In the article a vedic mathematics subsutra anurupyena vidhi has been extended to find the nth power of an integer of any number of digits and of a rational number with terminating condition by applying binomial theorem for positive integral index, which is an attempt to correlate vedic mathematics with modern mathematics. The article reveals one of the logic hidden in this vedic sutra. Thus an attempt has been made to explain the unconventional aspects of method. Some people may find it difficult at first reading to understand the arithmetical operations but would play and enjoy it after some exercises.

## Abstract

In the article a vedic mathematics subsutra anurupyena vidhi has been extended to find the nth power of an integer of any number of digits and of a rational number with terminating condition by applying binomial theorem for positive integral index, which is an attempt to correlate vedic mathematics with modern mathematics. The article reveals one of the logic hidden in this vedic sutra. Thus an attempt has been made to explain the unconventional aspects of method. Some people may find it difficult at first reading to understand the arithmetical operations but would play and enjoy it after some exercises.

**Key Words:** Sutra (Formula), Subsutra, Anurupyena Vidhi, Binomial Theorem.

**Mathematics Subject Classification 2010:** 00A22, 05A10, 11B65

## Introduction

The Vedas are ancient Indian texts containing a record of human experience and knowledge. It is believed as the fountainhead and illimitable store-house of all knowledge. The ** Vedas are Four ** in number named as Yajur, Sama, Atharva and Rgveda, but they have also

**: Ayurveda, Gandharvaveda, Dhanurveda, Sthapatyaveda and**

*Four Upavedas***. Mathematics is regarded to fall under the Upaveda Sthapatyaveda. From this, Vedic Mathematics was rediscovered by**

*Six Vedangas***Swami Bharati Krishna Tirthaji Maharaja**(1884-1960) of Govardhana Math, Puri during 1911 to 1918.

This system of mathematics is based on ** Sixteen Sutras **: Ekadhikena Purven, Nikhilam Navatascaramam Dasatah, Urdhva-tiryagbhyam, Paravartya Yojayet, Sunyam Samayasamuccaye, Anurupye Sunyamanyat, Sankalana-vyavakalanabhyam, Puranapuranabhyam, Calana-kalanabhyam, Yavadunam, Vyastisamastih, Sesanyankena Caramena, Sopantyadvayamantyam, Ekanyunena Purvena, Gunitasamuccayah, Gunakasamuccayah and

**: Anurupyena, Sisyate Sesasamjnah, Adyamadyenantyamantyena, Kevalaih Saptakam Gunyat, Vestanam, Yavadunam Tavadunam, Yavadunam Tavadunikrtya Varganca Yojayet, Antyayordasake’pi, Antyayoreva, Samuccayagunitah, Lopanasthapanabhyam, Vilokanam, Gunitasamuccayah Samuccayagunitah; which describes the natural thinking pattern of mind. It is a common belief that through vedic mathematics we use both part of our brain keeping us mentally fit. It deals mainly with various sutras and their applications for carrying out tedious and cumbersome arithmetical operations, and to a large extent, executing them mentally in short.**

*Thirteen Sub-sutras*As far as the binomial theorem is concerned, Greek mathematician **Euclid** knew it for exponent 2, whereas the theorem for cubes was known in India by 6th century. The binomial theorem can also be found in the work of 11th century Persian mathematician **Al-Karaji**. The binomial expansion of small degrees was known in the 13th century mathematical works of **Yang Hui** and **Chu Shih-Chieh**. In 1544, **Michael Stifel** introduced the term “binomial coefficient”. **Issac Newton** is credited with the generalized binomial theorem for any rational exponent.

## Preliminary Ideas

We should have the basic knowledge of the two terms one from vedic sutra and another from modern mathematics to understand the present article:

**Anurupyena Sub-Formula**

**Swami** states that the upasutra or sub-formula or sub-sutra ‘ *Anurupyena* ’ means ‘proportionately’. In actual application, it connotes that, in all cases where there is a rational ratio-wise relationship, the ratio should be taken into account and should lead to a proportionate multiplication or division as the case may be.

In other words, when neither the multiplicand nor the multiplier is sufficiently near a convenient power of 10 which can suitably serve as a base, we can take a convenient multiple or sub-multiple of a suitable base, as our ‘working base’, perform the necessary operation with its aid and then multiply or divide the result proportionately, i.e. in the same proportion as the original base may bear to the working base actually used by us. This is used in multiplying two numbers and so useful in finding square, cubes, etc.

**Sharma** (1996) states that this sub-formula is used to find the square of a number, which is neither near to base number nor near to sub-base number, for example 66. **Shashtri** (2011) states the limits of this upasutra that this is useful for all those numbers which are of two or three digits numbers.

