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- 1. VEDIC MATHEMATICS A quick way to square numbers that end in 5752 = 5625752 means 75 x 75.The answer is in two parts: 56 and 25.The last part is always 25.The first part is the first number, 7, multiplied by the number "one more", which is8:so 7 x 8 = 56Method for multiplying numbers where the first figures are the same and thelast figures add up to 10. 32 x 38 = 1216 Both numbers here start with 3 and the last figures (2 and 8) add up to 10. So we just multiply 3 by 4 (the next number up) to get 12 for the first part of the answer. And we multiply the last figures: 2 x 8 = 16 to get the last part of the answer. Diagrammatically:Find square of a number between 50 and 60for example......562 = 3136572 = 3249582 = 3364
- 2. there is a 2 steps trick to get the ans.1) add the digit at the units place to 25 and write the sum2) then calculate the square of units place digit and write iteg in case of 562we have 25+6=31and square of 6 is 36hence the result is 3136Square of numbers in 100s1. let the number be 108:2. Square the last two digits (no carry): 8 × 8 = 64: _ _ _ 643. Add the last two digits to the number: 108 + 08= 116:so 1 1 6 _ _4. So 108 × 108 = 11664.FIND SQUARE OF NUMBERS IN 200 TO 299Steps to find square of numbers in 200s1. let the number be 207:2. The first digit is 4so 4 _ _ _ _3. The next two digits are 4 times the last digit:4 × 7 = 28so _ 2 8 _ _Square the last digit: 7× 7 = 49so _ _ _ 49So finally we get 207 × 207 = 42849.Example1. If the number to be squared is 225:2. Square last two digits (keep carry):25x25 = 625 (keep 6): _ _ _ 2 53. 4 times the last two digits + carry:4x25 = 100; 100+6 = 106 (keep 1): _ 0 6 _ _4. Square the first digit + carry:2x2 = 4; 4+1 = 5: 5 _ _ _ _5. So 225 × 225 = 50625.Ekadhikena Purvena or "By one more than the previous one"
- 3. 1. Square of numbers ending in 565 x 65 = (6 x (6+1) ) 25 = (6x7) 25 = 422545 x 45 = (4 x (4+1) ) 25 = (4x5) 25 = 2025105 x 105 = (10 x (10+1) 25 = (10 x 11) 25 = 11025** if the number is greater than 10 take the surplussquare of 12 is (12+2) (2x2) = 144square of 13 is (13+3) (3x3) = 169Finding Square of an adjacent number: One upYou know the squares of 30, 40, 50, 60 etc. but if you are required to calculate square of 31 or say 61 then youwill scribble on paper and try to answer the question. Can it be done mentally? Some of you will say may beand some of you will say may not be. But if I give you a formula then all of you will say, yes! it can be. What isthat formula…..The formula is simple and the application is simpler.Say you know 602 = 3600Then 612 will be given by the following612 = 602 + (60 + 61) = 3600 + 121 = 3721or Say you know 252 = 625 then262 = 625 + (25 + 26) = 676Likewise, you can find out square of a number that is one less than the number whose square is known.Let me show it by taking an example:Say you know 602 = 3600Then 592 will be given by the following592 = 602 - (60 + 59) = 3600 - 119 = 3481or Say you know 252 = 625 then242 = 625 - (25 + 24) = 576Apply it to find square of a digit, which is one, less than the square of known digit. This works very well forthe complete range of numbers.Comparison Between Vedic and Conventional System ( indu thakur)
- 4. The sutra "vertically and crosswise" is often used in long multiplication. Suppose we wish tomultiply32 by 44. We multiply vertically 2x4=8.Then we multiply crosswise and add the two results: 3x4+4x2=20, so put down 0 and carry 2.Finally we multiply vertically 3x4=12 and add the carried 2 =14. Result: 1,408.for example, 96 by 92. 96 is 4 below the base and 92 is 8 below.We can cross-subtract either way: 96-8=88 or 92-4=88. This is the first part of the answer andmultiplying the "differences" vertically 4x8=32 gives the second part of the answer.The above problem has been done using Criss-cross technique of Vedic Mathematics.Use the formula ALL FROM 9 AND THE LAST FROM 10 to perform instantsubtractions.For example 1000 - 357 = 643We simply take each figure in 357 from 9 and the last figure from 10.So the answer is 1000 - 357 = 643And thats all there is to it!This always works for subtractions from numbers consisting of a 1 followed by noughts: 100;1000; 10,000 etc.Similarly 10,000 - 1049 = 8951
