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- You can learn Vedic Maths from the experts of Magical Methods. (Your questions will not go unanswered)
- You can learn Vedic Mathematics from the comfort of your drawing room.
- You can learn it from anywhere in the world.
- All you need is a computer connected with internet with mike and speaker.
- Click the link below to know the schedule.
Introduction
The origin of "Magical Methods" is from Vedic sutras. There are sixteen Sutras and thirteen Sub-Sutras scattered in the writings of "Atharva Vedas" the fourth Veda. "Atharva Veda" deals with the branches like Engineering, Mathematics, sculpture, Medicine, and all other sciences with which we are today aware of.
The first book on these wonderful methods was written by Late Swami Bharati Krishna Tirtha ji Maharaj, Shankaracharya of Goverdhan Peeth. He was the person who collected lost formulae from the writings of "Atharva Vedas" and wrote them in the form of Sixteen Sutras and thirteen sub-sutras.
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These Sutras and Sub-Sutras can be applied to Arithmetical computations, theory of numbers, compound multiplication, algebraic operations, factorisations, simple quadratic and higher order equations, simultaneous quadratic equations, partial fractions, calculus, squaring, cubing, square root, cube root, coordinate geometry etc.
Magical Methods will try its best to provide you the concepts of Vedic Mathematics in very simple and lucid language.
About the Father of Vedic Mathematics
The pioneer in this field was Jagadguru Swami Sri Bharati Krishna Tirthaji Maharaj, who compiled sixteen sutras and unleashed this wonderful science to the world. He wrote sixteen volumes encompassing some of the greatest Vedic sutras but all were lost and he finally wrote one book explaining the sixteen sutras before he passed away from this world.
The ancient system of Vedic Mathematics was rediscovered from the Vedas, between 1911 and 1918 by Sri Bharati Krishna Tirthaji (1884-1960). At the beginning of the twentieth century, when there was a great interest in the Sanskrit texts in Europe, Bharati Krishna tells us, some scholars ridiculed certain texts which were headed 'Ganita Sutras'- which means mathematics. They could find no mathematics in the translation and dismissed the texts as rubbish. Bharati Krishna, who himself was a scholar of Sanskrit, Mathematics, History and Philosophy, studied these texts and after lengthy and careful investigation was able to reconstruct the mathematics of the Vedas. According to his research all of mathematics is based on sixteen Sutras, or word-formulae.
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Bharati Krishna wrote sixteen volumes expounding the Vedic system but these were unaccountably lost and when the loss was confirmed in his final years he wrote a single book: Vedic Mathematics, currently available. It was published in 1965, five years after his death.
Vedic Mathematics Sutras
To know sutras and their meaning
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| S. No. | Sutras | Meaning |
| 1. | Ekadhikena Purvena | One more than the previous |
| 2. | Nikhilam Navatascharamam Dastah | All from nine and last from ten |
| 3. | Urdhwa-tiryagbhyam | Criss-cross |
| 4. | Paravartya Yojayet | Transpose and adjust |
| 5. | Sunyam Samyasamuchchaye | When the samuchchaya is the same, the samuchchaya is zero, i.e it should be equated to zero. |
| 6. | (Anurupye) Sunyamanyat | If one is in ratio, the other one is zero. |
| 7. | Sankalana-vyavkalanabhyam | By addition and by subtraction |
| 8. | Puranpuranabhyam | By completion or non-completion |
| 9. | Chalana-Kalanabhyam | Differential |
| 10. | Yavdunam | Double |
| 11. | Vyastisamastih | Use the average |
| 12. | Sesanyankena Charmena | The remainders by the last digit |
| 13. | Sopantyadyaymantyam | The ultimate & twice the penultimate |
| 14. | Ekanyunena Purvena | One less than the previous |
| 15. | Gunitasamuchachayah | The product of the sum of coefficients in the factors |
| 16. | Gunaksamuchchayah | When a quadratic expression is product of the binomials then its first differential is sum of the two factors |
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Vedic Mathematics Subsutras
To know subsutras and their meaning
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| S. No. | Sutras | Meaning |
| 1. | Anurupyena | Proportionately |
| 2. | Sisyate Sesasamjnah | Remainder remains constant |
| 3. | Adyamadyenantyamantyena | First by first and last by last |
