e., d of the multiplier and by the addition of all the three products i.e., af + be +cd :  iv) The coefficient of x is obtained by the cross wise multiplication of the second coefficient i.e., b of the multiplicand by the third one i.e., f of the multiplier, and conversely the third coefficient i.e., c of the multiplicand by the second coefficient i.e., e of the multiplier and by addition of the two products, i.e., bf + ce ;  v) And finally the last (independent of x) term is obtained by the vertical multiplication of the last coefficients c and f i.e., cf  Thus the process can be put symbolically as (from left to right)  Consider the following example 124 X 132. Proceeding from right to left i) 4 X 2 = 8. First digit = 8 ii) (2 X 2) + (3 X 4) = 4 + 12 = 16. The digit 6 is retained and 1 is carried over to left side. Second digit = 6. iii) (1 X 2) + (2 X 3) + (1 X 4) = 2 + 6 + 4 =12. The carried over 1 of above step is added i.e., 12 + 1 = 13. Now 3 is retained and 1 is carried over to left side. Thus third digit = 3. iv) ( 1X 3 ) + ( 2 X 1 ) = 3 + 2 = 5. the carried over 1 of above step is added i.e., 5 + 1 = 6 . It is retained. Thus fourth digit = 6 v) ( 1 X 1 ) = 1. As there is no carried over number from the previous step it is retained. Thus fifth digit = 1 124 X 132 = 16368. Let us work another problem by placing the carried over digits under the first row and proceed. 234x 316 ¯¯¯¯¯¯¯ 61724 1222 ¯¯¯¯¯¯¯ 73944 i) 4 X 6 = 24 : 2, the carried over digit is placed below the second digit. Note :ii) (3 X 6) + (4 x 1) = 18 + 4 = 22 ; 2, the carried over digit is placed below third digit. iii) (2 X 6) + (3 X 1) + (4 X 3) = 12 + 3 + 12 = 27 ; 2, the carried over digit is placed below fourth digit. iv) (2 X 1) + ( 3 X 3) = 2 + 9 = 11; 1, the carried over digit is placed below fifth digit. v) ( 2 X 3 ) = 6. vi) Respective digits are added. 1. We can carry out the multiplication in urdhva - tiryak process from left to right or right to left. Example 1 : Find the product of (a+2b) and (3a+b).2. The same process can be applied even for numbers having more digits. 3. urdhva –tiryak process of multiplication can be effectively used in multiplication regarding algebraic expressions.  Example 2 : 3a2 + 2a + 4 x 2a2 + 5a + 3 ¯¯¯¯¯¯¯¯¯¯¯¯¯¯ i) 4 X 3 = 12 Hence the product is 6a4 + 19a3 + 27a2 + 26a + 12ii) (2 X 3) + ( 4 X 5 ) = 6 + 20 = 26 i.e., 26a iii) (3 X 3) + ( 2 X 5 ) + ( 4 X 2 ) = 9 + 10 + 8 = 27 i.e., 27a2 iv) (3 X 5) + ( 2 X 2 ) = 15 + 4 = 19 i.e., 19 a3 v) 3 X 2 = 6 i.e., 6a4 Example 3 : Find (3x2 + 4x + 7) (5x +6) Now 3.x2 + 4x + 7 0.x2 + 5x + 6 ¯¯¯¯¯¯¯¯¯¯¯¯ i) 7 X 6 = 42 Hence the product is 15x3 + 38x2 + 59x + 42ii) (4 X 6) + (7 X 5) = 24 + 35 = 59 i.e., 59x iii) (3 X 6) + (4 X 5) + (7 X 0) = 18 + 20 + 0 = 38 i.e., 38x2 iv) (3 X 5) + (0 X 4) = 15 + 0 = 15 i.e., 15x3 v) 3 X 0 = 0 Find the products using urdhva tiryagbhyam process. 1) 25 X 16 2) 32 X 48 3) 56 X 56 4) 137 X 214 5) 321 X 213 6) 452 X 348 7) (2x + 3y) (4x + 5y) 8) (5a2 + 1) (3a2 + 4) 9) (6x2 + 5x + 2 ) (3x2 + 4x +7) 10) (4x2 + 3) (5x + 6) As per the statement it an used as a simple argumentation for division process particularly in algebra. Consider the division of (x3 + 5x2 + 3x + 7) by (x – 2) process by converse of urdhva – tiryak : i) x3 divided by x gives x2 . x3 + 5x2 + 3x + 7 It is the first term of the Quotient. ___________________ x – 2 Q = x2 + - - - - - - - - - - - ii) x2 X – 2 = - 2x2 . But 5x2 in the dividend hints 7x2 more since 7x2 – 2x2 = 5x2 . This ‘more’ can be obtained from the multiplication of x by 7x. Hence second term of Q is 7x. x3 + 5x2 + 3x + 7¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ gives Q = x2 + 7x + - - - - - - - - x – 2 iii)We now have – 2 X 7x = -14x. But the 3rd term in the dividend is 3x for which ‘17x more’ is required since 17x – 14x =3x. Now multiplication of x by 17 gives 17x. Hence third term of quotient is 17 Thus x3 + 5x2 + 3x + 7 _________________ gives Q= x2 + 7x +17 x – 2 iv) Now last term of Q, i.e., 17 multiplied by –2 gives 17X–2 = -34 but the relevant term in dividend is 7. So 7 + 34 = 41 ‘more’ is required. As there no more terms left in dividend, 41 remains as the remainder. x3 + 5x2 + 3x + 7 ________________ gives Q= x2 + 7x +17 and R = 41. x – 2 Find the Q and R in the following divisions by using the converse process of urdhva – tiryagbhyam method : 1) 3x2 – x – 6 2) 16x2 + 24x +9 ¯¯¯¯¯¯¯¯¯ ¯¯¯¯¯¯¯¯¯¯¯¯ 3x – 7 4x+3 3) x3+ 2x2 +3x + 5 4) 12x4 – 3x2 – 3x + 12 ¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ x - 3 x2 + 1 |
PARĀVARTYA – YOJAYET ‘Paravartya – Yojayet’ means 'transpose and apply' (i) Consider the division by divisors of more than one digit, and when the divisors are slightly greater than powers of 10.Example 1 : Divide 1225 by 12. Step 1 : (From left to right ) write the Divisor leaving the first digit, write the other digit or digits using negative (-) sign and place them below the divisor as shown. Example 2 : Divide 1697 by 14.12 -2 ¯¯¯¯ Step 2 : Write down the dividend to the right. Set apart the last digit for the remainder. i.e.,, 12 122 5 - 2 Step 3 : Write the 1st digit below the horizontal line drawn under the dividend. Multiply the digit by –2, write the product below the 2nd digit and add. i.e.,, 12 122 5 -2 -2 ¯¯¯¯¯ ¯¯¯¯ 10 Since 1 x –2 = -2 and 2 + (-2) = 0 Step 4 : We get second digits’ sum as ‘0’. Multiply the second digits’ sum thus obtained by –2 and writes the product under 3rd digit and add. 12 122 5 - 2 -20 ¯¯¯¯ ¯¯¯¯¯¯¯¯¯¯ 102 5 Step 5 : Continue the process to the last digit. i.e., 12 122 5 - 2 -20 -4 ¯¯¯¯¯ ¯¯¯¯¯¯¯¯¯¯ 102 1 Step 6: The sum of the last digit is the Remainder and the result to its left is Quotient. Thus Q = 102 and R = 1 14 1 6 9 7 - 4 -4–8–4 ¯¯¯¯ ¯¯¯¯¯¯¯ 1 2 1 3 Q = 121, R = 3. Example 3 : Divide 2598 by 123. Note that the divisor has 3 digits. So we have to set up the last two digits of the dividend for the remainder. 