Multiplication at your fingertips!

Here I am going to talk about "Ekadhikena Purveṇa" i.e "One more than the previous one" and very important Vedic Mathematics formula and how to use it for multiplication.

Application:

This formula can be applied to multiplication of numbers that satisfy both the following conditions

1. the same first digit
2. the sum of their last unit digits is 10.

We have already discussed about applying this formula for deriving squares for any number ending with 5, in previous post (Amazing way to derive squares within seconds). So here, I am going to talk about application of this formula for multiplication of numbers satisfying the conditions stated above.

Method:

Let us break the method into two parts:

1.  Deriving last two digits of answer
2. Deriving all previous digits of answer

1. Deriving last two digits of answer
These shall be multiplication of last digit of the numbers we are looking to multiply.

2. Deriving all previous digits of answer

All the digits in the multipliers other than the last would constitute base for us and let us call them B1.

Another base is B1 +1 (Because the formula is about one more than the previous one…)

So previous digits in the final answer would be simply [B1 x (B1+1)]

Example:

Let us understand this with an example: 24 x 26

Last two digit of the answer:

4 x 6 = 24 (If this answer is in single digit, then we need to put 0 before it)

For other digits, we need B1 and it goes as follows:

B1 = 2 (All the digits other than the last 5 in the number we want to derive square for)

B1 + 1 = 2 + 1 = 3

Therefore other digits of the answer preceding 25 are = 2 x 3 = 6

Hence the final answer would be 624.

Logic:

So what’s the logical explanation of this wonderful shortcut:

We all know that (a-b) x (a+b) = a²-b²

Now how does this help with multiplication of 24 & 26? It is like this:

24 x 26

(25-1) x (25+1)

25² – 1²

625 – 1 (25² derived from the same formula applied for squaring discussed in the previous post)

624 (Well this matches!)

Vedic mathematics rocks!

Amazing way to calculate squares within seconds

"Ekadhikena Purveṇa" as we discussed for division is the Sanskrit term for "One more than the previous one" and very important Vedic Mathematics formula. We already know from the previous post that this formula can be used for both multiplication and division, here let us have a look at how it helps for multiplication:
Application:
This formula can be applied to multiplication of numbers that satisfy both the following conditions

1. the same first digit
2. the sum of their last unit digits is 10.

Even though it might so appear that this formula is useful in limited cased but an interesting category of numbers that satisfy both above requirements are all the numbers ending with 5. So when we need to derive squares for any number ending with 5, this formula is applicable straightaway. In this post, I am going to talk about limited application of this formula for deriving squares of numbers with last digit as 5.

Method:

Let us break the method into two parts:

1. Deriving last two digits of answer
2. Deriving all previous digits of answer

1. Deriving last two digits of answer

No application of mind, last two digits are 25, always.

2. Deriving all previous digits of answer

All the digits in the number before the last five would constitute base for us and let us call them B1.

Another base is B1 +1 (Because the formula is about one more than the previous one…)

Digits previous of 25 in the final answer would be simply [B1 x (B1+1)]

Example:

Let us understand this with an example of squaring 25

Last two digit of the answer are 25

For other digits, we need B1 and it goes as follows:

B1 = 2 (All the digits other than the last 5 in the number we want to derive square for)

B1 + 1 = 2 + 1 = 3

Therefore other digits of the answer preceding 25 are = 2 x 3 = 6

Hence the final answer would be 625.

So now squaring is not a headache anymore having understood vedic mathematics, right?

Logic:

Learning just the shortcut method isn’t enough, so let us understand the logic how this operates. For explaining the logic we need to use the basic concepts of factorization:

We all know that (a+b)² = a² + 2ab + b² 

Now how does this square of numbers ending with 5 work in this case?

Let us continue with the above example:

25²

(20 + 5)²

20² + (2x20x5) + 5² (Factorization using the above rule)

20² + (20x10) + 5² (Simplified with multiplication of 5x2 for middle term)

20 (20 + 10) + 5² (taking 20 common from first two terms)

20 (30) + 5²

2 x 10 x 3 x 10 + 5²

(2x3) x (10x10) + 5²

(2x3) x (100) + 5²

Now this expression looks like this [B1 x (B1+1)] x 100 + 5²

Because in this expression two things are constant i.e. whatever is the B’s multiplication that shall at the end be multiplied by 100 and 25 shall be added so it is safe to make a rule that last two digits of the answer are going to be 25 and right preceding these digits we need to insert the multiplication of B1 & B1+1

Now you must feel awakened with the logical explanation of Vedic mathematics.