**Binomial Theorem**

We know that if a and b are two real numbers and n is a positive integer, then

illustration not visible in this excerpt

where [illustration not visible in this excerpt] and illustration not visible in this excerpt.

## Anurupyena Binomial Method

**Yadav** (2007) has propounded this method to find the nth power of any integer of any number of digits and a rational number with terminating condition. He observed certain pattern of calculation in finding the square and cube of integers using vedic sutra but some limits have also been found in the application of this sutra. No general rule has been mentioned in the vedic mathematics text or in any others to find the nth power of any integers. He combined the anurupyena sutra and binomial theorem for positive integral index to result out this method. That’s why it has been named as anurupyena binomial method. *Yadav* has discussed the following general method to find nth power:

**Nth POWER OF A POSITIVE INTEGER**

To find the nth power of any positive integer of any number of digits, we break the given number into two parts A starting from the left side digit and B the remaining digits of the given number. For this we first find the number of digits (let it be R) in the given number whose nth power is to be found. There may be two cases:

**Case I**: When the number of digits (R) is odd (let 2m +1), then part A has (m+1) digits from the left side and part B will have the remaining ‘m’ digits.

**Case II**: When the number of digits (R) is even (let 2m), then part A has ‘m’ digits from the left side and part B will have the remaining ‘m’ digits. In this case both parts A and B have the same number of digits.

It should be kept in mind that in finding A and B, there must not be any change in the order of digits of the given number.

Now we proceed as follows:

1. Make (n+1) columns and write the (n+1) terms of right hand side of the binomial theorem for positive integral index

illustration not visible in this excerpt

where illustration not visible in this excerpt, as follows:

illustration not visible in this excerpt

We must not change the order of the columns to avoid the wrong result.

2. Evaluate each column separately.

3. Divide the number of digits R of the given number by 2 and find the quotient Q. Q may also be found as: If R = 2m+1 or 2m, where m is a positive integer, then Q = m.

4. Now to find the required result we combine all n+1 parts as:

i. Take as many digits as the quotient value is from the right side of column-(n+1) (let it be Pn+1) and add the remaining digits (number) in its left side in column–n.

ii. Again take as many digits as the quotient value is from the right side of the column–n (let it be Pn) and write it in the left side of Pn+1 (like Pn Pn+1) and add the remaining digits (number) in its left side column-(n-1).

iii. Repeat (i) and (ii) up to column-2 till P2 and finally find the number P1 in column-1 after adding remaining digits (numbers) of column–2 and write P1 to the left of P2.

Thus we have already found the required nth power of the given positive integer as

illustration not visible in this excerpt.

Here between any two Pi, (where i = 1, 2, ., n+1), there is no operation like multiplication. It is just representation of numbers only. Here all P2 , P3,…., Pn+1 are numbers of m digits, where m is equal to the quotient Q.

From this method we can develop methods for different exponent values of n one by one as follows:

## TO FIND THE SQUARE OF A POSITIVE NUMBER

To find the square of any positive integer of any number of digits, we break the given number into two parts A starting from the left side digit and B the remaining digits of the given number using the process discussed above. Thereafter we proceed as follows:

1. Make 3 columns and write the 3 terms of right hand side of the binomial theorem for positive integral index

illustration not visible in this excerpt

as follows:

illustration not visible in this excerpt

2. Evaluate each column separately.

3. Divide the number of digits R of the given number by 2 and find the quotient Q. Q may be found as: If R = 2m+1 or 2m, where m is a positive integer, then Q = m.

4. Now to find the required result we combine all 3 parts as:

i. Take as many digits as the quotient value is from the right side of column-3 (let it be P3) and add the remaining digits (number) in its left side in column–2.

ii. Again take as many digits as the quotient value is from the right side of the column–2 (let it be P2) and write it in the left side of P3 (like P2 P3) and add the remaining digits (number) in its left side column-1.

iii. Finally find the number P1 in column-1 after adding remaining digits (numbers) of column–2 and write P1 to the left of P2.

Thus we find the square of the positive integer as illustration not visible in this excerpt.

Let us examine the method with following examples:

**Example 1**: Find the square of 13.

The given number is of 2 digit, therefore R = 2, A = 1, B = 3. We make the three columns as:

illustration not visible in this excerpt

Thus 13[2] = 169. Since here R = 2, Q = 1, we take only one digit from right side of column-3 & column-2 respectively.