- 5. For 1000 - 83, in which we have more zeros than figures in the numbers being subtracted, wesimply suppose 83 is 083.So 1000 - 83 becomes 1000 - 083 = 917The easy way to add and subtract fractions.Use VERTICALLY AND CROSSWISE to write the answer straight down!Multiply crosswise and add to get the top of the answer:2 x 5 = 10 and 1 x 3 = 3. Then 10 + 3 = 13.The bottom of the fraction is just 3 x 5 = 15.You multiply the bottom number together.Subtracting is just as easy: multiply crosswise as before, but the subtract:1 divided by 19, 29, 39,..............Consider 1/19 since 19 is not divisible by 2 or 5 it is a purely a recurring decimaltake last digit 1multiply this with 1+1 (one more) i.e 2 (this is the key digit) ==>21multiply 2 by 2 ==> 421 multiplying 4 by 2 ==> 8421multiply 8 by 2 ==> 68421 carry 1multiply 6 by 2 =12 + carry 1= 13 ==> 368421 carry 1continuing (till 18 digits =denominator-numerator)the result is 0.052631578947368421 1/19 using divisionsdivide 1 by 2, answer is 0 with remainder 1 ==> .0next 10 divided by 2 is 5 ==> .05next 5 divided by 2 is 2 remainder 1 ==> 0.052next 12 (remainder 2) divided by 2 is 6 ==> 0.0526next 6 divided by 2 is 3 ==> 0.05263next 3 divided by 2 is 1 remainder 1 ==> 0.052631next 11 divided by 2 is 5 remainder 1 ==> 0.0526315and so on...
- 6. 1/7 = 7/49 previous digit is 4 so multiply by 4+1 i.e. by 57-> 57 -> 857 -> 42857 -> 0.142857 (stop after 7-1= 6 digits)Method for diving by 9. 23 / 9 = 2 remainder 5 The first figure of 23 is 2, and this is the answer. The remainder is just 2 and 3 added up! 43 / 9 = 4 remainder 7 The first figure 4 is the answer and 4 + 3 = 7 is the remainder - could it be easier? 134 / 9 = 14 remainder 8 The answer consists of 1,4 and 8. 1 is just the first figure of 134. 4 is the total of the first two figures 1+ 3 = 4, and 8 is the total of all three figures 1+ 3 + 4 = 8.Vulgar fractions whose denominators are numbers ending in NINE :We now take examples of 1 / a9, where a = 1, 2, -----, 9. In the conversion of suchvulgar fractions into recurring decimals, Ekadhika process can be effectively usedboth in division and multiplication.a) Division Method : Value of 1 / 19.The numbers of decimal places before repetition is the difference of numerator anddenominator, i.e.,, 19 -1=18 places.For the denominator 19, the purva (previous) is 1.Step. 1 : Divide numerator 1 by 20. i.e.,, 1 / 20 = 0.1 / 2 = .10 ( 0 times, 1 remainder)Step. 2 : Divide 10 by 2
- 7. i.e.,, 0.005( 5 times, 0 remainder )Step. 3 : Divide 5 by 2 i.e.,, 0.0512 ( 2 times, 1 remainder )Step. 4 : Divide 12 i.e.,, 12 by 2 i.e.,, 0.0526 ( 6 times, No remainder )Step. 5 : Divide 6 by 2 i.e.,, 0.05263 ( 3 times, No remainder )Step. 6 : Divide 3 by 2 i.e.,, 0.0526311(1 time, 1 remainder )Step. 7 : Divide 11 i.e.,, 11 by 2 i.e.,, 0.05263115 (5 times, 1 remainder )Step. 8 : Divide 15 i.e.,, 15 by 2 i.e.,, 0.052631517 ( 7 times, 1 remainder )Step. 9 : Divide 17 i.e.,, 17 by 2 i.e.,, 0.05263157 18 (8 times, 1 remainder )Step. 10 : Divide 18 i.e.,, 18 by 2 i.e.,, 0.0526315789 (9 times, No remainder )Step. 11 : Divide 9 by 2 i.e.,, 0.0526315789 14 (4 times, 1 remainder )Step. 12 : Divide 14 i.e.,, 14 by 2 i.e.,, 0.052631578947 ( 7 times, No remainder )Step. 13 : Divide 7 by 2 i.e.,, 0.05263157894713 ( 3 times, 1 remainder )