1 2 3 25 98 Step ( 1 ) & Step ( 2 ) -2-3 ¯¯¯¯¯ ¯¯¯¯¯¯¯¯ Now proceed the sequence of steps write –2 and –3 as follows : 1 2 3 2 5 9 8 -2-3 -4 -6 ¯¯¯¯¯ -2–3 ¯¯¯¯¯¯¯¯¯¯ 2 1 1 5 Since 2 X (-2, -3)= -4 , -6; 5 – 4 = 1 and (1 X (-2,-3); 9 – 6 – 2 = 1; 8 – 3 = 5. Hence Q = 21 and R = 15. Example 4 : Divide 239479 by 11213. The divisor has 5 digits. So the last 4 digits of the dividend are to be set up for Remainder. 1 1 2 1 3 2 3 9 4 7 9 -1-2-1-3 -2 -4-2-6 with 2 ¯¯¯¯¯¯¯¯ -1-2-1-3 with 1 ¯¯¯¯¯¯¯¯¯¯¯¯¯ 2 1 4 0 0 6 Hence Q = 21, R = 4006. Example 5 : Divide 13456 by 1123 1 1 2 3 1 3 4 5 6 -1–2–3 -1-2-3 ¯¯¯¯¯¯¯ -2-4 –6 ¯¯¯¯¯¯¯¯¯¯¯¯¯ 1 2 0–2 0 Note that the remainder portion contains –20, i.e.,, a negative quantity. To over come this situation, take 1 over from the quotient column, i.e.,, 1123 over to the right side, subtract the remainder portion 20 to get the actual remainder. Thus Q = 12 – 1 = 11, and R = 1123 - 20 = 1103. Find the Quotient and Remainder for the problems using paravartya – yojayet method. 1) 1234 ÷ 112 2) 11329 ÷ 1132 3) 12349 ÷ 133 4) 239479 ÷ 1203 Example 1 : Divide 6x2 + 5x + 4 by x – 1 X - 1 6x2 + 5x + 4 ¯¯¯¯¯¯ 1 6 + 11 ¯¯¯¯¯¯¯¯¯¯¯¯ 6x + 11 + 15 Thus Q = 6x+11, R=15. Example 2 : Divide x3 – 3x2 + 10x – 4 by x - 5 X - 5 x3 – 3x2 + 10x – 4 ¯¯¯¯¯ 5 5 + 10 100 ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ x2 + 2x + 20, + 96 Thus Q= x2 + 2x + 20, R = 96. The procedure as a mental exercise comes as follows : i) x3 / x gives x2 i.e.,, 1 the first coefficient in the Quotient. Example 3:ii) Multiply 1 by + 5,(obtained after reversing the sign of second term in the Quotient) and add to the next coefficient in the dividend. It gives 1 X( +5) = +5, adding to the next coefficient, i.e.,, –3 + 5 = 2. This is next coefficient in Quotient. iii) Continue the process : multiply 2 by +5, i.e.,, 2 X +5 =10, add to the next coefficient 10 + 10 = 20. This is next coefficient in Quotient. Thus Quotient is x2 + 2x + 20 iv) Now multiply 20 by + 5 i.e.,, 20 x 5 = 100. Add to the next (last) term, 100 + (-4) = 96, which becomes R, i.e.,, R =9. x4 – 3x3 + 7x2 + 5x + 7 ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ x + 4 Now thinking the method as in example ( 1 ), we proceed as follows. x + 4 x4 - 3x3 + 7x2 + 5x + 7 ¯¯¯¯¯ -4 - 4 + 28 - 140 + 540 ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ x3 - 7x2 + 35x - 135 547 Thus Q = x3 – 7x2 + 35x – 135 and R = 547. or we proceed orally as follows: x4 / x gives 1 as first coefficient. i) -4 X 1 = - 4 : add to next coefficient – 4 + (-3) = - 7 which gives next coefficient in Q. Thus Q = x3 – 7x2 + 35x – 135 , R = 547.ii) – 7 X - 4 = 28 : then 28 + 7 = 35, the next coefficient in Q. iii) 35 X - 4 = - 140 : then – 140 + 5 = - 135, the next coefficient in Q. iv) - 135 X - 4 = 540 : then 540 + 7 = 547 becomes R. Note : 1. We can follow the same procedure even the number of terms is more. Now consider the divisors of second degree or more as in the following example.2. If any term is missing, we have to take the coefficient of the term as zero and proceed. Example :4 2x4 – 3x3 – 3x + 2 by x2 + 1. Here x2 term is missing in the dividend. Hence treat it as 0 . x2 or 0 . And the x term in divisor is also absent we treat it as 0 . x. Now x2 + 1 2x4 - 3x3 + 0 . x2 - 3x + 2 x2 + 0 . x + 1 0 - 2 ¯¯¯¯¯¯¯¯¯¯¯¯ 0 - 1 0 + 3 0 + 2 ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ 2 - 3 - 2 0 4 Thus Q = 2x2 - 3x - 2 and R = 0 . x + 4 = 4. Example 5 : 2x5 – 5x4 + 3x2 – 4x + 7 by x3 – 2x2 + 3. We treat the dividend as 2x5 – 5x4 + 0. x3 + 3x2 – 4x + 7 and divisor as x3 – 2x2 + 0 . x + 3 and proceed as follows : x3 – 2x2 + 0 . x + 3 2x5 – 5x4 + 0.x3 + 3x2 – 4x + 7 ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ 2 0 - 3 4 0 - 6 -2 0 + 3 - 4 0 + 6 ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ 2 - 1 - 2 - 7 - 1 +13 Thus Q = 2x2 – x – 2, R = - 7 x2 – x + 13. You may observe a very close relation of the method paravartya in this aspect with regard to REMAINDER THEOREM and HORNER PROCESS of Synthetic division. And yet paravartya goes much farther and is capable of numerous applications in other directions also. Apply paravartya – yojayet to find out the Quotient and Remainder in each of the following problems. 1) (4x2 + 3x + 5) ÷ (x+1) 2) (x3 – 4x2 + 7x + 6) ÷ (x – 2) 3) (x4 – x3 + x2 + 2x + 4) ÷ (x2 - x – 1) 4) (2x5 + x3 – 3x + 7) ÷ (x3 + 2x – 3) 5) (7x6 + 6x5 – 5x4 + 4x3 – 3x2 + 2x – 1) ÷ (x-1) Recall that 'paravartya yojayet' means 'transpose and apply'. The rule relating to transposition enjoins invariable change of sign with every change of side. i.e., + becomes - and conversely ; and X becomes ÷ and conversely. Further it can be extended to the transposition of terms from left to right and conversely and from numerator to denominator and conversely in the concerned problems. Type ( i ) : Consider the problem 7x – 5 = 5x + 1 7x – 5x = 1 + 5 i.e.,, 2x = 6 x = 6 ÷ 2 = 3. Observe that the problem is of the type ax + b = cx + d from which we get by ‘transpose’ (d – b), (a – c) and d - b. x = ¯¯¯¯¯¯¯¯ a - c In this example a = 7, b = - 5, c = 5, d = 1 Hence 1 – (- 5) 1+5 6 x = _______ = ____ = __ = 3 7 – 5 7-5 2 Example 2: Solve for x, 3x + 4 = 2x + 6 d - b 6 - 4 2 x = _____ = _____ = __ = 2 a - c 3 - 2 1 Type ( ii ) : Consider problems of the type (x + a) (x+b) = (x+c) (x+d). By paravartya, we get cd - ab x = ______________ (a + b) – (c + d) It is trivial form the following steps (x + a) (x + b) = (x + c) (x + d) x2 + bx + ax + ab = x2 + dx + cx + cd bx + ax – dx – cx = cd – ab