| 4. | Kevalaih Saptakam Gunyat | In case of seven our multiplicand should be 143 |
| 5. | Vestanam | Osculation |
| 6. | Yavdunam Tavdunam | Whatever the extent of its deficiency, lessen it still further to that very extent |
| 7. | Yavdunam Tavdunam Varganchya Yojayet | Whatever the extent of its deficiency, lessen it still further to that very extent; and also set up the square of that deficiency. |
| 8. | Antyayordasakepi | Whose last digits together total 10 and whose previous part is exactly the same |
| 9. | Antyayoreva | Only the last terms |
| 10. | Samuchchyagunitah | The sum of the coefficients in the product |
| 11. | Lopanasthapanabhyam | By alternate elimination and retention |
| 12. | Vilokanam | By observation |
| 13. | Gunitsamuchchayah Samuchchayagunitah | The product of sum of the coefficients in the factors is equal to the sum of the coefficients in the product. |
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Test Your Understanding
Multiplication
Multiplication is supposedly the toughest of all four operations namely multiplication, division, addition and subtraction. Students feel threatened by multiplication.
I have covered this topic in detail.
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First Formula
I have called this “First Formula” because in my opinion a person willing to learn “Magical methods" should start from here. Formula will be explained by taking various examples. The sutra involved here is "Ekadhikena Purvena"
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Two digit number multiplied by two digit number
To learn the special formula : When sum of the last digits is equal to 10 and first digits are same.
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Let us start with an example:-
How would you multiply this in conventional way ?
Let us solve it :-
What are the steps you took here ?
- First you multiplied 65 by 5 and wrote it below the line (325)
- Then you multiplied 65 by 6 and wrote it below the first row leaving one space from right (390)
- You added the numbers in first row with the numbers in the second row by first putting right most digit down and adding other digits thereafter conventionally.
- You got 4225 as answer.
Now let us do it by magical method :-
What did we do here ?
- We multiplied 5 by 5 and put 25 as right hand side of the answer.
- We added 1 to the top left digit 6 to make it 7.
- We then multiplied it (7) by bottom left digit 6 and get 42, this is left hand side of the answer.
- We arrived at our desired answer 4225.
Did you get it ?
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Let us do some more by the method learned just now !
Let me explain the method again !!
- We multiplied 5 by 5 and put 25 on the right hand side.
- We added 1 to the top left digit 7 to make it 8.
- We then multiplied 8 by bottom left digit 7 and kept 56 on left hand side.
- We arrived at our desired answer 5625.
Now the method should be crystal clear to you.
In the same manner we can multiply the following :-
15 by 15, 25 by 25, 35 by 35, 45 by 45, 55 by 55, etc.
I understand, you are getting inquisitive here and planning to ask a loaded question.
Your question is whether the applicability of the formula is limited to a number ending with 5 only?
My answer is no, its not like that.
Let us expand the formula……………….
We can apply this formula to find multiplication of a good amount of two digit, three digit numbers.
Preconditions is :-
Let us take an example:-
In this example left hand digits are same i.e. 6 and addition of right hand digits are 10. So we can apply this formula here.
Can we apply the same formula to the following :-
Yes, We can apply the same formula to all these since their left hand digits are same and addition or right hand digits are 10.
Here another question may creep into your mind that in the third one above when 9 is multiplied by 1 then it gives 9, but how come we are putting 09 there. The answer is simple, from all above examples we have learned that the right hand side should have two digits but we are getting only one digit i.e. 9 so what to do? How can we use this harmlessly without changing its value ? You know it, add 0 to the left. Now see whether your formula is applicable to the given examples:-
I know that, your answer is affirmative and you can write the answers as 2024, 2021, 2016 and 2009.
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Three digit number multiplied by three digit number
How the First formula would be applied to three digit numbers.