**Example 2**: Find the square of 312.

Here R = 3, A = 31, B = 2. We make the three columns as:

illustration not visible in this excerpt

Thus 31[2] = 97344. Here R = 3, Q = 1, therefore we have taken one digit from right side. The underlined number in column-2 has been added in its left side in column-1.

**Example 3**: Find the square of 8394.

Here R = 4, A = 83, B = 94. We make the three columns as:

illustration not visible in this excerpt

Thus 8394[2] = 70459236. Here R = 4 and Q = 2. Therefore we have taken two digits from right side.

**Example 4**: Find the square of 33333.

Here R = 5, A = 333, B = 33. We make the three columns as:

illustration not visible in this excerpt

Thus 33333[2] = 1111088889. Here R = 5 and Q = 2, Therefore we have taken two digits from right side.

**Example 5**: Find the square of 389456.

Here R = 6, A = 389, B = 456. We make the three columns as:

illustration not visible in this excerpt

Thus 389456[2] = 151675975936. Here R = 6 and Q = 3, Therefore we have taken three digits from right side.

**Example 6**: Find the square of 1111111111.

Here R = 10, A = 11111, B = 11111. We make the three columns as:

illustration not visible in this excerpt

Thus 1111111111[2] = 1234567900987654321. Here R = 10 and Q = 5, therefore we have taken five digits from right side.

**Example 7**: Find the square of 1000.

Here R = 4, A = 10, B = 00. We make the three columns as:

illustration not visible in this excerpt

Thus (1000)[2] = 1000000. Here R = 4 and Q = 2, therefore we have taken two digits from right side of column-3 and column-2 respectively. Since we have to take two digits from the right side of column-3 and column-2, therefore we can write 0 = 00.

**Example 8**: Find the square of 6.

Here the given number is of only one digit and to apply the method, we must have at least two digits. Therefore we write it as 6 = 06. Now R = 1, A = 0, B = 6. We make the three columns as:

illustration not visible in this excerpt

Thus (6)[2] = (06)[2] = 36. Here A = 0 and Q = 1, therefore we have taken one digit from the right side of column-3 and column-2 respectively.

**TO FIND THE THIRD POWER OF A POSITIVE INTEGER**

To find the third power or cube of any positive integer of any number of digits, we break the given number into two parts A starting from the left side digit and B the remaining digits of the given number using the process discussed earlier. Thereafter we proceed as follows:

1. Make 4 columns and write the 4 terms of right hand side of the binomial theorem for positive integral index

illustration not visible in this excerpt

as follows:

illustration not visible in this excerpt

2. Evaluate each column separately.

3. Divide the number of digits R of the given number by 2 and find the quotient Q. Q may also be found as: If R = 2m+1 or 2m, where m is a positive integer, then Q = m.

4. Now to find the required result we combine all 4 parts as:

i. Take as many digits as the quotient value is from the right side of column-4 (let it be P4) and add the remaining digits (number) in its left side in column–3.

ii. Again take as many digits as the quotient value is from the right side of the column–3 (let it be P3) and write it in the left side of P4 (like P3 P4) and add the remaining digits (number) in its left side column-2.

iii. Repeat (i) and (ii) up to column-2 till P2 and finally find the number P1 in column-1 after adding remaining digits (numbers) of column–2 and write P1 to the left of P2.

Thus we have already found the required nth power of the given positive integer as illustration not visible in this excerpt.

Let us see the method working in finding the cube of the numbers as:

**Example 9**: Find the cube of 68.

Here the given number is of two digits, therefore R=2, A=6, B=8. Now we make the four columns as follows:

illustration not visible in this excerpt

Thus 68[3] = 314432. Here R=2, Q=1. Therefore we have taken one digit from right side of column-4, column-3, and column-2 respectively.

**Example 10**: Find the cube of 668.

Here the given number is of three digits, therefore R = 3, A = 66, B = 8. Now we make the four columns as follows:

illustration not visible in this excerpt

Thus 668[3] = 298077632. Here R = 3 and Q=1, therefore we have taken one digit from right side.

Example 11: Find the cube of 3333.

Here the given number is of four digits, therefore R = 4, A = 33, B = 33. Now we make the four columns as follows:

illustration not visible in this excerpt

Thus 3333[3] = 37025927037. Here R=4 and Q=2, therefore we have taken two digits from right side.

Example 12: Find the cube of 1000.

Here the given number is of four digits, therefore R = 4, A = 10, B = 00. Now we make the four columns as follows:

illustration not visible in this excerpt

Thus (1000)[3] = 1000000000. Here R = 4 and Q=2, therefore we have taken two digits from right side of column-4, column-3, column-2 respectively. Since here we have to take two digits from the right side and in last three columns we have only one digit, so we can write 0=00 as it is necessary to follow this rule according to the method discussed earlier.