x( a + b – c – d) = cd – ab cd – ab cd - ab x = ____________ x = _________________ a + b – c – d ( a + b ) – (c + d.) Example 1 : (x – 3) (x – 2 ) = (x + 1 ) (x + 2 ). By paravartya cd – ab 1 (2) – (-3) (-2) x = __________ = ______________ a + b – c –d - 3 – 2 – 1 – 2 2 - 6 - 4 1 = _______ = ___ = __ - 8 - 8 2 Example 2 : (x + 7) (x – 6) = (x +3) (x – 4). Now cd - ab (3) (-4) – (7) (-6) x = ___________ = ________________ a + b – c – d 7 + (-6) – 3 - (-4) - 12 + 42 30 = ____________ = ___ = 15 7 – 6 – 3 + 4 2 Note that if cd - ab = 0 i.e.,, cd = ab, i.e.,, if the product of the absolute terms be the same on both sides, the numerator becomes zero giving x = 0. For the problem (x + 4) (x + 3) = (x – 2 ) ( x – 6 ) Solution is x = 0 since 4 X 3 = - 2 X - 6. = 12 Type ( iii) : Consider the problems of the type ax + b m ______ = __ cx + d n By cross – multiplication, n ( ax + b) = m (cx + d) nax + nb = mcx + md nax - mcx = md – nb x( na – mc ) = md – nb md - nb x = ________ na - mc. Now look at the problem once again ax + b m _____ = __ cx + d n paravartya gives md - nb, na - mc and md - nb x = _______ na - mc Example 1: 3x + 1 13 _______ = ___ 4x + 3 19 md - nb 13 (3) - 19(1) 39 - 19 20 x = ______ = ____________ = _______ = __ na - mc 19 (3) - 13(4) 57 - 52 5 = 4 Example 2: 4x + 5 7 ________ = __ 3x + 13/2 8 (7) (13/2) - (8)(5) x = _______________ (8) (4) - (7)(3) (91/2) - 40 (91 - 80)/2 11 1 = __________ = _________ = ______ = __ 32 – 21 32 – 21 2 X 11 2 Type (iv) : Consider the problems of the type m n _____ + ____ = 0 x + a x + b Take L.C.M and proceed. m(x+b) + n (x+a) ______________ = 0 (x + a) (x +b) mx + mb + nx + na ________________ = 0 (x + a)(x + b) (m + n)x + mb + na = 0  (m + n)x = - mb - na-mb - na x = ________ (m + n) Thus the problem m n ____ + ____ = 0, by paravartya process x + a x + b gives directly -mb - na x = ________ (m + n) Example 1 : 3 4 ____ + ____ = 0 x + 4 x – 6 gives -mb - na x = ________ Note that m = 3, n = 4, a = 4, b = - 6 (m + n) -(3)(-6) – (4) (4) 18 - 16 2 = _______________ = ______ = __ ( 3 + 4) 7 7 Example 2 : 5 6 ____ + _____ = 0 x + 1 x – 21 gives -(5) (-21) - (6) (1) 105 - 6 99 x = ________________ = ______ = __ = 9 5 + 6 11 11 I . Solve the following problems using the sutra Paravartya – yojayet. 1) 3x + 5 = 5x – 3 6) (x + 1) ( x + 2) = ( x – 3) (x – 4) 2) (2x/3) + 1 = x - 1 7) (x – 7) (x – 9) = (x – 3) (x – 22) 3) 7x + 2 5 8) (x + 7) (x + 9) = (x + 3 ) (x + 21) ______ = __ 3x - 5 8 4) x + 1 / 3 _______ = 1 3x - 1 5) 5 2 ____ + ____ = 0 x + 3 x – 4 1. Show that for the type of equations m n p ____ + ____ + ____ = 0, the solution is x + a x + b x + c - mbc – nca – pab x = ________________________ , if m + n + p =0. m(b + c) + n(c+a) + p(a + b) 2. Apply the above formula to set the solution for the problem Problem 3 2 5 ____ + ____ - ____ = 0 x + 4 x + 6 x + 5 m n m + n ____ + ____ = _____ x + a x + b x + c Now this can be written as, m n m n ____ + ____ = _____ + _____ x + a x + b x + c x + c m m n n ____ - ____ = _____ - _____ x + a x + c x + c x + b m(x +c) – m(x + a) n(x + b) – n(x + c) ________________ = ________________ (x + a) (x + c) (x + c) (x + b) mx + mc – mx – ma nx + nb – nx – nc ________________ = _______________ (x + a) (x + c) (x +c ) (x + b) m (c – a) n (b –c) ____________ = ___________ x +a x + b m (c - a).x + m (c - a).b = n (b - c). x + n(b - c).a x [ m(c - a) - n(b - c) ] = na(b - c) – mb (c - a) or x [ m(c - a) + n(c - b) ] = na(b - c) + mb (a - c) Thus mb(a - c) + na (b - c) x = ___________________ m(c-a) + n(c-b). By paravartya rule we can easily remember the formula. Example 1 : solve 3 4 7 ____ + _____ = ____ x + 1 x + 2 x + 3 In the usual procedure, we proceed as follows. 3 4 7 ____ + ____ = ____ x + 1 x + 2 x + 3 3(x + 2) + 4(x + 1) 7 ________________ = _____ (x + 1) (x + 2) x + 3 3x + 6 + 4x + 4 7 _____________ = ____ x2 + 2x + x + 2 x + 3 7x + 10 7 _________ = ____ x2 + 3x + 2 x + 3 (7x + 10) (x + 3) = 7(x2 + 3x + 2) 7x2 + 21x + 10x + 30 = 7x2 + 21x + 14. 31x + 30 = 21x + 14 31 x – 21 x = 14 – 30 i.e.,, 10x = - 16 x = - 16 / 10 = - 8 / 5 Now by Paravartya process 3 4 7 ____ + ____ = ____ ( ... N1 + N2 = 3+4 = 7 = N3) x + 1 x + 2 x + 3 mb( a – c ) + na ( b – c ) x = _____________________ here N1 = m = 3 , N2 = n = 4 ; m ( c – a ) + n ( c – b ) a = 1, b = 2, c = 3 3 . 2 ( 1 – 3 ) + 4 . 1 . ( 2 – 3) = __________________________ 3 ( 3 – 1 ) + 4 ( 3 – 2 ) 6 ( -2)+ 4 (-1) - 12 – 4 - 16 - 8 = _____________ = _______ = ____ = ___ 3 (2) + 4(1) 6 + 4 10 5 Example 2 : 3 5 8 ____ + ____ = _____ Here N1 + N2 = 3 + 5 = 8. x - 2 x – 6 x + 3 mb ( a – c ) + na ( b – c) x = _____________________ m ( c – a ) + n ( c – b ) 3 . ( -6 ) ( - 2 - 3 ) + 5 .( -2 ) ( -6 – 3 ) = __________________________________ 3 ( 3 – ( -2 ) ) + 5 ( 3 – ( - 6 ) ) 3 ( - 6 ) ( - 5 ) + 5 ( - 2 ) ( - 9 ) = ____________________________ 3( 3 + 2 ) + 5 ( 3 + 6 ) 90 + 90 = _______ = 180 / 60 = 3. 15 + 45 1) 2 3 5 ____ + ____ = ____ x + 2 x + 3 x + 5 2) 4 6 10 ____ + ____ = ____ x + 1 x + 3 x + 4 3) 5 2 3 ____ + ___ = ____ x - 2 3 - x x – 4 4) 4 9 15 _____ + _____ = _____ 2x + 1 3x + 2 3x + 1 But 3 (4) 2 (9) 2(15) ________ + ________ = _______ gives 3(2x + 1) 2( 3x + 2) 2(3x + 1) 12 18 30 _____ + _____ = _____ Now proceed. 