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Having done this for two digit numbers, can we extend the same formula to the three digit numbers ? The answer is yes, we can do that.
Let us take an example:-
In the case mentioned above, first two digits on the left hand side are same and addition of the right most digits come to 10 so we can apply our formula here.
The steps will be :-
- Multiply 5 by 5 and keep 25 on the right hand side.
- Add 1 to 11 to make it 12.
- Multiply 12 by 11 and put 132 on the left hand side. Our operation is complete.
- The answer is 13225.
You can apply this technique to these also:-
Answers :-
| 1. | 13224 | 2. | 13221 | 3. | 13216 | 4. | 13209 |
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Applications
How to apply this formula when numbers in question does not strictly follow the conditions laid down in the formula.
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Utility of First Formula is very wide. You can use this formula to multiply two digit numbers when first digit are same but the addition of last digits does not come to ten. Say 67
65. What will you do in this case?
67
65 can be written as (65 + 2)
65
by our First Formula we know (65
65 = 4225).
Further we are required to add 2
65 = 130 to 4225.
The answer comes to 4355.
Can you apply the technique used above to find out 68 × 64 ?
Yes you can. How? Let us see:-
You can break down 68 × 64 in two ways.
Let us solve these:-
The way, mentioned above you can find out multiplication of range of numbers. Let us take some more examples to clarify:-
Examples:-
Upto this point we have worked with those numbers whose first digits were same and addition of last digits exceeded 10. Now let us solve some of the examples where first digit remains same but addition of last digits is less than 10.
Let me take an example :-
47 × 42
In this case our first digits are same but the addition of last two digits are less than 10.
Let us take few more examples.
Examples:-
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Test Your Understanding
Squaring Technique
You would be learning various methods of finding squares of a number.
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Squares of a number ending with 5
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You have already learned this technique while going through multiplication. This technique has been covered in the beginning named first formula.
The steps revisited :-
- Multiply 5 by 5 and put composite digit 25 on the RHS.
- Add 1 to upper left hand digit 8 i.e. 8 + 1 = 9.
- Multiply 9 to lower left hand digit 8 i.e. 9 x 8 = 72 , put this on LHS.
- Our answer is 7225.
This way we can find out square of any two digit number ending with 5.
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Finding squares of an adjacent number
You would learn to find out squares of adjacent numbers if you know square of a number without much of effort.
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Forward method
To learn to find square of the next number if the square of a number is known.
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We know squares of a number say (752 = 5625); How can we find out square of 76?
Steps :-
Its very easy, the format shown above is self explanatory. But I am explaining it for the benefit of the students :-
- 752 = 5625 is known
- Add (75 + 76 = 151) to this to get 762
- 762 = 5776.
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Reverse Method
To learn to find square of the previous number if the square of a number is known.
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Did you like the forward method? now you will be able to find out square of a number which is 1 more than the given number whose square is known.
Now let me explain the Reverse Method through which you will be able to find out squares of a number which is one less than the given number.
Let me explain with an example :-
We know square of a number say 70; How can find out square of 69?
Steps :-
Do you want me to explain the steps (???)
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Mental Formula for finding Squares
You would learn to find out squares of any number from 11 to 99 without much of hassle
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Let us first find square of 11 using the formula :-
The formula is self explanatory. However, Let me explain it for more clarification:-
- Slash used here is just a separator.
- Our operating zone is 10 x 1 or simply 10.
- 11 is one more than 10.
- We added 1 to 11 to make it 12.
- Number of digits after the slash can be only one.
- If the number of digits after the slash exceeds one then we place only the right most digit after the slash and the remaining digits gets added to left hand side of the slash.
Will you be able to find squares of other numbers in a similar manner ? Try…..
You can work like this up to 192. What about numbers above 20 ?
Formula remains same with a slight change. The change you will appreciate:-
This change is because now we are operating in 10 x 2 zone. Is it going to hold good for the whole range from 21 to 29 ? Let us check :-
Having learnt this can you find square of the numbers 31 to 39 ? Hopefully yes (?).
By the method explained above you should be able to memorize the squares up to 99 without much hassle.
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