## TO FIND THE FOURTH POWER OF A POSITIVE INTEGER

To find the 4th power of any positive integer of any number of digits, we break the given number into two parts A starting from the left side digit and B the remaining digits of the given number using the procedures discussed earlier. Thereafter we proceed as follows:

1. Make 5 columns and write the 5 terms of right hand side of the binomial theorem for positive integral index

illustration not visible in this excerpt

as follows:

illustration not visible in this excerpt.

2. Evaluate each column separately.

3. Divide the number of digits R of the given number by 2 and find the quotient Q. Q may also be found as: If R = 2m+1 or 2m, where m is a positive integer, then Q = m.

4. Now to find the required result we combine all five parts as:

i. Take as many digits as the quotient value is from the right side of column-5 (let it be P5) and add the remaining digits (number) in its left side in column–4.

ii. Again take as many digits as the quotient value is from the right side of the column–4 (let it be P4) and write it in the left side of P5 (like P4 P5) and add the remaining digits (number) in its left side column-3.

iii. Repeat (i) and (ii) up to column-2 till P2 and finally find the number P1 in column-1 after adding remaining digits (numbers) of column–2 and write P1 to the left of P2.

Thus we find the required 4th power of the given positive integer as illustration not visible in this excerpt.

Let us see the method working on some examples:

**Example 13**: Find the fourth power of 12.

Since the given number is of two digits, therefore R = 2, A = 1, B = 2. Now we make the five columns as follows:

illustration not visible in this excerpt

Thus 12[4] = 20736. Here R = 2 and Q = 1, therefore we have taken one digit from right side.

Example 14: Find the fourth power of 111.

Here the given number is of three digits, therefore R = 3, A = 11, B = 1. Now we make the five columns as follows:

illustration not visible in this excerpt

Thus 111[4] = 151807041. Here R = 3 and Q = 1, therefore we have taken one digit from right side.

Example 15: Find the fourth power of 1111.

Here the given number is of four digits, therefore R = 4, A = 11, B = 11. Now we make the five columns as follows:

illustration not visible in this excerpt

Thus (1111)[4] = 1523548331041. Here R = 4 and Q = 2, therefore we have taken two digits from right side.

**Example 16**: Find the fourth power of 1302.

Here R = 4, A = 13, B = 02. We make the five columns as:

illustration not visible in this excerpt

Thus (1302)[4] = 2873716601616. Here R = 4 and Q = 2. Therefore we have taken two digits from the right side of column-5, column-4, column-3, column-2 respectively.

## TO FIND THE FIFTH POWER OF A POSITIVE INTEGER

To find the 5th power of any positive integer of any number of digits, we break the given number into two parts A starting from the left side digit and B the remaining digits of the given number using the procedures discussed earlier. Thereafter we proceed as follows:

1. Make 6 columns and write the 6 terms of right hand side of the binomial theorem for positive integral index

illustration not visible in this excerpt

as follows:

illustration not visible in this excerpt

2. Evaluate each column separately.

4. Now to find the required result we combine all six parts as:

i. Take as many digits as the quotient value is from the right side of column-6 (let it be P6) and add the remaining digits (number) in its left side in column–5.

ii. Again take as many digits as the quotient value is from the right side of the column–5 (let it be P5) and write it in the left side of P6 (like P5 P6) and add the remaining digits (number) in its left side column-4.

iii. Repeat (i) and (ii) up to column-2 till P2 and finally find the number P1 in column-1 after adding remaining digits (numbers) of column–2 and write P1 to the left of P2.

Thus we have already found the required 5th power of the given positive integer as illustration not visible in this excerpt.

**Example 17**: Find the 5th power of 23.

Since the given number is of two digits therefore R = 2, A = 2, B = 3. We have six columns:

illustration not visible in this excerpt

Thus illustration not visible in this excerpt. Since Q = 1, we have taken only one digit from right side.

## TO FIND THE SIXTH POWER OF A POSITIVE INTEGER

## To find the 6th power of any positive integer of any number of digits, we break the given number into two parts A starting from the left side digit and B the remaining digits of the given number using the method discussed earlier. Thereafter we proceed as follows:

1. Make 7 columns and write the 7 terms of right hand side of the binomial theorem for positive integral index

illustration not visible in this excerpt

as follows:

illustration not visible in this excerpt

2. Evaluate each column separately.

4. Now to find the required result we combine all seven parts as:

i. Take as many digits as the quotient value is from the right side of column-7 (let it be P7) and add the remaining digits (number) in its left side in column–6.

ii. Again take as many digits as the quotient value is from the right side of the column–6 (let it be P6) and write it in the left side of P7 (like P6 P7) and add the remaining digits (number) in its left side column-5.

iii. Repeat (i) and (ii) up to column-2 till P2 and finally find the number P1 in column-1 after adding remaining digits (numbers) of column–2 and write P1 to the left of P2.