6x + 3 6x + 4 6x + 2 Simultaneous simple equations: By applying Paravartya sutra we can derive the values of x and y which are given by two simultaneous equations. The values of x and y are given by ration form. The method to find out the numerator and denominator of the ratio is given below. Example 1: 2x + 3y = 13, 4x + 5y = 23. i) To get x, start with y coefficients and the independent terms and cross-multiply forward, i.e.,, right ward. Start from the upper row and multiply across by the lower one, and conversely, the connecting link between the two cross-products being a minus. This becomes numerator. Example 2: 5x – 3y = 11i.e.,, 2x + 3y = 13 4x + 5y = 23 Numerator of the x – value is 3 x 23 – 5 x 13 = 69 – 65 = 4 ii) Go from the upper row across to the lower one, i.e.,, the x- coefficient but backward, i.e.,, leftward. Denominator of the x – value is 3 x 4 – 2 x 5 = 12 – 10 = 2 Hence value of x = 4 ÷ 2 = 2. iii) To get y, follow the cyclic system, i.e.,, start with the independent term on the upper row towards the x–coefficient on the lower row. So numerator of the y–value is 13 x 4 – 23 x 2 = 52 – 46 = 6. iv) The denominator is the same as obtained in Step(ii) i.e.,, 2. Hence value of y is 6÷2=3. Thus the solution to the given equation is x = 2 and y = 3. 6x – 5y = 09 Now Nr. of x is (-3) (9) – (5) (11) = - 27 + 55 = 28 Dr. of x is (-3) (6) – (5) (-5) = - 18 + 25 = 07 x = Nr ÷ Dr = 28 ÷ 7 = 4 and for y, Nr is (11) (6) – (9)(5) = 66 – 45 = 21 Dr is 7 Hence y = 21 ÷ 7 = 3. Example 3: solve 3x + y = 5 4x – y = 9 Now we can straight away write the values as follows: (1)(9) – (-1)(5) 9 + 5 14 x = _____________ = _____ = ___ = 2 (1)(4) – (3)(-1) 4 + 3 7 (5)(4) – (9)(3) 20 – 27 -7 y = ____________ = _______ = ___ = -1 (1)(4) – (3)(-1) 4 + 3 7 Hence x = 2 and y = -1 is the solution. Algebraic Proof: ax + by = m ……… ( i ) cx + dy = n ………. ( ii ) Multiply ( i ) by d and ( ii ) by b, then subtract adx + bdy = m.d cbx + dby = n.b ____________________ ( ad – cb ) .x = md – nb md - nb bn - md x = ______ = ______ ad - cb bc - ad Multiply ( i ) by c and ( ii ) by a, then subtract acx + bcy = m.c cax + day = n.a _____________________ ( bc – ad ) . y = mc - na mc - na y = ______ bc - ad You feel comfort in the Paravartya process because it avoids the confusion in multiplication, change of sign and such other processes. 1. 2x + y = 5 2. 3x – 4y = 7 3x – 4y = 2 5x + y = 4 3. 4x + 3y = 8 4. x + 3y = 7 6x - y = 1 2x + 5y = 11 |
| SŨNYAM SĀMYASAMUCCAYE The Sutra 'Sunyam Samyasamuccaye' says the 'Samuccaya is the same, that Samuccaya is Zero.' i.e., it should be equated to zero. The term 'Samuccaya' has several meanings under different contexts.i) We interpret, 'Samuccaya' as a term which occurs as a common factor in all the terms concerned and proceed as follows. Example 1: The equation 7x + 3x = 4x + 5x has the same factor ‘ x ‘ in all its terms. Hence by the sutra it is zero, i.e., x = 0. Otherwise we have to work like this: 7x + 3x = 4x + 5x 10x = 9x 10x – 9x = 0 x = 0 This is applicable not only for ‘x’ but also any such unknown quantity as follows. Example 2: 5(x+1) = 3(x+1) No need to proceed in the usual procedure like 5x + 5 = 3x + 3 5x – 3x = 3 – 5 2x = -2 or x = -2 ÷ 2 = -1 Simply think of the contextual meaning of ‘ Samuccaya ‘ Now Samuccaya is ( x + 1 ) x + 1 = 0 gives x = -1 ii) Now we interpret ‘Samuccaya ‘as product of independent terms in expressions like (x+a) (x+b) Example 3: ( x + 3 ) ( x + 4) = ( x – 2) ( x – 6 )Here Samuccaya is 3 x 4 = 12 = -2 x -6 Since it is same , we derive x = 0 This example, we have already dealt in type ( ii ) of Paravartya in solving simple equations. iii) We interpret ‘ Samuccaya ‘ as the sum of the denominators of two fractions having the same numerical numerator. Consider the example.1 1 ____ + ____ = 0 3x-2 2x-1 for this we proceed by taking L.C.M. (2x-1)+(3x–2) ____________ = 0 (3x–2)(2x–1) 5x–3 __________ = 0 (3x–2)(2x–1) 5x – 3 = 0 5x = 3 3 x = __ 5 Instead of this, we can directly put the Samuccaya i.e., sum of the denominators i.e., 3x – 2 + 2x - 1 = 5x - 3 = 0 giving 5x = 3 x = 3 / 5 It is true and applicable for all problems of the type m m ____ + _____ = 0 ax+b cx+d Samuccaya is ax+b+cx+d and solution is ( m ≠ 0 ) - ( b + d ) x = _________ ( a + c ) iii) We now interpret ‘Samuccaya’ as combination or total. If the sum of the numerators and the sum of the denominators be the same, then that sum = 0. Consider examples of type ax + b ax + c _____ = ______ ax + c ax + b In this case, (ax+b) (ax+b) = (ax+c) (ax+c) a2x2 + 2abx + b2 = a2x2 + 2acx + c2 2abx – 2acx = c2 – b2 x ( 2ab – 2ac ) = c2 – b2 c2–b2 (c+b)(c-b) -(c+b) x = ______ = _________ = _____ 2a(b-c) 2a(b-c) 2a As per Samuccaya (ax+b) + (ax+c) = 0 2ax+b+c = 0 2ax = -b-c -(c+b) x = ______ 2a Hence the statement. Example 4: 3x + 4 3x + 5 ______ = ______ 3x + 5 3x + 4 Since N1 + N2 = 3x + 4 + 3x + 5 = 6x + 9 , And D1 + D2 = 3x + 4 + 3x + 5 = 6x + 9 We have N1 + N2 = D1 + D2 = 6x + 9 Hence from Sunya Samuccaya we get 6x + 9 = 0 6x = -9 -9 -3 x = __ = __ 6 2 Example 5: 5x + 7 5x + 12 _____ = _______ 5x +12 5x + 7 Hence N1 + N2 = 5x + 7 + 5x + 12 = 10x + 19 And D1 + D2 = 5x + 12 + 5x + 7 = 10x + 19 N1 + N2 = D1 + D2 gives 10x + 19 = 0 10x = -19 -19 x = ____ 10 Consider the examples of the type, where N1 + N2 = K ( D1 + D2 ), where K is a numerical constant, then also by removing the numerical constant K, we can proceed as above. Example 6: 2x + 3 x + 1 _____ = ______ 4x + 5 2x + 3 Here N1 + N2 = 2x + 3 + x + 1 = 3x + 4 D1 + D2 = 4x + 5 + 2x + 3 = 6x + 8 = 2 ( 3x + 4 ) Removing the numerical factor 2, we get 3x + 4 on both sides. 