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Easier Solution for Complex Divisions

Now let us have a look at a sutra - "Ekadhiken Purven" out of a total of sixteen such sutras of Vedic Mathematics. Though this deals with multiplication and division both. Will talk about multiplication using this later, lets concentrate on division for the time being.

So first things first, when this sutra would apply

Values like 1/x9 (i.e. 1/19, 1/29, 1/39, etc.)

There are two methods by which we can approach this sutra I shall explain both in detail:

Prob: 1/19
First Conclusion:

19 is not a factor of either 2 or 5 which means that the result of this division is a purely circulating decimal.

Therefore, we must find out what are going to be the number of digits in this circulating decimal and the answer is 18 (divisor -1).

These 18 digits form two equal parts of 9 each and they complement each other to sum up to 9 as shown below: (1 denotes the left most digit of the result and 18 denotes the right most or last digit of the circulating sequence we shall receive out of this result)

1     2    3     4     5    6    7    8    9
10  11  12  13  14  15  16  17  18

This shows that the sum of digits at 1st place and 10th place in the circulating sequence should be 9 and so with sum of 2nd and 11th place and for all the rest of the pairs.

Thus, we effectively need to find out either 1st to 9th or 10th to 18th digits complement them respectively with 9 to get the other part of the answer.

There are two method in which the sutra can be used as shown below:

Method 1: Multiplication - To find out digits at places 10th to 18th
Method 2: Division - To find out digits at the places 1st to 9th

1. Multiplication (Deriving digits at places 10th to 18th of the answer)

This method gives answer from the right most digit (i.e. 18th) onwards to the left most digit (i.e. 10th)

18th Digit: Simply put the dividend (or numerator of the problem 1/19) here i.e. 1

17th to 10th Digits:

Just look at the problem as 1/x9 which will give us value of x=1, now add one to x which will make it 2 and thats our multiplier for the deriving the rest of the answers. so

Place  Digit
18th      1
17th      2 (i.e. 2 multiplier as explained above is multiplied with 1 i.e. 18th digit)
16th      4 (i.e. 2 multiplied with 17th digit)
15th      8 (i.e. 2 multiplied with 16th digit)
14th      6 (i.e. 2 multiplied with 15th digit, carry over 1)
13th      3 (i.e. 2 multiplied with 14th digit + carry over 1 = 13, so 1 carried over)
12th      7 (i.e. 2 multiplied with 13th digit)
11th      4 (i.e. 2 multiplied with 12th digit, carry over 1)
10th      9 (i.e. 2 multiplied with 16th digit + carry over 1 = 9)

So the set of digits from 10th to 18th place looks like this

9 4 7 3 6 8 4 2 1

Now lets subtract all of them from 9 to get 1st to 9th digits respectively;

9 4 7 3 6 8 4 2 1
0 5 2 6 3 1 5 7 8

So the final answer is:

1/19 = 0.052631578947368421 (circulating sequence)

2. Division (Deriving digits at places 1st to 9th of the answer)

This method gives answer from the left most digit (i.e. 1st) onwards to the right most digit (i.e. 9th)

Again, let us just look at the problem as 1/x9 which will give us value of x=1, now add one to x which will make it 2 and thats our magic divisor for the deriving the answer.

1st Place: a / b where a = dividend of the problem i.e. 1 and b = the magic divisor = 2

So, 1 / 2 is 0 with a remainder of 1. Hence, 1st place = 0

2nd place onwards: (in terms of a/b for each place)

Thus our 1st to 9th digits are

0 5 2 6 3 1 5 7 8

Now lets subtract all of them from 9 to get 1st to 9th digits respectively;

0 5 2 6 3 1 5 7 8

9 4 7 3 6 8 4 2 1

So the final answer is:

1/19 = 0.052631578947368421 (circulating sequence)

So with blessings of Vedic Mathematic, a little practice on this and you shall be quick like a gun!!

Will come back with application of this Vedic Maths sutra for multiplication which is even more interesting till then stay tuned and visit GyanCircle.com

Multiplication Made Easy Using The Line System

Now that we have seen the simpler way of multiplication using vedic mathematics for two digit numbers at Basics of Vedic Mathematics, I will show you an alternate method which even doesn't require multiplication of two single digit numbers to derive product of any large numbers.

Keep connected to www.gyancircle.com for more updates.

