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Mathematical Calculations in the Mind

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Ranvir Rahul Deshmukh; Applied Sciences; Fall 2020

Source: Wikimedia Commons

Introduction 

Speeding techniques involve calculating large numbers by mental math using Vedic mathematics. Vedic mathematics is based on various techniques that help enhance the speed with which mental calculations can be done. In this paper, two such Vedic mathematics techniques will be explored with algebraic proofs, and their relevance to the real world will be assessed. The formulas I will explore in this paper are: Ekadhikena Purvena, Urdhva – tiryagbhyam, Yavadunam Tavadunikrtya Varganca Yojayet, and Antyayor Dasakepi {these are their names in the Hindu Sanskrit Language}.

The Sanskrit word Veda comes from the root Vid, meaning to know without limit. The word Veda covers all Veda-sakhas known to mankind. The Veda-sakha is a vault of all information, holding deep knowledge about ancient Indian calculating skills. Vedic Mathematics is an ancient system of calculations which was “rediscovered” from the Vedas between 1911 and 1918 by Sri Bharati Krishna Tirthaji Maharaj. Tirthaji stated that all of Vedic arithmetic is based on 16 Sutras (aphorisms) and 13 sub sutra (Corollaries). These formulas are supposed to depict the manner in which the brain normally works, and are consequently expected to be an extraordinary assistance in guiding the understudy to the suitable strategy for mind calculation. However, none of these sutras has ever been found in Vedic writing, nor are its techniques consistent with known scientific information from the Vedic period. Thus, many believe that they were invented by Tirthaji himself

Vedic mathematics encourages students to apply the sutras to precise issues including calculations with many digits. Quick mental calculation strategies alongside brisk, cross-checking systems converts a repetitive subject into a less tedious one. The speeding technique is innovative, and more efficient in calculations . This method ensures both speed and precision, and is dependent on balanced and coherent thinking, as well as utilization of the sutras to prompt improvement in the computational aptitudes of students in a wide region of mathematical issues. The Sutras can be applied to essentially every part of mathematics, even highly complex problems.

 

1. Ekadhikena Purvena 

This Sutra (Formula) means “by more than the previous one.” This formula can be used to solve two types of calculations: squaring a number ending in five, and using fractions whose denominators end in nine.

1.1 Squares  numbers ending in five

Here the number is 35. The task is to find out the square of the number. For the number 35, the last digit is 5 and the previous digit is 3. Hence, ‘one more than the previous one’, that is, 3+1=4. The Sutra, in this context, gives the procedure to multiply the previous digit 3 by one more than itself, that is, by 4. It becomes the L.H.S(Left Hand side in the calculation) of the result, that is, 3 x 4 = 12. The R.H.S(Right hand side) of the result is 52, that is, 25.

  • Thus 352 = (3*4) | 25 = 1225.
  • In the same way, 652 =6*7 | 25=4225;
  • 1052 = 10*11 | 25 = 11025;
  • 1352 = 13*14 | 25 = 18225

Algebraic proof:

10a + 5 in this equation represents the two digit number 35, 45, 55, …, 85 for the values a = 1, 2, 3, … , 8 respectively.

  • Consider (ax + b)2 = a2x2+ 2abx + b2.
  • (10a + 5)2 = a2*102 + 2*10a*5 + 52
  • = a2*102 + a*102 + 52
  • = 102(a2 + a) + 52
  • = a(a + 1)*102 + 25

 

2. Urdhva – tiryagbhyam

It is the general recipe pertinent to all instances of increase and furthermore in the division of a huge number by another huge number.

Ex.1: Find the product of 12X13

  1. Take the right hand-most digit of the multiplicand (the first number, 12), in this case the digit 2. Multiply it by the right hand-most digit of the multiplier (the second number 13), in this case the digit 3. The product 2 * 3 = 6 produces the number in the ones place of the solution.
  2. Now, diagonally multiply the second digit of the multiplicand (2) and first digit of the multiplier (1), giving the answer 2 * 1=2; then, multiply the first digit of the multiplicand (1) and second digit of the multiplier (3), giving the answer 1 * 3 = 3; adding these two answers (2+3= 5) gives the tens place of the answer.
  3. Now, multiply the first digit of the multiplicand (1) and first digit of the multiplier (1), giving the answer 1 * 1 = 1. It gives the hundreds place of the answer. The final answer is therefore 156.

