Sutras of Vedic Mathematics

Merely 16 Sutras and 13 Sub-Sutra form the basis of Vedic Mathematics. These sutras can be used to solve all types of Mathematical Problems!!!

These sutras had been long lost in the cryptic Vedas where the mathematical solutions were hidden in Sanskrit Verses for centuries. Sri Bharati Krsna Tirthaji's "Vedic Mathematics" recently discovered these techniques again and given it to all of us to use.

The 16 sutras, or rules, and 14 sub-sutras can be used to solve complex mathematical problems mentally, leaving your fellow students and colleagues in awe.

The 16 Sutras of Vedic Mathematics:

Sutra 1 : By one more than the one before

Sutra 2 : All from 9 and the last from 10

Sutra 3 : Vertically and crosswise

Sutra 4 : Transpose and apply

Sutra 5 : If the Samuccaya is the same it is zero

Sutra 6 : If one is in ratio the other is zero

Sutra 7: By addition and by subtraction

Sutra 8: By the completion or non-completion

Sutra 9: Differential calculus

Sutra 10: By the deficiency

Sutra 11: Specific and general

Sutra 12: The remainders by the last digit

Sutra 13: The ultimate and twice the penultimate

Sutra 14: By one less than the one before

Sutra 15: The product of the sum

Sutra 16: All the multipliers

The 13 Sub-sutras:

SS 1: Proportionately

SS 2: The remainder remains constant

SS 3: The first by the first and the last by the last

SS 4: For 7 the multiplicand is 143

SS 5: By osculation

SS 6: Lessen by the deficiency

SS 7: Whatever the deficiency lessen by that amount and set up the square of the deficiency

SS 8: Last totaling 10

SS 9: Only the last terms

SS 10: The sum of the products

SS 11: By alternative elimination and retention

SS 12: By mere observation

SS 13: The product of the sum is the sum of the products

We will work through these Sutras in the next few pages and also will give you lot of practice to get used to the techniques.

Check out the blog on the link here to see the problems in the present method of teaching mathematics in schools.

Sutra 1: "By one more than the one before" provides a simple way of multiplication of numbers with the same first digit and the sum of their last unit digits is 10.

An interesting application of this formula is in computing squares of numbers ending in five. Consider:

35 × 35 = ((3 × 3) + 3)*10 + 25 = 1225 or
= (3 x 4) *10 +(5 x 5) = 1225
125 x 125 = (12 x 13)*10 + 25 = 15625
Explanation: The latter portion is multiplied by itself (5 by 5) and the previous portion is multiplication of first digit and the digit higher to the first digit resulting in the answer 1225

This is a simple application of (a+b)² = a² + 2ab + b² ....Here 2b = 10

It can also be applied in multiplications when the last digit is not 5 but the sum of the last digits is the base (10) and the previous parts are the same. Consider:

37 × 33 = (3 × 4)*100 + (7 × 3) = 1221
29 × 21 = (2 × 3)*100 + (9 × 1) = 609
This uses (a + b)(a − b) = a² − b² twice combined with the previous result to produce: (10c + 5 + d)(10c + 5 − d) = (10c + 5)2 − d2 = 100c(c + 1) + 25 − d2 = 100c(c + 1) + (5 + d)(5 − d).

The Sutra is very useful in its application to convert fractions into their equivalent decimal form. Consider fraction 1/19. Using this formula, this can be converted into a decimal form in a single step. This can be done by applying the formula for either a multiplication or division operation, thus yielding two methods.


Method 1: Using multiplication. The sutra "one more than the one before" provides a simple way of calculating values like 1/x9 (e.g 1/19, 1/29, etc). Let's take one 1/x9 and calculate e.g. 1/19. In this case, x=1. To convert 1/19 to decimals, since 19 is not divisible by 2 or 5, the fractional result is a purely circulating (recurring) decimal, i.e the digits would repeat themselves after some time. (If the denominator contains only factors 2 and 5 it is a purely non-circulating decimal i.e it is a perfect decimal, else it is a mixture of the two.).

So to get the decimals of 1/19, we start with the numerator digit i.e 1 in this case and Multiply this by "one more" (1+x), in this case: 1+1 = 2.
21 Multiplying 2 by 2, followed by multiplying 4 by 2
421 → 8421 Now, multiplying 8 by 2, sixteen
68421 1 ← carry
If the result of the multiplication is greater than 10, keep (value - 10) and keep the "1" as "carry over" which you'll add to the next digit.
multiplying 6 by 2 is 12 plus 1 carry gives 13
368421 1 ← carry
Continuing
7368421 → 47368421 → 947368421

Now we have 9 digits of the answer. There are a total of 18 digits (= denominator − numerator i.e 19-1 = 18) in the answer. The last 9 digits can be computed by complementing the lower half (with its complement from nine i.e number + complement = 9):

052631578
947368421
Thus the result is 1/19 = 0.052631578,947368421 repeating.