3x + 4 = 0 3x = -4 x = - 4 / 3. v) ‘Samuccaya‘ with the same meaning as above, i.e., case (iv), we solve the problems leading to quadratic equations. In this context, we take the problems as follows; If N1 + N2 = D1 + D2 and also the differences N1 ~ D1 = N2 ~ D2 then both the things are equated to zero, the solution gives the two values for x. Example 7: 3x + 2 2x + 5 _____ = ______ 2x + 5 3x + 2 In the conventional text book method, we work as follows : 3x + 2 2x + 5 _____ = ______ 2x + 5 3x + 2 ( 3x + 2 ) ( 3x + 2 ) = ( 2x + 5 ) ( 2x + 5 ) 9x2 + 12x + 4 = 4x2 + 20x + 25 9x2 + 12x + 4 - 4x2 - 20x – 25 = 0 5x2 – 8x – 21 = 0 5x2 – 15x + 7x – 21 = 0 5x ( x – 3 ) + 7 ( x – 3 ) = 0 (x – 3 ) ( 5x + 7 ) = 0 x – 3 = 0 or 5x + 7 = 0 x = 3 or - 7 / 5 Now ‘Samuccaya’ sutra comes to help us in a beautiful way as follows : Observe N1 + N2 = 3x + 2 + 2x + 5 = 5x + 7 D1 + D2 = 2x + 5 + 3x + 2 = 5x + 7 Further N1 ~ D1 = ( 3x + 2 ) – ( 2x + 5 ) = x – 3 N2 ~ D2 = ( 2x + 5) – ( 3x + 2 ) = - x + 3 = - ( x – 3 ) Hence 5x + 7 = 0 , x – 3 = 0 5x = -7 , x = 3 i.e., x = -7 / 5 , x = 3 Note that all these can be easily calculated by mere observation. Example 8: 3x + 4 5x + 6 ______ = _____ 6x + 7 2x + 3 Observe that N1 + N2 = 3x + 4 + 5x + 6 = 8x + 10 and D1 + D2 = 6x + 7 + 2x + 3 = 8x + 10 Further N1 ~ D1 = (3x + 4) – (6x + 7) = 3x + 4 – 6x – 7 = -3x – 3 = -3 ( x + 1 ) N2 ~ D2 = (5x + 6) – (2x + 3) = 3x + 3 = 3( x + 1) By ‘Sunyam Samuccaye’ we have 8x + 10 = 0 3( x + 1 ) = 0 8x = -10 x + 1 = 0 x = - 10 / 8 x = -1 = - 5 / 4 vi)‘Samuccaya’ with the same sense but with a different context and application . Example 9: 1 1 1 1 ____ + _____ = ____ + ____ x - 4 x – 6 x - 2 x - 8 Usually we proceed as follows. x–6+x-4 x–8+x-2 ___________ = ___________ (x–4) (x–6) (x–2) (x-8) 2x-10 2x-10 _________ = _________ x2–10x+24 x2–10x+16 ( 2x – 10 ) ( x2 – 10x + 16 ) = ( 2x – 10 ) ( x2 – 10x + 24) 2x3–20x2+32x–10x2+100x–160 = 2x3–20x2+48x–10x2+100x-240 2x3 – 30x2 + 132x – 160 = 2x3 – 30x2 + 148x – 240 132x – 160 = 148x – 240 132x – 148x = 160 – 240 – 16x = - 80 x = - 80 / - 16 = 5 Now ‘Samuccaya’ sutra, tell us that, if other elements being equal, the sum-total of the denominators on the L.H.S. and their total on the R.H.S. be the same, that total is zero. Now D1 + D2 = x – 4 + x – 6 = 2x – 10, and D3 + D4 = x – 2 + x – 8 = 2x – 10 By Samuccaya, 2x – 10 gives 2x = 10 10 x = __ = 5 2 Example 10: 1 1 1 1 ____ + ____ = ____ + _____ x - 8 x – 9 x - 5 x – 12 D1 + D2 = x – 8 + x – 9 = 2x – 17, and D3 + D4 = x – 5 + x –12 = 2x – 17 Now 2x – 17 = 0 gives 2x = 17 17 x = __ = 8½ 2 Example 11: 1 1 1 1 ____ - _____ = ____ - _____ x + 7 x + 10 x + 6 x + 9 This is not in the expected form. But a little work regarding transposition makes the above as follows. 1 1 1 1 ____ + ____ = ____ + _____ x + 7 x + 9 x + 6 x + 10 Now ‘Samuccaya’ sutra applies D1 + D2 = x + 7 + x + 9 = 2x + 16, and D3 + D4 = x + 6 + x + 10 = 2x + 16 Solution is given by 2x + 16 = 0 i.e., 2 x = - 16. x = - 16 / 2 = - 8. 1. 7 ( x + 2 ) + 3 ( x + 2 ) = 6 ( x + 2 ) + 5 ( x + 2 ) 2. ( x + 6 ) ( x + 3 ) = ( x – 9 ) ( x – 2 ) 3. ( x - 1 ) ( x + 14 ) = ( x + 2 ) ( x – 7 ) 1 1 4. ______ + ____ = 0 4 x - 3 x – 2 4 4 5. _____ + _____ = 0 3x + 1 5x + 7 2x + 11 2x+5 6. ______ = _____ 2x+ 5 2x+11 3x + 4 x + 1 7. ______ = _____ 6x + 7 2x + 3 4x - 3 x + 4 8. ______ = _____ 2x+ 3 3x - 2 1 1 1 1 9. ____ + ____ = ____ + _____ x - 2 x - 5 x - 3 x - 4 1 1 1 1 10. ____ - ____ = _____ - _____ x - 7 x - 6 x - 10 x - 9 Consider the problem ( x – 4 )3 + ( x – 6 )3 = 2 ( x – 5 )3. For the solution by the traditional method we follow the steps as given below: ( x – 4 )3 + ( x – 6 )3 = 2 ( x – 5 )3 x3 – 12x2 + 48x – 64 + x3 – 18x2 + 108x – 216 = 2 ( x3 – 15x2 + 75x – 125 ) 2x3 – 30x2 + 156x – 280 = 2x3 – 30x2 + 150x – 250 156x – 280 = 150x – 250 156x – 150x = 280 – 250 6x = 30 x = 30 / 6 = 5 But once again observe the problem in the vedic sense We have ( x – 4 ) + ( x – 6 ) = 2x – 10. Taking out the numerical factor 2, we have ( x – 5 ) = 0, which is the factor under the cube on R.H.S. In such a case “Sunyam samya Samuccaye” formula gives that x – 5 = 0. Hence x = 5 Think of solving the problem (x–249)3 + (x+247)3 = 2(x–1)3 The traditional method will be horrible even to think of. But ( x – 249 ) + ( x + 247 ) = 2x – 2 = 2 ( x – 1 ). And x – 1. on R.H.S. cube, it is enough to state that x – 1 = 0 by the ‘sutra’. x = 1 is the solution. No cubing or any other mathematical operations. Algebraic Proof : Consider ( x – 2a )3 + ( x – 2b )3 = 2 ( x – a – b )3 it is clear that x – 2a + x – 2b = 2x – 2a – 2b = 2 ( x – a – b ) Now the expression, x3 - 6x2a + 12xa2 – 8a3 + x3 – 6x2b + 12xb2 – 8b3 = 2(x3–3x2a–3x2b+3xa2+3xb2+6axb–a3–3a2b–3ab2–b3) = 2x3–6x2a–6x2b+6xa2+6xb2+12xab–2a3–6a2b–6ab2–2b3 cancel the common terms on both sides 12xa2+12xb2–8a3–8b3 = 6xa2+6xb2+12xab–2a3–6a2b–6ab2–2b3 6xa2 + 6xb2 – 12xab = 6a3 + 6b3 – 6a2b – 6ab2 6x ( a2 + b2 – 2ab ) = 6 [ a3 + b3 – ab ( a + b )] x ( a – b )2 = [ ( a + b ) ( a2 + b2 –ab ) – ( a + b )ab] = ( a + b ) ( a2 + b2 – 2ab ) = ( a + b ) ( a – b )2 since x = a + b 1. ( x – 3 )3 + ( x – 9 )3 = 2 ( x – 6 )3 2. ( x + 4 )3 + ( x – 10 )3 = 2 ( x – 3 )3 3. ( x + a + b – c )3 + ( x + b + c – a )3 = 2 ( x + b )3 (x + 2)3 x + 1 ______ = _____ (x + 3)3 x + 4 with the text book procedures we proceed as follows x3 + 6x2 + 12x +8 x + 1 _______________ = _____ x3 + 9x2 + 27x +27 x + 4 Now by cross multiplication, ( x + 4 ) ( x3 + 6x2 + 12x + 8 ) = ( x + 1 ) ( x3 + 9x2 + 27x + 27 ) x4 + 6x3 + 12x2 + 8x + 4x3 + 24x2 + 48x + 32 = x4 + 9x3 + 27x2 + 27x + x3 + 9x2 + 27x + 27 x4 + 10x3 + 36x2 + 56x + 32 = x4 + 10x3 + 36x2 + 54x + 27 56x + 32 = 54x + 27 56x – 54x = 27 – 32 2x = - 5 x = - 5 / 2 Observe that ( N1 + D1 ) with in the cubes on L.H.S. is x + 2 + x + 3 = 2x + 5 and N2 + D2 on the right hand side is x + 1 + x + 4 = 2x + 5. By vedic formula we have 2x + 5 = 0 x = - 5 / 2. 1. (x + 3)3 x+1 ______ = ____ (x + 5)3 x+7 2. (x - 5)3 x - 3 ______ = ____ (x - 7)3 x - 9 |