Thus we find the required 6th power of the given positive integer as illustration not visible in this excerpt.

**Example 18**: Find the 6th power of 21.

Since the given number is of two digits therefore R = 2, A = 2, B = 1. Here we have to make seven columns:

illustration not visible in this excerpt

Thus 21[6] = 85766121. Here R = 2, Q = 1, therefore we have taken one digit from the right side of each column up to column-2.

## TO FIND THE Nth POWER OF A NEGATIVE INTEGER

To find the nth power of negative integers, we use the fact (-k)n = (-1)n.(k)n, where k is any positive integer. Now we find (k)n as the method discussed before and multiply it by +1 or –1 according to (–1)n when ‘n’ is even or odd. Thus we get the result.

Example 19: Find the nth power of (–21) if n=6.

Since we can write (–21)[6] = (-1)[6] (21)[6]. But 21[6] = 85766121 and (-1)[6] = 1.

Therefore (-21)[6] = 85766121

Example 20: Find the nth power of (-12) if n=3.

Since we can write (-12)[3] = (-1)[3] 12[3]. But 12[3] = 1728 and (-1)[3] = -1.

Therefore (-12)[3] = -1728

## TO FIND THE Nth POWER OF A RATIONAL NUMBER

Let the given rational number [illustration not visible in this excerpt] be terminating and in terminating decimal form is represented by

illustration not visible in this excerpt

where ‘.’ between ai and b1 is a decimal point.

To find the nth power of illustration not visible in this excerpt, we proceed as follows:

1. Find the nth power of (a1a2a3aib1b2b3br) according to the method discussed above without considering the decimal point between ai and b1.

2. Now put the decimal point in the result (1) before (n.r) digits from the right side.

3. If the given rational number is negative, take positive or negative sign according to ‘n’ is even or odd respectively.

Thus we get the required result.

Example 21: Find the nth power of [illustration not visible in this excerpt] if n=3.

Since we have illustration not visible in this excerpt

Therefore to find the cube of it, we first find the cube of 3875 without considering the decimal point. Here we have R = 4, A = 38, B = 75.

Now we make the table of four columns as follows:

illustration not visible in this excerpt

Thus 3875[3] = 58185546875. Since here n = 3, r = 2, therefore n.r = 3.2 = 6. Hence we put decimal point before six digits from the right side. Therefore illustration not visible in this excerpt

Note: This method is also applicable for n = 1. But in this case no extra operation is required.

Applications

If we know the nth power of 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 and have simple knowledge of multiplication then we can find the nth power of any integer or rational numbers in terminating decimal form by A-B Method.

Limits of the Method

This method is not applicable for finding the nth power of irrational numbers and rational numbers (in non-terminating decimal forms).

## References

*Maharaja S. B. K. T.*, Vedic Mathematics, Motilal Banarsi Dass Publishers Pvt Ltd, 17-22, 1992, Delhi, India

*Murthy T. S. B.*, A Modern Introduction to Ancient Indian Mathematics, Wiley Eastern Ltd, 75-82, 1992, Delhi, India

*Sharma S. S.*, Application of Vedic Mathematics in Competitive Arithmetic, Series-1, Sam Samyiki Ghatna Chakra, 32-33, Dec.1996, Allahabad, India

*Shashtri R.*, Vedic Mathematics Made Easy, Arihant Publications (I) Pvt Ltd, 42-43, 51, 2011, Meerut, India

*Shrivastava C. M.*, Vedic Ganita Paddati, Manoj Publications, 184-185, 2011, Delhi, India

*Yadav D. K.*, Aanuruppen Binomial Method to find the nth power of Integers and Rational Numbers in Terminating Decimal Forms, Acta Ciencia Indica, 33M (2), 647-655, 2007, Meerut, India

*Knnojiya D. S.* & *Yadav D. K.*, Algorithm of Aanuruppen Binomial Method, Int. J of Math. Sci. & Engg. Appls, 2(4), 101-112, 2008, Pune, India https://en.wikipedia.org/wiki/Binomial_theorem

- Quote paper
- Dr. Dharmendra Kumar Yadav (Author), 2007, Anurupyena Binomial Method, Munich, GRIN Verlag, https://www.hausarbeiten.de/document/418900