Download "VEDIC MATHEMATICS" by H.H. Jagadguru Swami Sri Bharati

"VEDIC MATHEMATICS"
by H.H. Jagadguru Swami Sri Bharati Krishna Tirthaji Maharaj.
Publishers Motilal Banarasidass, Bunglow Road, Jawahar Nagar, Delhi –110 007; or
Chowk, Varanasi (UP); or Ashok Raj Path, Patna, (Bihar)
Is it useful today?
Given the initial training in modern maths in today's schools, students will be able to comprehend the logic of Vedic mathematics
after they have reached the 8th standard. It will be of interest to every one but more so to younger students keen to make their
mark in competitive entrance exams.
India's past could well help them make it in today's world.
It is amazing how with the help of 16 Sutras and 16 sub-sutras, the Vedic seers were able to mentally calculate complex
mathematical problems.
Some of the Sutras (Phrases) used in Vedic Mathematics :
1. Ekadhikena Purvena (One More than the Previous) is useful in solving Special Multiplications like 25X25, 95X95, 105X105
etc Special Divisons like 1 divided by 19, 29, 39, …. 199 etc.
2. Eka Nunena Purvena (One less than the Previous) is useful in solving Special Multiplications like 777 X 999, 123456789X
999 999 999
3. Urdhva Tiryak bhyam (Vertically and Cross-wise) is useful in General Multiplication of any number by any number.
4. Paravartya Yojayet (P-64) (Transpose and Apply) is useful in solving Algebraic factors and divisions of some numbers etc.
5. Anurupyena (P-87) (Proportionately)
6. Adhyam-Adhyena, Antyam-antyena (P-87) (first by the first and the last by the last) are useful in solving Quadratics
7. Lopana-stapana-bhyam (P-90) (by (alternate) Elimination and Retention) is useful in factorizing long and harder quadratics..
8. Gunita-Samuchhayah Samuchhaya-gunitah which means
"The product of the sum of the coefficients in the factors is equal to
the sum of the coefficients in the product"
is a Sub-sutra of immense utility for the purpose of verifying the correctness of our answers in multiplications, divisions and
factorisations:
9. Sunyam Samya samuccaye P-107 (when Samuccaya is the same, that Samuccaya is zero) Samuccaya is a technical term
which has several meanings. This is useful in solving many complex factors and equations.
1st Example - 1 Divided by 19, 29, 39, …. 129 etc
1.
To Divide 1 by numbers ending in 9 like 1 divided by 19, 29, 39, ….. 119 etc.
Some of these numbers like 19, 29, 59 are prime numbers and so cannot be factorised and division becomes
all the more difficult and runs into many pages in the present conventional method and the chances of making
mistakes are many.
The Vedic Solution is obtained by applying the Sutra (theorem) Ekadhikena Purvena which when translated
means "One more than the Previous"
Take for example 1 divided by 19. In the divisor 19, the previous is 1 and the factor is obtained by adding 1
to it which is 2. Similarly when we have to divide by 29, 39, … 119 the factors shall be 3,4,… 12
respectively. (Add 1 to the previous term in the divisor). After this divide 1 by the factor in a typical Vedic
way
and
the
answer
is
obtained
in
1
step.
Thus
1 divided by 19 = 0.0 5 2 6 3 1 5 7 8 9 4 7 3 6 8 4 2 1
1 divided by 29 = 0.0 3 4 4 8 2 7 5 8 6 2 0 6 8 9 6 5 5 1 7 2 4 1 3 7 9 3 1
2nd Example - Square of Numbers ending in 5
Squares of 25, 35, 45, 85, 95, 105, 195 etc can be worked out mentally
Again the Sutra used here is Ekadhikena Purvena which means, "One more than the previous."
The last term is always 5 and the Previous terms are 2, 3, 4, 8, 9, 10, 19 etc and we have to add 1 to them.
Square of the last term 5 is always 25.
Thus the Square of 25 is 2x3/25 = 625
the Square of 35 is 3x4/25 = 1225
the Square of 45 is 4x5/25 = 2025
the Square of 85 is 8x9/25 = 7225
the Square of 95, 105, 195 can be obtained in the same way.
Use the above formula to find the products of
23 multiplied by 27; 44 multiplied by 46; 192 multiplied by 198 and so on.
3rd Example - Multiplier-digits consist entirely of nines
The Sutra (theorem) used here is Ekanyunena Purvena, sound as if it were the converse of Ekadhik Sutra ie
"one less"
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
Such multiplications come up in advanced astronomy.