ĀNURŨPYE ŚŨNYAMANYAT The Sutra Anurupye Sunyamanyat says : 'If one is in ratio, the other one is zero'.We use this Sutra in solving a special type of simultaneous simple equations in which the coefficients of 'one' variable are in the same ratio to each other as the independent terms are to each other. In such a context the Sutra says the 'other' variable is zero from which we get two simple equations in the first variable (already considered) and of course give the same value for the variable. Example 1: 3x + 7y = 2 4x + 21y = 6 Observe that the y-coefficients are in the ratio Example 2: 323x + 147y = 1615 969x + 321y = 4845 The very appearance of the problem is frightening. But just an observation and anurupye sunyamanyat give the solution x = 5, because coefficient of x ratio is 323 : 969 = 1 : 3 and constant terms ratio is 1615 : 4845 = 1 : 3. y = 0 and 323 x = 1615 or 969 x = 4845 gives x = 5. 1. 12x + 78y = 12 2. 3x + 7y = 24 16x + 96y = 16 12x + 5y = 96 3. 4x – 6y = 24 4. ax + by = bm 7x – 9y = 36 cx + dy = dm Example 3 : Solve for x and y x + 4y = 10 x2 + 5xy + 4y2 + 4x - 2y = 20 x2 + 5xy + 4y2 + 4x - 2y = 20 can be written as ( x + y ) ( x + 4y ) + 4x – 2y = 20 10 ( x + y ) + 4x – 2y = 20 ( Since x + 4y = 10 ) 10x + 10y + 4x – 2y = 20 14x + 8y = 20 Now x + 4y = 10 14x + 8y = 20 and 4 : 8 :: 10 : 20 from the Sutra, x = 0 and 4y = 10, i.e.,, 8y = 20 y = 10/4 = 2½ Thus x = 0 and y = 2½ is the solution. |
SAŃKALANA – VYAVAKALANĀBHYAM This Sutra means 'by addition and by subtraction'. It can be applied in solving a special type of simultaneous equations where the x - coefficients and the y - coefficients are found interchanged.Example 1: 45x – 23y = 113 23x – 45y = 91 In the conventional method we have to make equal either the coefficient of x or coefficient of y in both the equations. For that we have to multiply equation ( 1 ) by 45 and equation ( 2 ) by 23 and subtract to get the value of x and then substitute the value of x in one of the equations to get the value of y or we have to multiply equation ( 1 ) by 23 and equation ( 2 ) by 45 and then subtract to get value of y and then substitute the value of y in one of the equations, to get the value of x. It is difficult process to think of. From Sankalana – vyavakalanabhyam add them, i.e., ( 45x – 23y ) + ( 23x – 45y ) = 113 + 91 i.e., 68x – 68y = 204 x – y = 3subtract one from other, i.e., ( 45x – 23y ) – ( 23x – 45y ) = 113 – 91 i.e., 22x + 22y = 22 x + y = 1and repeat the same sutra, we get x = 2 and y = - 1 Very simple addition and subtraction are enough, however big the coefficients may be. Example 2: 1955x – 476y = 2482 476x – 1955y = -4913 Oh ! what a problem ! And still just add, 2431( x – y ) = - 2431 x – y = -1subtract, 1479 ( x + y ) = 7395 x + y = 5once again add, 2x = 4 x = 2subtract - 2y = - 6 y = 31. 3x + 2y = 18 2x + 3y = 17 2. 5x – 21y = 26 21x – 5y = 26 3. 659x + 956y = 4186 956x + 659y = 3889 |
PŨRANĀPŨRAŅĀBHYĀM The Sutra can be taken as Purana - Apuranabhyam which means by the completion or non - completion. Purana is well known in the present system. We can see its application in solving the roots for general form of quadratic equation.We have : ax2 + bx + c = 0 x2 + (b/a)x + c/a = 0 ( dividing by a ) x2 + (b/a)x = - c/a completing the square ( i.e.,, purana ) on the L.H.S. x2 + (b/a)x + (b2/4a2) = -c/a + (b2/4a2) [x + (b/2a)]2 = (b2 - 4ac) / 4a2 ________ - b ± √ b2 – 4ac Proceeding in this way we finally get x = _______________ 2a Now we apply purana to solve problems. Example 1. x3 + 6x2 + 11 x + 6 = 0. Since (x + 2 )3 = x3 + 6x2 + 12x + 8 Add ( x + 2 ) to both sides We get x3 + 6x2 + 11x + 6 + x + 2 = x + 2 i.e.,, x3 + 6x2 + 12x + 8 = x + 2 i.e.,, ( x + 2 )3 = ( x + 2 ) this is of the form y3 = y for y = x + 2 solution y = 0, y = 1, y = - 1 i.e.,, x + 2 = 0,1,-1 which gives x = -2,-1,-3 Example 2: x3 + 8x2 + 17x + 10 = 0 We know ( x + 3 )3 = x3 + 9x2 + 27x + 27 So adding on the both sides, the term ( x2 + 10x + 17 ), we get x3 + 8x2 + 17x + x2 + 10x + 17 = x2 + 10x + 17 i.e.,, x3 + 9x2 + 27x + 27 = x2 + 6x + 9 + 4x + 8 i.e.,, ( x + 3 )3 = ( x + 3 )2 + 4 ( x + 3 ) – 4 y3 = y2 + 4y – 4 for y = x + 3 y = 1, 2, -2. Hence x = -2, -1, -5 Thus purana is helpful in factorization. Further purana can be applied in solving Biquadratic equations also. 1. x3 – 6x2 + 11x – 6 = 0 2. x3 + 9x2 + 23x + 15 = 0 3. x2 + 2x – 3 = 0 4. x4 + 4x3 + 6x2 + 4x – 15 = 0 |
CALANA - KALANĀBHYĀM In the book on Vedic Mathematics Sri Bharati Krishna Tirthaji mentioned the Sutra 'Calana - Kalanabhyam' at only two places. The Sutra means 'Sequential motion'.i) In the first instance it is used to find the roots of a quadratic equation 7x2 – 11x – 7 = 0. Swamiji called the sutra as calculus formula. Its application at that point is as follows. Now by calculus formula we say: 14x–11 = ±√317 A Note follows saying every Quadratic can thus be broken down into two binomial factors. An explanation in terms of first differential, discriminant with sufficient number of examples are given under the chapter ‘Quadratic Equations’. ii) At the Second instance under the chapter ‘Factorization and Differential Calculus’ for factorizing expressions of 3rd, 4th and 5th degree, the procedure is mentioned as 'Vedic Sutras relating to Calana – Kalana – Differential Calculus'. Further other Sutras 10 to 16 mentioned below are also used to get the required results. Hence the sutra and its various applications will be taken up at a later stage for discussion. But sutra – 14 is discussed immediately after this item. Now the remaining sutras : 10. YĀVADŨNAM ( The deficiency ) 11. VYAŞŢISAMAŞŢIH ( Whole as one and one as whole ) 12. ŚEŞĀNYAŃ KENA CARAMEŅA ( Remainder by the last digit ) 13. SOPĀNTYADVAYAMANTYAM ( Ultimate and twice the penultimate ) 15. GUŅITASAMUCCAYAH ( The whole product is the same ) 16. GUŅAKA SAMUCCAYAH ( Collectivity of multipliers ) The Sutras have their applications in solving different problems in different contexts. Further they are used along with other Sutras. So it is a bit of inconvenience to deal each Sutra under a separate heading exclusively and also independently. Of course they will be mentioned and also be applied in solving the problems in the forth coming chapter wherever necessary. This decision has been taken because up to now, we have treated each Sutra independently and have not continued with any other Sutra even if it is necessary. When the need for combining Sutras for filling the gaps in the process arises, we may opt for it. Now we shall deal the fourteenth Sutra, the Sutra left so far untouched. |