- 8. Step. 14 : Divide 13 i.e.,, 13 by 2 i.e.,, 0.052631578947316 ( 6 times, 1 remainder )Step. 15 : Divide 16 i.e.,, 16 by 2 i.e.,, 0.052631578947368 (8 times, No remainder )Step. 16 : Divide 8 by 2 i.e.,, 0.0526315789473684 ( 4 times, No remainder )Step. 17 : Divide 4 by 2 i.e.,, 0.05263157894736842 ( 2 times, No remainder )Step. 18 : Divide 2 by 2 i.e.,, 0.05263157 8947368421 ( 1 time, No remainder )Now from step 19, i.e.,, dividing 1 by 2, Step 2 to Step. 18 repeats thus giving 0 __________________ . . 1 / 19 = 0.052631578947368421 or 0.052631578947368421Note that we have completed the process of division only by using ‘2’. Nowherethe division by 19 occurs.b) Multiplication Method: Value of 1 / 19First we recognize the last digit of the denominator of the type 1 / a9. Here the lastdigit is 9.For a fraction of the form in whose denominator 9 is the last digit, we take the caseof 1 / 19 as follows:For 1 / 19, previous of 19 is 1. And one more than of it is 1 + 1 = 2.Therefore 2 is the multiplier for the conversion. We write the last digit in thenumerator as 1 and follow the steps leftwards.Step. 1 : 1Step. 2 : 21(multiply 1 by 2, put to left)Step. 3 : 421(multiply 2 by 2, put to left)
- 9. Step. 4 : 8421(multiply 4 by 2, put to left)Step. 5 : 168421 (multiply 8 by 2 =16, 1 carried over, 6 put to left)Step. 6 : 1368421 ( 6 X 2 =12,+1 [carry over] = 13, 1 carried over, 3 put to left )Step. 7 : 7368421 ( 3 X 2, = 6 +1 [Carryover] = 7, put to left)Step. 8 : 147368421 (as in the same process)Step. 9 : 947368421 ( Do – continue to step 18)Step. 10 : 18947368421Step. 11 : 178947368421Step. 12 : 1578947368421Step. 13 : 11578947368421Step. 14 : 31578947368421Step. 15 : 631578947368421Step. 16 : 12631578947368421Step. 17 : 52631578947368421Step. 18 : 1052631578947368421Now from step 18 onwards the same numbers and order towards left continue.Thus 1 / 19 = 0.052631578947368421It is interesting to note that we have i) not at all used division process ii) instead of dividing 1 by 19 continuously, just multiplied 1 by 2 and continuedto multiply the resultant successively by 2.Observations :
- 10. a) For any fraction of the form 1 / a9 i.e.,, in whose denominator 9 is the digit inthe units place and a is the set of remaining digits, the value of the fraction is inrecurring decimal form and the repeating block’s right most digit is 1. b) Whatever may be a9, and the numerator, it is enough to follow the saidprocess with (a+1) either in division or in multiplication. c) Starting from right most digit and counting from the right, we see ( in thegiven example 1 / 19) Sum of 1st digit + 10th digit = 1 + 8 = 9 Sum of 2nd digit + 11th digit = 2 + 7 = 9 - - - - - - - - -- - - - - - - - - - - - - - - - - - - Sum of 9th digit + 18th digit = 9+ 0 = 9 From the above observations, we conclude that if we find first 9 digits, further digits can be derived as complements of 9.(i) Consider the division by divisors of more than one digit, and when the divisors are slightlygreater 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 ordigits using negative (-) sign and place them below the divisor as shown. 12 -2 ¯¯¯¯Step 2 : Write down the dividend to the right. Set apart the last digit for the remainder. i.e.,, 12 122 5 -2Step 3 : Write the 1st digit below the horizontal line drawn under the dividend. Multiply thedigit 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
- 11. 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 5Step 5 : Continue the process to the last digit. i.e., 12 122 5 -2 -20 -4 ¯¯¯¯¯ ¯¯¯¯¯¯¯¯¯¯ 102 1Step 6: The sum of the last digit is the Remainder and the result to its left is Quotient.Thus Q = 102 and R = 1Example 2 : Divide 1697 by 14. 14 169 7 -4 -4–8–4 ¯¯¯¯ ¯¯¯¯¯¯¯ 121 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 dividendfor the remainder. 123 25 98 Step ( 1 ) & Step ( 2 ) -2-3 ¯¯¯¯¯ ¯¯¯¯¯¯¯¯ Now proceed the sequence of steps write –2 and –3 as follows : 123 25 98 -2-3 -4 -6 ¯¯¯¯¯ -2–3 ¯¯¯¯¯¯¯¯¯¯ 21 15 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 thedividend are to be set up for Remainder.