4th Example General Multiplication of any number by any number
The Sutra used is Urdhva-Tiryagbhyam which means, "Vertically and cross-wise" (P-39)
To multiply 12 by 13 mentally multiply 1 by 1, 3 by 1 and 2 by 1 and finally 2 by 3 and write the answer as 1 56
To multiply 37 by 42 mentally multiply 3 by 4, 3 by 2 and 7 by 4 and finally 7 by 2 and the answer is 12/34/14 = 1554
To multiply 1021 by 2103 mentally multiply as follows 1 by2 1by1+0by2 1by0+2by2+1by0 1by3+0by0+2by1+1by2
0by3+2by0+1by1 2by3+1by0 1by3 = 2147163
Multiplying 8 7 2 6 5 by 3 2 1 1 7 gives 2 8 0 2 6 9 0 0 0 5
Sixteen Simple Mathematical Sutras (Phrases) From The Vedas
Sixteen Sutras and Their Corollaries
Sutras or Formulae
Sub-Sutras or Corollaries
Ekadhikena Purvena
Anurupyena
(One more than the previous) P-2
(Proportionately) P-20, 87 Multiplications
Division eg. 1/19
Nikhilam Navatascaram Dashtah
Sishyate Seshasanjnah
(All from 9 last from 10) P-13
Multiplication eg 9927 X 9999
Urdhva-tiryaghyam P-
Adhyam-adhyena antyam-antyena P-87
(Vertically and cross-wise)
(First by the first and last by the last)
Paravartya Yojayet P-
Kevalaih Saptakam Gunyat
(Transpose and apply)
Sunyam Samya samuccaye P-107
Veshtanam
(when Samuccaya is the same, that Samuccaya is zero)
(Anurupye) Sunyamanyat
Yavadunam Tavadunam
Sankalana-vyavakalana
Yavadunam Tavadunikrtya Vargamca Yojayet
Purna purnabhyam
Antyayor Dasakepi (also for two numbers whose last digits together total 10)
Calana-Kalanabhyam
Antyayoreva
Yavadunam
Samuccayagunitah
Vyashti samashti
Lopana Sthapanabhyam
"by (alternate) Elimination and Retention"
very useful in HCF, Solid Geometry, Coordinate Geometry,
Seshanyankena caramena
Vikolanam
Sopantya dvayamantyam
Gunita samuccayah Samucchaya gunitah P-89 *13
Ekanyunena purvena
(one less than the previous) P-35
one of the multiplier-digits is all 9
Gunita Samuccayah P16.
Gunaka Samuccayah
The remarkable system of Vedic maths was rediscovered from ancient Sanskrit texts early last century. The system is based on 16 sutras or
aphorisms, such as: “by one more than the one before” and “all from nine and the last from 10″. These describe natural processes in the mind and
ways of solving a whole range of mathematical problems. For example, if we wished to subtract 564 from 1,000 we simply apply the sutra “all from nine
and the last from 10″. Each figure in 564 is subtracted from nine and the last figure is subtracted from 10, yielding 436.
This can easily be extended to solve problems such as 3,000 minus 467. We simply reduce the first figure in 3,000 by one and then apply the sutra, to
get the answer 2,533. We have had a lot of fun with this type of sum, particularly when dealing with money examples, such as £10 take away £2. 36.
Many of the children have described how they have challenged their parents to races at home using many of the Vedic techniques - and won. This
particular method can also be expanded into a general method, dealing with any subtraction sum.
The sutra “vertically and crosswise” has many uses. One very useful application is helping children who are having trouble with their tables above 5×5.
For example 7×8. 7 is 3 below the base of 10, and 8 is 2 below the base of 10.
The sutra “vertically and crosswise” is often used in long multiplication. Suppose we wish to multiply 32 by 44. We multiply vertically 2×4=8.
Then we multiply crosswise and add the two results: 3×4+4×2=20, so put down 0 and carry 2. Finally we multiply vertically 3×4=12 and add the carried
2 =14. Result: 1,408.
We can extend this method to deal with long multiplication of numbers of any size. The great advantage of this system is that the answer can be
obtained in one line and mentally. By the end of Year 8, I would expect all students to be able to do a “3 by 2″ long multiplication in their heads. This
gives enormous confidence to the pupils who lose their fear of numbers and go on to tackle harder maths in a more open manner. Multiplication can
also be carried out starting from the left, which can be better because we write and pronounce numbers from left to right. Here is an example of doing
this in a special method for long multiplication of numbers near a base (10, 100, 1,000 etc), 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 and multiplying the “differences” vertically 4×8=32
gives the second part of the answer.