EKANYŨŅENA PŨRVENA The Sutra Ekanyunena purvena comes as a Sub-sutra to Nikhilam which gives the meaning 'One less than the previous' or 'One less than the one before'.1) The use of this sutra in case of multiplication by 9,99,999.. is as follows . Method : a) The left hand side digit (digits) is ( are) obtained by applying the ekanyunena purvena i.e. by deduction 1 from the left side digit (digits) . e.g. ( i ) 7 x 9; 7 – 1 = 6 ( L.H.S. digit ) b) The right hand side digit is the complement or difference between the multiplier and the left hand side digit (digits) . i.e. 7 X 9 R.H.S is 9 - 6 = 3. c) The two numbers give the answer; i.e. 7 X 9 = 63. Example 1: 8 x 9 Step ( a ) gives 8 – 1 = 7 ( L.H.S. Digit ) Step ( b ) gives 9 – 7 = 2 ( R.H.S. Digit ) Step ( c ) gives the answer 72 Example 2: 15 x 99 Step ( a ) : 15 – 1 = 14 Step ( b ) : 99 – 14 = 85 ( or 100 – 15 ) Step ( c ) : 15 x 99 = 1485 Example 3: 24 x 99 Answer :  Example 4: 356 x 999 Answer :  Example 5: 878 x 9999 Answer :  Note the process : The multiplicand has to be reduced by 1 to obtain the LHS and the rightside is mechanically obtained by the subtraction of the L.H.S from the multiplier which is practically a direct application of Nikhilam Sutra. Now by Nikhilam 24 – 1 = 23 L.H.S. x 99 – 23 = 76 R.H.S. (100–24) _____________________________ 23 / 76 = 2376 Reconsider the Example 4: 356 – 1 = 355 L.H.S. x 999 – 355 = 644 R.H.S. ________________________ 355 / 644 = 355644 and in Example 5: 878 x 9999 we write 0878 – 1 = 877 L.H.S. x 9999 – 877 = 9122 R.H.S. __________________________ 877 / 9122 = 8779122 Algebraic proof : As any two digit number is of the form ( 10x + y ), we proceed ( 10x + y ) x 99 = ( 10x + y ) x ( 100 – 1 ) = 10x . 102 – 10x + 102 .y – y = x . 103 + y . 102 – ( 10x + y ) = x . 103 + ( y – 1 ) . 102 + [ 102 – ( 10x + y )] Thus the answer is a four digit number whose 1000th place is x, 100th place is ( y - 1 ) and the two digit number which makes up the 10th and unit place is the number obtained by subtracting the multiplicand from 100.(or apply nikhilam). Thus in 37 X 99. The 1000th place is x i.e. 3 100th place is ( y - 1 ) i.e. (7 - 1 ) = 6 Number in the last two places 100-37=63. Hence answer is 3663. Apply Ekanyunena purvena to find out the products 1. 64 x 99 2. 723 x 999 3. 3251 x 9999 4. 43 x 999 5. 256 x 9999 6. 1857 x 99999 i) When the multiplicand and multiplier both have the same number of digits ii) When the multiplier has more number of digits than the multiplicand. In both the cases the same rule applies. But what happens when the multiplier has lesser digits? i.e. for problems like 42 X 9, 124 X 9, 26325 X 99 etc., For this let us have a re-look in to the process for proper understanding. Multiplication table of 9. a b 2 x 9 = 1 8 3 x 9 = 2 7 4 x 9 = 3 6 - - - - - - - - - - 8 x 9 = 7 2 9 x 9 = 8 1 10 x 9 = 9 0 Observe the left hand side of the answer is always one less than the multiplicand (here multiplier is 9) as read from Column (a) and the right hand side of the answer is the complement of the left hand side digit from 9 as read from Column (b) Multiplication table when both multiplicand and multiplier are of 2 digits. a b 11 x 99 = 10 89 = (11–1) / 99 – (11–1) = 1089 12 x 99 = 11 88 = (12–1) / 99 – (12–1) = 1188 13 x 99 = 12 87 = (13–1) / 99 – (13–1) = 1287 ------------------------------------------------- 18 x 99 = 17 82 ---------------------------- 19 x 99 = 18 81 20 x 99 = 19 80 = (20–1) / 99 – (20–1) = 1980 The rule mentioned in the case of above table also holds good here Further we can state that the rule applies to all cases, where the multiplicand and the multiplier have the same number of digits. Consider the following Tables. (i) a b 11 x 9 = 9 9 12 x 9 = 10 8 13 x 9 = 11 7 ---------------------- 18 x 9 = 16 2 19 x 9 = 17 1 20 x 9 = 18 0 (ii) 21 x 9 = 18 9 22 x 9 = 19 8 23 x 9 = 20 7 ----------------------- 28 x 9 = 25 2 29 x 9 = 26 1 30 x 9 = 27 0 (iii) 35 x 9 = 31 5 46 x 9 = 41 4 53 x 9 = 47 7 67 x 9 = 60 3 -------------------------so on. From the above tables the following points can be observed: 1) Table (i) has the multiplicands with 1 as first digit except the last one. Here L.H.S of products are uniformly 2 less than the multiplicands. So also with 20 x 9 2) Table (ii) has the same pattern. Here L.H.S of products are uniformly 3 less than the multiplicands. 