- 12. 11213 23 9 479 -1-2-1-3 -2 -4-2-6 with 2 ¯¯¯¯¯¯¯¯ -1-2-1-3 with 1 ¯¯¯¯¯¯¯¯¯¯¯¯¯ 21 4006 Hence Q = 21, R = 4006.Example 5 : Divide 13456 by 1123 112 3 134 5 6 -1–2–3 -1-2-3 ¯¯¯¯¯¯¯ -2-4 –6 ¯¯¯¯¯¯¯¯¯¯¯¯¯ 1 2 0–2 0Note that the remainder portion contains –20, i.e.,, a negative quantity. To over come thissituation, take 1 over from the quotient column, i.e.,, 1123 over to the right side, subtract theremainder portion 20 to get the actual remainder.Thus Q = 12 – 1 = 11, and R = 1123 - 20 = 1103.Sutra : Yaavadunam Taavaduunikruthya vargam cha yogayetMeaning : "Whatever the extent of its deficiency, lessen it further to that very extent;and also set up the square of that deficiency".This sutra is a corollary of the Nikhilam sutra.1. Consider a simple example 92Step 1 : Consider the nearest base (here 10).Step 2 : As 9 has a deficiency of 1 (10 - 9 = 1), we should decrease it further by 1, and setdown our LHS of the Answer as 8.Step 3 : On the RHS put the square of the deficiency (here 1).we get 92 = 81.2. consider 1021) Base is 1002) Deficiency is -2 (100 - 102 = -2) Therefore we subtract -2 from 102
- 13. 102 - (-2) = 104This is our RHS3) Our LHS now becomes (-2)2 which is 4Since the base is 100 we write it as 04, so that we get 1022 = 10404If we have multiples or sub multiples of a base, we employ the same technique as inAanurupyena. (See Nikhilam Multiplication)3. Consider 2821) Let 20 be the Working Base and 10 as the Main Base.Therefore x = (Main Base)/(Working Base) = 10/20 = 1/22) Here the deficiency = 20 - 28 = -8Therefore RHS = 28 - (-8) = 36Divide by x i.e. by (1/2).We get 36/(1/2) = 72. This is the required RHS.3) LHS = (-8)2 = 64Since Main Base is 10, we put only 4 on the LHS and carry over 6 to the RHSTherefore we get282 = 72+6 | 4 == 784 Compute: 8 x 7 8 is 2 below 10 and 7 is 3 below 10.You subtract crosswise 8-3 or 7 - 2 to get 5,the first figure of the answer.And you multiply vertically: 2 x 3 to get 6,the last figure of the answerThe answer is 56.Multiply 88 by 98Both 88 and 98 are close to 100.88 is 12 below 100 and 98 is 2 below 100As before the 86 comes from
- 14. subtracting crosswise: 88 - 2 = 86(or 98 - 12 = 86: you can subtracteither way, you will always getthe same answer).And the 24 in the answer isjust 12 x 2: you multiply vertically.So 88 x 98 = 8624Multiply 103 x 104 = 10712The answer is in two parts: 107 and 12,107 is just 103 + 4 (or 104 + 3),and 12 is just 3 x 4.The Sutra Ekanyunena purvena comes as a Sub-sutra to Nikhilam which givesthe 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. bydeduction 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 lefthand 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 72Example 2: 15 x 99 Step ( a ) : 15 – 1 = 14 Step ( b ) : 99 – 14 = 85 ( or 100 – 15 ) Step ( c ) : 15 x 99 = 1485Example 3: 24 x 99 Answer :Example 4: 356 x 999 (indu thakur)