This works equally well for numbers above the base: 105×111=11,655. Here we add the differences. For 205×211=43,255, we double the first part of
the answer, because 200 is 2×100.
Tutorial 1
Use the formula ALL FROM 9 AND THE LAST FROM 10 to perform instant subtractions. For example 1000 - 357 = 643
We simply take each figure in 357 from 9 and the last figure from 10.
So the answer is 1000 - 357 = 643
And 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
For 1000 - 83, in which we have more zeros than figures in the numbers being subtracted, we simply suppose 83 is 083.
So 1000 - 83 becomes 1000 - 083 = 917
Exercise 1 Tutorial 1
Try some yourself:
1) 1000 - 777
=
2) 1000 - 283
=
3) 1000 - 505
=
8) 1000 - 57
=
9) 10,000 - 321
=
10) 10,000 - 38
=
4) 10,000 - 2345 =
5) 10,000 - 9876 =
6) 10,000 - 1011 =
7) 100 –57=
Using VERTICALLY AND CROSSWISE you do not need the multiplication tables beyond 5 X 5. Suppose you need 8 x 7. 8 is 2 below 10 and 7 is 3
below 10. Think of it like this:
The answer is 56. The diagram below shows how you get it.
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 answer. That’s
all you do:
See how far the numbers are below 10, subtractone number’s deficiency from the other number, and multiply the deficiencies together.
7 x 6 = 42
Here there is a carry: the 1 in the 12 goes over to make 3 into 4.
Exercise 1 Tutorial 2
Multply These:
1) 8 x 8 =
2) 9 x 7 =
3) 8 x 9 =
4) 7 x 7 =
5) 9 x 9 =
6) 6 x 6 =
Here’s how to use VERTICALLY AND CROSSWISE
for multiplying numbers close to 100. Suppose you want to multiply 88 by 98.Not easy, you might think. But with VERTICALLY AND CROSSWISE you
can give the answer immediately, using the same method as above Both 88 and 98 are close to 100. 88 is 12 below 100 and 98 is 2 below 100.
You can imagine the sum set out like this:
As before the 86 comes from subtracting crosswise: 88 - 2 = 86 (or 98 - 12 = 86: you can subtract either way,
you will always get the same answer). And the 24 in the answer is just 12 x 2: you multiply vertically. So 88 x 98 = 8624
Exercise 2 Tutorial 2
1) 87 x 98 =
2) 88 x 97 =
3) 77 x 98 =
4) 93 x 96 =
5) 94 x 92 =
6) 64 x 99 =
7) 98 x 97 =
Multiplying numbers just over 100.
103 x 104 = 10712
The answer is in two parts: 107 and 12, 107 is just 103 + 4 (or 104 + 3), and 12 is just 3 x 4.
Similarly 107 x 106 = 11342
107 + 6 = 113 and 7 x 6 = 42
Exercise 3 Tutorial 2
1) 102 x 107 =
2) 106 x 103 =
3) 104 x 104 =
4) 109 x 108 =
5) 101 x123 =
6) 103 x102 =
Tutorial 3
The 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. So:
Subtracting is just as easy: multiply crosswise as before, but the subtract:
Tutorial 4
A quick way to square numbers that end in 5 using the formula BY ONE MORE THAN THE ONE BEFORE.
752 = 5625
75² 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 is 8: so 7 x 8 = 56
Similarly 852 = 7225 because 8 x 9 = 72.
Exercise 1 Tutorial 4
1) 452 = 2) 652 = 3) 952 = 4) 352 = 5) 152 =
Method for multiplying numbers where the first figures are the same and the last 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.
And 81 x 89 = 7209
We put 09 since we need two figures as in all the other examples.
Exercise 2 Tutorial 4
1) 43 x 47 =
2) 24 x 26 =
3) 62 x 68 =
4) 17 x 13 =
5) 59 x 51 =
6) 77 x 73 =
Tutorial 5
An elegant way of multiplying numbers using a simple pattern
21 x 23 = 483
This is normally called long multiplication but actually the answer can be written straight down using the VERTICALLY AND CROSSWISE formula. We
first put, or imagine, 23 below 21:
There are 3 steps:
a) Multiply vertically on the left: 2 x 2 = 4. This gives the first figure of the answer.
b) Multiply crosswise and add: 2 x 3 + 1 x 2 = 8. This gives the middle figure.
c) Multiply vertically on the right: 1 x 3 = 3. This gives the last figure of the answer. And that’s all there is to it.