This is the fun of vedic mathematics where you can multiply without multiplying!

Basics of Vedic Mathematics

After introduction to Vedic Mathematics, let us start with basics thereof.
It would be good to take multiplication, and for that any 2 digit number with any other 2 digit number.

Basic Definitions:

Vedic mathematics is based on the concept of placing the numbers either at the unit place or tenths or hundredth and so on. So for ease of understanding, let us use this legend:

Unit Place: UP

Tenth Place: XP

Hundredth Place: HP

Thousand Place: TP

Ten thousandth Placce: TXP

Hundred thousandth Place: THP (and so on..)

Let us take an example of 23 x 45:

So now we need to place this numbers under each other in such a way that UP of both multipliers are in one column and XP numbers in one column:

XP UP

2 3

4 5

1. Now we have to start with UP and multiply both the numbers there. (i.e. 3 x 5) and the answer is 15. Out of this answer 5 would be placed on the UP of the product of two multipliers and 1 would be a carry over. So we now know that our final product has a UP of 5.

2. Second step is then to go for cross multiplication which means:

"XP of first multiplier" x "UP of Second multiplier" = 2 x 5 = 10

"XP of Second multiplier" x "UP of First multiplier" = 4 x 3 = 12

3. Now the sum of above two products added with carry over if any, would give us the XP of our final answer which is to be derived as follows:

Sum of Cross Multiplication (from Step 2) = 10 + 12 = 22

Carry over from the UP (from Step 1) = 1

XP for the final product = 23

As we did in step 1, number 3 shall occupy the XP of final product and 2 shall be a carry over. so the answer constructed by us so far look like this _ _ 3 5. Now we shall go into final steps to identify TP and TXP of the final product.

4. Multiplication of XP of both multipliers i.e. 2 x 4 = 8.

We need to add the carry over of two from step 3 into this which would give us 8 + 2 = 10. So 0 is the TXP and 1 is TP of the final product.

Thus, the final product looks like this 1035.

Benefits of this method

Now let us revisit and see what we have actually done to achieve this product of two digit numbers:

1. Multiplication of single digit number

2. Sum of two digit numbers.

Thus, the requirement of multiplying two digit numbers has become very very easy using this Vedic Maths technique and with a little bit of practice on this lines would make you very quick in deriving products.

Concept

Also, to understand the concept better, let us take this pictorial demonstration of steps. one this is understood, you shall be able to apply the concept to even larger digit numbers as well.


So Vedic Maths Rocks!

I hope this would help you master the trick for multiplication of two digit numbers and shortly I shall come up with multiplication of larger digit numbers using wonders of Vedic Mathematics.

As I mentioned previously, in this attempt to simplify the study and make it more interesting, I am supported by Gyancircle.com. Do visit them.

Mathematics Redefined (Introduction of Vedic Maths)

Here I am going to talk about Vedic Mathematics, so lets first have an introduction as to what is Vedic Mathematics.


We all know that for higher academic achievements, successful business venture or even more so for day to day transactions, we need to master maths. With the gadgets to the likes of calculators, computers and even now tables, its easier to get through with basic number crunching requirements. However, what if we are shown a way of crunching numbers which is even faster than the use of these gadgets, easier and completely reliable!

Yes, that's right, we I am talking about a revolutionary methodology which would not just make the number crunching easy but would make maths fun for students. And I am not just talking about simplifying multiplication or division, also about solving complex equations just by a glimpse.

This method comes from the ancient Indian literature, written in Sanskrit language in forms of sutras and it contains the very logic behind currently used mathematics approach and shows a much simplified version.

Let me just give you an example of basic mathematics operation (i.e. multiplication):

Multiplication of 11 with any two digit number

Ex. 11 x 23

For this, without doing anything or even going into the table of two and three we have to go as follows:

"The sum of two digits (i.e. 2+3= 5) should be placed in between the two digits (i.e. 2 & 3)"

Thus, your answer is = 2 5 3 = 253.

Isn't it fun when you don't even have to use anything when faced with such requirements and before your friends could take out their gadgets punch in all the numbers, you have the answer ready with you.

This trick is not even equal to a drop if we consider Vedic Mathematics to be an Ocean!!!

Well I guess this in enough introduction and I shall continue to add more of this wonderful insights to be helpful to everyone coming across this.

Those, who are eager to find out more on revolutionary concepts like Vedic Mathematics, please feel free to write to me and also visit www.gyancircle.com wherein there are many more such things to find out.

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