3) Table (iii) is of mixed example and yet the same result i.e. if 3 is first digit of the multiplicand then L.H.S of product is 4 less than the multiplicand; if 4 is first digit of the multiplicand then, L.H.S of the product is 5 less than the multiplicand and so on. 4) The right hand side of the product in all the tables and cases is obtained by subtracting the R.H.S. part of the multiplicand by Nikhilam. Keeping these points in view we solve the problems: Example1 : 42 X 9 i) Divide the multiplicand (42) of by a Vertical line or by the Sign : into a right hand portion consisting of as many digits as the multiplier. i.e. 42 has to be written as 4/2 or 4:2 ii) Subtract from the multiplicand one more than the whole excess portion on the left. i.e. left portion of multiplicand is 4. one more than it 4 + 1 = 5. We have to subtract this from multiplicand i.e. write it as 4 : 2 :-5 --------------- 3 : 7 This gives the L.H.S part of the product. This step can be interpreted as "take the ekanyunena and sub tract from the previous" i.e. the excess portion on the left. iii) Subtract the R.H.S. part of the multiplicand by nikhilam process. i.e. R.H.S of multiplicand is 2 its nikhilam is 8 It gives the R.H.S of the product i.e. answer is 3 : 7 : 8 = 378. Thus 42 X 9 can be represented as 4 : 2 :-5 : 8 ------------------ 3 : 7 : 8 = 378. Example 2 : 124 X 9 Here Multiplier has one digit only . We write 12 : 4 Now step (ii), 12 + 1 = 13 i.e. 12 : 4 -1 : 3 ------------ Step ( iii ) R.H.S. of multiplicand is 4. Its Nikhilam is 6 124 x 9 is 12 : 4 -1 : 3 : 6 ----------------- 11 : 1 : 6 = 1116 The process can also be represented as 124 x 9 = [ 124 – ( 12 + 1 ) ] : ( 10 – 4 ) = ( 124 – 13 ) : 6 = 1116 Example 3: 15639 x 99 Since the multiplier has 2 digits, the answer is [15639 – (156 + 1)] : (100 – 39) = (15639 – 157) : 61 = 1548261 1. 58 x 9 2. 62 x 9 3. 427 x 99 4. 832 x 9 5. 24821 x 999 6. 111011 x 99 Thus we have a glimpse of majority of the Sutras. At some places some Sutras are mentioned as Sub-Sutras. Any how we now proceed into the use of Sub-Sutras. As already mentioned the book on Vedic Mathematics enlisted 13 Upa-Sutras. But some approaches in the Vedic Mathematics book prompted some serious research workers in this field to mention some other Upa-Sutras. We can observe those approaches and developments also. |
ĀNURŨPYENA The upa-Sutra 'anurupyena' means 'proportionality'. This Sutra is highly useful to find products of two numbers when both of them are near the Common bases i.e powers of base 10 . It is very clear that in such cases the expected 'Simplicity ' in doing problems is absent.Example 1: 46 X 43 As per the previous methods, if we select 100 as base we get 46 -54 This is much more difficult and of no use. 43 -57 ¯¯¯¯¯¯¯¯ Now by ‘anurupyena’ we consider a working base In three ways. We can solve the problem. Method 1: Take the nearest higher multiple of 10. In this case it is 50. Treat it as 100 / 2 = 50. Now the steps are as follows: i) Choose the working base near to the numbers under consideration. i.e., working base is 100 / 2 = 50 ii) Write the numbers one below the other i.e. 4 6 4 3 ¯¯¯¯¯¯¯ iii) Write the differences of the two numbers respectively from 50 against each number on right side i.e. 46 -04 43 -07 ¯¯¯¯¯¯¯¯¯ iv) Write cross-subtraction or cross- addition as the case may be under the line drawn.  v) Multiply the differences and write the product in the left side of the answer. Thus 39 ÷ 2 gives 19½ where 19 is quotient and 1 is remainder . This 1 as Reminder gives one 50 making the L.H.S of the answer 28 + 50 = 78(or Remainder ½ x 100 + 28 )46 -04 43 -07 ____________ 39 / -4 x –7 = 28 vi) Since base is 100 / 2 = 50 , 39 in the answer represents 39X50. Hence divide 39 by 2 because 50 = 100 / 2 i.e. R.H.S 19 and L.H.S 78 together give the answer 1978 We represent it as 46 -04 43 -07 ¯¯¯¯¯¯¯¯¯ 2) 39 / 28 ¯¯¯¯¯¯¯¯¯ 19½ / 28 = 19 / 78 = 1978 Example 2: 42 X 48. With 100 / 2 = 50 as working base, the problem is as follows: 42 -08 48 -02 ¯¯¯¯¯¯¯¯¯ 2) 40 / 16 ¯¯¯¯¯¯¯¯¯ 20 / 16 42 x 48 = 2016 Method 2: For the example now  (195 + 2) / 8 = 1978 [Since we operate with 10, the R.H.S portion shall have only unit place .Hence out of the product 28, 2 is carried over to left side. The L.H.S portion of the answer shall be multiplied by 5, since we have taken 50 = 5 X 10.] Now in the example 2: 42 x 48 we can carry as follows by treating 50 = 5 x 10  Method 3: We take the nearest lower multiple of 10 since the numbers are 46 and 43 as in the first example, We consider 40 as working base and treat it as 4 X 10.  Since 10 is in operation 1 is carried out digit in 18. Since 4 X 10 is working base we consider 49 X 4 on L.H.S of answer i.e. 196 and 1 carried over the left side, giving L.H.S. of answer as 1978. Hence the answer is 1978. We proceed in the same method for 42 X 48  Let us see the all the three methods for a problem at a glance Example Method - 1: Working base = 100 / 5 = 20 24 04 23 03 ¯¯¯¯¯¯¯¯ 5) 27 / 12 ¯¯¯¯¯¯¯¯ 5 2/5 / 12 = 5 / 52 = 552 [Since 2 / 5 of 100 is 2 / 5 x 100 = 40 and 40 + 12 = 52] Method - 2: Working base 2 X 10 = 20  Now as 20 itself is nearest lower multiple of 10 for the problem under consideration, the case of method – 3 shall not arise. Let us take another example and try all the three methods. Example 4: 492 X 404 Method - 1 : working base = 1000 / 2 = 500 492 -008 404 -096 ¯¯¯¯¯¯¯¯¯¯¯ 2) 396 / 768 since 1000 is in operation ¯¯¯¯¯¯¯¯¯¯¯ 198 / 768 = 198768 Method 2: working base = 5 x 100 = 500  Method - 3. Since 400 can also be taken as working base, treat 400 = 4 X 100 as working base. Thus  No need to repeat that practice in these methods finally takes us to work out all these mentally and getting the answers straight away in a single line. Example 5: 3998 X 4998 Working base = 10000 / 2 = 5000 3998 -1002 4998 -0002 ¯¯¯¯¯¯¯¯¯¯¯¯ 2) 3996 / 2004 since 10,000 is in operation 1998 / 2004 = 19982004 or taking working base = 5 x 1000 = 5,000 and  What happens if we take 4000 i.e. 4 X 1000 as working base? _____ 3998 0002 4998 0998 Since 1000 is an operation ¯¯¯¯¯¯¯¯¯¯¯¯ 4996 / 1996 ___ ___ As 1000 is in operation, 1996 has to be written as 1996 and 4000 as base, the L.H.S portion 5000 has to be multiplied by 4. i. e. the answer is  Example 6: 58 x 48 Working base 50 = 5 x 10 gives  Since 10 is in operation. Find the following product. 1. 46 x 46 2. 57 x 57 3. 54 x 45 4. 18 x 18 5. 62 x 48 6. 229 x 230 7. 47 x 96 8. 87965 x 99996 9. 49x499 10. 389 x 512 |