If you picked up 1/29, you'll have to do it till 28 digits (i.e. 29-1). You'll get the following
1/29 = 03448275862068,
96551724137931
Run this on your calculator and check the result!

I do not like this method a lot as it seems academic. Why would someone want to find all the numbers in the ratio, until he is doing some research. For all practial purposes 4-5 digits of the decimal (from the left) are enough. So the next technique is my favoured method.

Method 2: Using division. The earlier process can also be done using division instead of multiplication.

For A/X9, We divide A by (1+X). Incase of 1/19, we divide 1/(1+1), the answer is 0 (lets say N) with remainder 1 (lets say D)

0.0: D (in this case 1) carries forward and become (D*10 + N) i.e. 10. This is then divided by 2 for N = 5

0.05: Next 05 divided by 2 with answer as 2 and remainder 1

0.052: Next 12 (1 from earlier remiander and 2 from the answer) is divided by 2 for answer 6

0.0526: and so on.

So the asnwer of 1/19 = 0.0526.

Consider another example, 1/7, this same as 7/49 which has last digit of the denominator as 9. The previous digit to 9 in 49 is 4. So by the sutra we take one more i.e. 5. So we divide the numerator by 5, that is,

...7/5: 1 R=2; 21/5: 4 R=1; 14/5: 2 R=4, 42/5: 8 R=2 and so on...so the answer is 1/7 = 0.1428

Now try to convert 18/19 into decimals.
(Numerator 18 to be divided by x+1=2)
...18/2:9 R=0 =>9/2:4 R:1 =>14/2: 7 R=0 =>7/2: 3 R=1 and so on.

So the answer for 18/19 is 0.9473...

Now try 18/29.
Can you do it mentally? The answer is 0.6206...

Note this technique can be used for conversion of vulgar fractions ending in 1, 3, 7 including 9 such as 1 / 11, 1 / 21, 1 / 31 - - -- ,1 / 13, 1 / 23, - - - -, 1 / 7, 1 / 17, - - - - - by multiplying appropriate factors to the numerator and denominator to make them ed with 9.
1/11 = 9/99 --> 0.09090909
1/13 = 3/39 --> 0.07692307

Amazed at the speed at which you can calculate these sums MENTALLY!!!

People say there are so many small things that we have to remember, instead of one general method of division or multiplication. My answer is once you get used to it the effort in remembering such an useful method is far surpassed by the ease and speed.


Note: 1/17 = 7/119 --> This technique becomes a bit difficult when you try to solve with 1/x9, where x is greater than 9. Will discuss other techniques for these type of big fractions.

Interested in learning how to convert any decimal to fractions??

Sutra 2: "All from nine and the last from 10" comes handy when we wish to subtract a number such as 578 from a number having zeros equivalent or more than the number of number being subtracted e.g. 1,000. Each figure in 578 is subtracted from nine and the last figure is subtracted from 10, yielding 422. The interesting thing is that this problem can be solved from left to right the same way we write and read text.

This sutra can be extended to solve problems such as 60,000 minus 987. We simply reduce the number left after taking out the 0s to account for the number being subtracted (i.e 60 left after removing the 3 left 0s) by 1 and then apply the sutra, to get the answer 59,013.

The concept gives rise to Sub-Sutra 7: "Whatever the extent of its deficiency, lessen it still further to that very extent; and also set up the square of that deficiency". Let me explain this further:

For instance, in computing 9² we go through the following steps:
Let us take 10 as our base. To get 9 to the base 10 we have to subtract 1 from 10.
Since decrease 9 further by 1 to 8. This is the left side of our answer.
On the right hand side put the square of the deficiency, which is 1². Hence the answer is 81.
Similarly, 8²-->(8-2)*10+ 2² --> 64, 7² = 49.

For numbers above 10,we take the surplus. For example:
11² = (11+1)*10 + 1² = 121
13² = (13+3)*10 + 3² = 169
and so on.

This is based on the fact that a²= (a + b)(a − b) + b² where a is the number whose square is to be found and b is the deficit (or surplus) from nearest factor of 10.



398 X 11 =



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