- 15. Answer :Example 5: 878 x 9999 Answer :Example (i) : Find the square of 195.The Conventional method : 1952 = 195 x 195 ______ 975 1755 195 _______ 38025 ¯¯¯¯¯¯¯(ii) By Ekadhikena purvena, since the number ends up in 5 we write the answer split up intotwo parts.The right side part is 52 where as the left side part 19 X (19+1) (Ekhadhikena)Thus 1952 = 19 X 20/52 = 380/25 = 38025(iii) By Nikhilam Navatascaramam Dasatah; as the number is far from base 100, we combinethe sutra with the upa-sutra ‘anurupyena’ and proceed by taking working base 200.a) Working Base = 200 = 2 X 100. Now 1952 = 195 X 195 (indu thakur)
- 16. iv) By the sutras "yavadunam tavadunikritya vargamca yojayet" and "anurupyena" 1952, base 200 treated as 2 X 100 deficit is 5.Example 2 : 98 X 92i) ‘Nikhilam’ sutra 98 -2 x 92 -8 ______________ 90 / 16 = 9016ii) by vinculum method _ 98 = 100 – 2 = 102 _ 92 = 100 – 8 = 108 now _ 102 _ 108 ______ _ 10006 _ 1 1 _______ __ 11016 = 9016Example 3: 493 X 497. a) Working base is 500, treated as 5 X 100 (indu thakur)
- 17. b) Working base is 500, treated as 1000 / 2 493 -7 497 -3 _________ 2) 490 / 021 _________ 245 / 021 = 2450212) Since end digits sum is 3+7 = 10 and remaining part 49 is same in both the numbers,‘antyayordasakepi’ is applicable. Further Ekadhikena Sutra is also applicable.Thus 493 x 497 = 49 x 50 / 3x7 = 2450 / 21 = 245021Example 4: 99 X 991) Now by urdhva - tiryak sutra. 99 X 99 _______ 8121 168 _______ 98012) By vinculum method _ 99 = 100 - 1 = 101 Now 99 X 99 is _ 101 _ x 101 ______ _ 10201 = 98013) By Nikhilam method (indu thakur)
- 18. 99 -1 99 -1 _________ 98 / 01 = 9801.4) ‘Yadunam’ sutra : 992 Base = 100Deficiency is 1 : It indicates 992 = (99 – 1) / 12 = 98 / 01 = 9801.SIMPLE TRICKS TO MULTIPLYMultiply by 5: Multiply by 10 and divide by 2.Multiply by 6: Sometimes multiplying by 3 and then 2 is easy.Multiply by 9: Multiply by 10 and subtract the original number.Multiply by 12: Multiply by 10 and add twice the original number.Multiply by 13: Multiply by 3 and add 10 times original number.Multiply by 14: Multiply by 7 and then multiply by 2Multiply by 15: Multiply by 10 and add 5 times the original number, as above.Multiply by 16: You can double four times, if you want to. Or you can multiply by 8 and then by2.Multiply by 17: Multiply by 7 and add 10 times original number.Multiply by 18: Multiply by 20 and subtract twice the original number (which is obvious fromthe first step).Multiply by 19: Multiply by 20 and subtract the original number.Multiply by 24: Multiply by 8 and then multiply by 3.Multiply by 27: Multiply by 30 and subtract 3 times the original number (which is obvious fromthe first step).Multiply by 45: Multiply by 50 and subtract 5 times the original number (which is obvious fromthe first step).Multiply by 90: Multiply by 9 (as above) and put a zero on the right.Multiply by 98: Multiply by 100 and subtract twice the original number.Multiply by 99: Multiply by 100 and subtract the original number.