Ex.2: 32 X 24

  • Step (i) : 2 * 4=8
  • Step (ii) : 3 * 4=12; 2 * 2=4; 12+4=16; Here 6 is to be retained. 1 is to be carried out to left side.
  • Step (iii) : 3 * 2=6.The digits 1 of 16 is to be added. i.e., 6 + 1 = 7.
  • Thus 32 * 24 = 768

 

3. Yavadunam Tavadunikrtya Varganca Yojayet

This Sutra is pertinent for acquiring the squares of numbers that are near bases with factors of 10 (i.e. 10, 100, 1000, etc.)

3.1.Values which are near or less than the bases

Ex. 1: 8

  • The answer is separated in to two parts by a ‘|’
  • Note that the deficit is 10 – 8 = 2
  • Square the deficit (22 = 4)
  • Also subtract the deficit from the number (8–2 = 6)
  • Now put 6 on the left and 4 on the right side of the vertical line (6 | 4)
  • Hence, 64 is answer.
  • Here, the base is 100.

Ex. 2: 98

  • Note that deficit is 100 – 98 = 2
  • square the deficit (22 = 4)
  • Also subtract the deficit from the number (98-2 = 96)
  • Now put 96 on the left and 04 on the right side of the vertical line or slash (96 | 04). 04 is used because the base is 100)
  • Hence, 9604 is answer.

3.1.2 Numbers which are greater than the base power of 10

 Ex: 162

  • Rather than subtracting the deficiency from the number we must add and proceed as in Method 1.
  • for 162 , the base is 10 and the surplus is 6.
  • The surplus added to the number is 16+6 = 22. The square of the surplus equals 36
  • The Answer is 25 | 6 = 256 (always remember that if you get a value like 36 so you will add the digit at the hundred place (3) to the 22 which makes it 25).

3.2 Cubing of the number

 Ex. 1: 106

  • For 106, Base is 100. The surplus is 6. Here we add double of the surplus i.e. 106+12 = 118. (remember, in squaring, we immediately  add the surplus).This makes the left-hand -most part of the answer. For example, answer proceeds like 116 / – – – – –
  • Put down the new surplus i.e. 118-100=18 multiplied by the initial surplus.i.e. 6=108.Since base is 100, we write 108 in carried over form 108 .As this is middle portion of the answer, the answer proceeds like 118 / 108 /- –  – –
  • Write down the cube of initial surplus i.e. 63 = 216 as the last portion i.e. right hand side last portion of the answer.Since base is 100, write 216 as 216 as 2 is to be carried over. Answer is 118 / 108 / 216.
  • Now proceeding from right to left and adjusting the carried over, 119 / 10 / 16 = 1191016.

Ex. 2: 1023 = (102+4) /6X2/ 23

  • 106 =12 =08
  • 1061208; Observe initial surplus = 2, next surplus =6 and base = 100.

 

4. Antyayor Dasakepi

The Antyayor Dasakepi means  numbers of which the last digits added up give 10, For eg. the formula works in multiplication of numbers for example: 25 and 25, 47 and 43, 62 and 68, 116 and 114. Note that in each case the sum of the last digit of first number to the last digit of second number is 10. Further the portion of digits or numbers left wards to the last digits remain the same. At that instant use Ekadhikena on left hand side digits. Multiplication of the last digits gives the right hand part of the answer.

Ex.1:  47X43

  • See the end digits sum 7 + 3 = 10 ; then by the sutras antyayor dasakepi and ekadhikena we have the answer.
  • 47 x 43 = ( 4 + 1 ) x 4/ 7 x 3 = 20 / 21 = 2021.

Ex.2: 62 x 68

  • 2 + 8 = 10, L.H.S. portion remains the same i.e.,, 6. Ekadhikena of 6 gives 7.
  • 62 x 68 = ( 6 x 7 )/ ( 2 x 8 ) = 42 / 16 = 4216.

 

Conclusion

This research paper has compiled some efficient ways to do arithmetic which can be used in day-to-day calculations. These techniques are not only useful for students but also for adults who need to do quick arithmetic on the job. The methods discussed are intended for any reader with some basic mathematical background. With these simple operations explained, it is easy to see the beauty, charm, and resourcefulness in Vedic Mathematics systems.

 

References

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Vedic Mathematics/Techniques/Addition And Subtraction. (n.d.). Retrieved July 02, 2020, from https://en.wikibooks.org/wiki/Vedic_Mathematics/Techniques/ Addition_And_Subtraction

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