Similarly 61 x 31 = 1891
6 x 3 = 18; 6 x 1 + 1 x 3 = 9; 1 x 1 = 1
Exercise 1 Tutorial 5
1) 14 x 21 2) 22 x 31 3) 21 x 31 4) 21 x 22 5) 32 x 21
Multiply any 2-figure numbers together by mere mental arithmetic!
If you want 21 stamps at 26 pence each you can easily find the total price in your head. There were no carries in the method given above. However,
there only involve one small extra step.
21 x 26 = 546
The method is the same as above except that we get a 2-figure number, 14, in the middle step, so the 1 is carried over to the left (4 becomes 5). So 21
stamps cost £5.46.
Practise a few:
1) 21 x 47
2) 23 x 43
3) 32 x 53
4) 42 x 32
5) 71 x 72
6) 32 x 56
7) 32 x 54
8) 31 x 72
9) 44 x 53
10) 54 x 64
Exercise 2b Tutorial 5
33 x 44 = 1452
There may be more than one carry in a sum:
Vertically on the left we get 12. Crosswise gives us 24, so we carry 2 to the left and mentally get 144. Then vertically on the right we get 12 and the 1
here is carried over to the 144 to make 1452.
Any two numbers, no matter how big, can be multiplied in one line by this method.
Tutorial 6
Multiplying a number by 11.
To multiply any 2-figure number by 11 we just put the total of the two figures between the 2 figures.
26 x 11 = 286
Notice that the outer figures in 286 are the 26 being multiplied. And the middle figure is just 2 and 6 added up. So 72 x 11 = 792
Exercise 1 Tutorial 6
Multiply by 11:
1) 43 =
2) 81 =
3) 15 =
4) 44 =
5) 11 =
77 x 11 = 847
This involves a carry figure because 7 + 7 = 14 we get 77 x 11 = 7147 = 847.
Exercise 2 Tutorial 6
Multiply by 11:
1) 11 x 88 =
2) 11 x 84 =
3) 11 x 48 =
4) 11 x 73 =
5) 11 x 56 =
234 x 11 = 2574
We put the 2 and the 4 at the ends. We add the first pair 2 + 3 = 5. and we add the last pair: 3 + 4 = 7.
Exercise 3 Tutorial 6
Multiply by 11:
1) 151 =
2) 527 =
3) 333 =
4) 714 =
5) 909 =
Tutorial 7
Method for dividing 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 answerand 4 + 3 = 7 is the remainder - could it be easier?
Exercise 1a Tutorial 7
Divide by 9:
1) 61 / 9 =
remainder
2) 33 / 9 =
remainder
3) 44 / 9 =
remainder
4) 53 / 9 =
remainder
5) 80 / 9 =
remainder
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.
Exercise 1b Tutorial 7
Divide by 9:
6) 232 =
remainder
7) 151 =
remainder
8) 303 =
remainder
9) 212 =
remainder
10) 2121 =
remainder
842 / 9 = 812 remainder 14 = 92 remainder 14
Actually a remainder of 9 or more is not usually permitted because we are trying to find how many 9’s there are in 842. Since the remainder, 14 has
one more 9 with 5 left over the final answer will be 93 remainder 5
Exercise 2 Tutorial 7
Divide these by 9:
1) 771 / 9 = remainder
2) 942 / 9 = remainder
3) 565 / 9 =
remainder
4) 555 / 9 = remainder
5) 2382 / 9 = remainder Answers
Multiplying by 12 - shortcut
12 X 7
The first thing is to always multiply the 1 of the twelve by the number we are multiplying by, in this case 7. So 1 X 7 = 7. Multiply this 7 by 10 giving 70.
(Why? We are working with BASES here. Bases are the fundamentals to easy calculations for all multiplication tables. Now multiply the 7 by the 2 of
twelve giving 14. Add this to 70 giving 84. Therefore 7 X 12 = 84
17 X 12
Remember, multiply the 17 by the 1 in 12 and multiply by 10 (Just add a zero to the end) 1 X 17 = 17, multiplied by 10 giving 170. Multiply 17 by 2 giving
34. Add 34 to 170 giving 204. So 17 X 12 = 204