- 19. Simple trick to remember table of 19LET THE NUMBER BE ABCSSimple trick to remember table of 19The first digit is increamenting by 2 and second is decrementing by 119 * 01 = 019 19 * 02 = 038 19 * 03 = 057 19 * 04 = 07619 * 05 = 095 19 * 06 = 114 19 * 07 = 133 19 * 08 = 15219 * 09 = 171 19 * 10 = 190tTrick to remember table of 29for 29 units digit decreases by 1 and tens digit increases by 3.29 * 1 = 029 29 * 2 = 058 29 * 3 = 087 29 * 4 = 116 29 * 5 = 14 29 * 6 = 17429 * 7 = 203 29 * 8 = 232 29 * 9 = 26 29*10=290eShunyam Saamyasamuccaye or "When the samuccaya is thesame, that samuccaya is zero"This sutra is useful in solution of several special types of equations that can besolved visually. The word samuccaya has various meanings in differentapplicatins.1: It is a term which occurs as a common factor in all the terms concernedThus 12x + 3x = 4x + 5x x is common, hence x = 0Or 9 (x+1) = 7 (x+1) here (x+1) is common; hence x +1= 02: Here Samuccaya means "the product of the independent terms"Thus, (x +7) (x +9) = (x +3) (x +21)Here 7 x9 = 3 x 21. Therefore x = 03: Samuccaya thirdly means the sum of the Denominators of two fractions havingthe same numerical numeratorThus, 1/(2x –1) + 1/(3x –1) = 0 Hence 5x – 2 =0 or x = 2/54: Here Samuccaya means combination (or TOTAL).If the sum of the Numerators and the sum of the Denominators be the same, thenthat sum = 0(2x +9)/ (2x +7) = (2x +7)/ (2x +9)N1 + N2 = D1 + D2 = 2x + 9 + 2x + 7 = 0
- 20. Hence 4x + 16 = 0 hence x = -4Note: If there is a numerical factor in the algebraic sum, that factor should beremoved.(3x +4)/ (6x +7) = (x +1)/ (2x +3)Here N1 +N2 = 4x +5; D1 +D2 = 8x + 10; 4x +5 =0 x= -5/45: Here Samuccaya means TOTAL ie Addition & subtractionThus, (3x +4)/ (6x +7) = (5x +6)/ (2x +3)Here N1+N2 = D1 + D2 = 8x + 10 =0 hence x = - 5/4D1 – D2 = N2 – N1 = 2x + 2 = 0 x = -16: Here Samuccaya means TOTAL; used in Harder equationsThus, 1/ (x-7) + 1/(x-9) = 1/(x-6) + 1/(x-10)Vedic Sutra says, (other elements being equal), the sum-total of the denominatorson LHS and the total on the RHS are the same, then the total is zero.Here, D1 + D2 = D3 + D4 = 2x-16 =0 hence x = 8Examples 1/(x+7) + 1/(x+9) = 1/(x+6) + 1/(x+10) x = - 81/(x-7) + 1(x+9) = 1/(x+11) + 1/(x-9) x = - 11/(x-8) + 1/(x-9) = 1/(x-5) + 1/(x-12) x = 8-1/21/(x-b) - 1/(x-b-d) = 1/(x-c+d) - 1/(x-c) x = 1/2(b+c)Special Types of seeming Cubics (x- 3)3 + (x –9)3 = 2(x –6)3current method is very lengthy, but Vedic method says, (x-3) + (x-9) = 2x – 12Hence x = 6(x-149)3 + (x-51)3 = 2(x-100)3 Hence 2x-200 =0 & x = 100(x+a+b-c)3 + (x+b+c-a)3 = 2(x+b)3 x = -b(Anurupye) Shunyamanyat or "If one is in ratio, the other one is zero"This sutra is often used to solve simultaneous simple equations which
- 21. may involve big numbers. But these equations in special cases can bevisually solved because of a certain ratio between the coefficients.Consider the following example:6x + 7y = 819x + 14y = 16Here the ratio of coefficients of y is same as that of the constant terms.Therefore, the "other" is zero, i.e., x = 0. Hence the solution of theequations is x = 0 and y = 8/7.This sutra is easily applicable to more general cases with any numberof variables. For instanceax + by + cz = abx + cy + az = bcx + ay + bz = cwhich yields x = 1, y = 0, z = 0.A corollary (upsutra) of this sutra says Sankalana-Vyavakalanaabhyam or By addition and bysubtraction. It is applicablein case of simultaneous linear equations where the x- and y-coefficients are interchanged. For instance:45x - 23y = 11323x - 45y = 91By addition: 68x - 68 y = 204 => 68(x-y) = 204 => x - y = 3By subtraction: 22x + 22y = 22 => 22(x+y) = 22 => x + y = 1Yaavadunam-"Whatever the extent of its deficiency" 1. Compute 133 Step 1 : Consider nearest base (here 10). Step 2 : As 13 has a excess of 3 (13 -10 = 3), we double the excess and add the original number (13) to it, and put it onthe LHS. Therefore we get 13 + 6 = 19 Step 3 : Now find the new excess. In thiscase it is 19-10 = 9. Now multiply this with the original excess to get the middlepart of the answer. Therefore we get 9 * 3 = 27 Step 4 : Now cube the originalexcess and put it as the last part Carry over any big numbers and total to get the answer. 19 7 7 2 2 21 9 7 Therefore 133 = 2197
- 22. 2. 473 As in Nikhilam and Squaring, we use Aanurupyena here. 1) Let the main base be 10 and the working base be 50 therefore the ratio x = (Main Base)/(Working Base) = 10/50 = 1/5 2) Excess is -3 (47 - 50 = -3). Double the excess and add the original number(here 47) to it. We get 47 - 6 = 41. The Base correction for this part is achieved by dividing by x2 . therefore we get 41/(1/25) = 41 * 25 = 1025 3) Excess in the new uncorrected number (41 - 50 = -9) is multiplied by theoriginal excess(-3) to obtain the second part. Therefore we get -9 * -3 = 27 The Base correction for this part is achieved by dividing by x . therefore we get 27 * 5 = 135 4) The third part is obtained by cubing the excess. (-3)3 = -27 5) Carry over the extra numbers and total to obtain the final answer 1025 0 0 13 5 0 -2 7 1038 2 3 Therefore the final answer is 103823
- 23. Vyashtisamanstih- "Part and Whole"Corollary : LopanasthapanabhyamIt is very difficult to factorise the long quadratic (2x2 + 6y2 + 3z2 + 7xy + 11yz + 7zx)But "Lopana-Sthapana" removes the difficulty. Eliminate z by putting z = 0.Hence the given expression E = 2x2 + 6y2 + 7xy = (x+2y) (2x+3y)Similarly, if y=0, then E = 2x2 + 3z2 + 7zx = (x+3z) (2x+z) Hence E = (x+2y+3z) (2x+3y+z) Factorise 2x2 + 2y2 + 5xy + 2x- 5y –12 = (x+3) (2x-4) and (2y+3) (y- 4) Hence, E = (x+2y+3) (2x+y-4) Sopaantyadvayamantyam- "The ultimate and twice the penultimate"Corollary : Gunitasamuccayah Samuccayagunitah -"The product of the sum ofthe coefficients in the factors is equal to the sum of the coefficients in the product"Sc of the product = Product of the Sc in the factorsFor example (x+7) (x+9) = (x2 + 16x + 63)(1+7) (1+9) = (1 + 16 + 63) = 80or (x+1) (x+2) (x+3) = (x3 + 6X2 + 11x + 6)(1+1) (1+2) (1+3) = (1 + 6 + 11 + 6) = 24 Ekanyunena Purvena-"By one less than the previous one"777 multiplied by 999 = 776,223(776 is one less than multiplicand 777 and 223 is the compliment of 776 from 9)120 35 79 multiplied by 999 99 99 = 120 35 78, 879 64 21 1234 5678 09 multiplied 9999 9999 99 = 1234 5678 08 8765 4321 91
- 24. The Ekanyunena sutra can be used to derive the following results: Kevalaih Saptakam Gunyaat, or in the case of seven the multiplicand should be 143 Kalau Kshudasasaih, or in the case of 13 the multiplicand should be 077 Kamse Kshaamadaaha-khalairmalaih, or in the case of 17 the multiplicand should be 05882353 (by the way, the literal meaning of this result is "In king Kamsas reign famine, and unhygenic conditions prevailed." -- not immediately obvious what it had to do with Mathematics. These multiple meanings of these sutras were one of the reasons why some of the early translations of Vedas missed discourses on vedaangas.)These are used to correctly identify first half of a recurring decimal number, and thenapplying Ekanyuna to arrive at the complete answer mechanically. Consider for examplethe following visual computations:1/7 = 143x999/999999 = 142857/999999 = 0.1428571/13 = 077x999/999999 = 076923/999999 = 0.0769231/17 = 05882353x99999999/9999999999999999 = 0.05882352 94117647Note that7x142857 = 99999913x076923 = 99999917x05882352 94117647 = 9999999999999999which says that if the last digit of the denominator is 7 or 3 then the last digit of theequivalent decimal fraction is 7 or 3 respectively.digit decreases by 1 and tens digit increases by 329 * 1 = 02929 * 2 = 05829 * 3 = 08729 * 4 = 11629 * 5 = 14529 * 6 = 17429 * 7 = 20329 * 8 = 23229 * 9 = 261
- 25. 29 *10 = 290p 4.> Choose a number over 100 (START WITH SMALLER NUMBER).> The last two places will be the square ofthe last two digits (keep if any carry

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