Vedic Math - Cube Roots

Today, we are taking a next level of topic i.e. finding the cube root. With normal approach, finding cube root is bit complex. However, using Vedic Math techniques, it becomes interesting and fast too. This amazing technique will help you to find out the cube root of a  4 or 5 or 6 digits number quickly and all using mind power only. Technique specified in this article will work for perfect cubes only, not for other numbers (that we shall discuss in forthcoming articles). Lets start learning.

We know that, cube of a 2-digit number will have at max 6 digits (99³ = 970,299). This implies that if you are given with a 6 digit number, its cube root will have 2 digits. Further, following are the points to remember for speedy calculation of cube roots (of perfect cubes).

1. The lowest cubes (i.e. the cubes of the fist nine natural numbers) are 1, 8, 27, 64, 125, 216, 343, 512 and 729.
2. They all have their own distinct endings; with no possibility of over-lapping (as in the case of squares).
3. The last digit of the cube root of an exact cube is obvious:
• 1³ = 1    > If the last digit of the perfect cube = 1, the last digit of the cube root = 1
• 2³ = 8    > If the last digit of the perfect cube = 8, the last digit of the cube root = 2
• 3³ = 27  > If the last digit of the perfect cube = 7, the last digit of the cube root = 3
• 4³ = 64  > If the last digit of the perfect cube = 4, the last digit of the cube root = 4
• 5³ = 125 > If the last digit of the perfect cube = 5, the last digit of the cube root = 5
• 6³ = 216 > If the last digit of the perfect cube = 6, the last digit of the cube root = 6
• 7³ = 343 > If the last digit of the perfect cube = 3, the last digit of the cube root = 7
• 8³ = 512 > If the last digit of the perfect cube = 2, the last digit of the cube root = 8
• 9³ = 729 > If the last digit of the perfect cube = 9, the last digit of the cube root = 9
4. In other words,
• 1, 4, 5, 6, 9 and 0 repeat themselves as last digit of cube.
• Cube of 2, 3, 7 and 8 have complements from 10 (e.g. 10's complement of 3 is 7 i.e. 3+7=10) as last digit.
5. Also consider, that 
• 8's cube ends with 2 and 2's cube ends with 8 
• 7's cube ends with 3 and 3's cube ends with 7

If we observe the properties of numbers, Mathematics becomes very interesting subject and fun to learn. Following same, let’s now see how we can actually find the cube roots of perfect cubes very fast.

Example 1:  Find Cube Root of 13824

Step 1:
Identify the last three digits and make groups of three digits from right side. That is 13824 can be written as          
   13  ,   824
 
Step 2: 
Take the last group which is 824.  The last digit of 824 is 4.
Remember point 3, If the last digit of the perfect cube = 4, the last digit of the cube root = 4
Hence the right most digit of the cube root  = 4

Step 3:
Take the next group which is 13.
From point 3, we see that 13 lies between 8 and 27 which are cubes of 2 and 3 respectively. So we will take the cube root of the smaller number i.e. 8 which is 2.
So 2 is the tens digit of the answer.

We are done and the answer is '24'

Isn't that easy and fun..

Example 2:  Find Cube Root of 185193

Step 1:
 185193 can be written as          
   185  ,   193
 
Step 2:
Take the last group which is 193.  The last digit of 193 is 3.
Remember point 3, If the last digit of the perfect cube = 3, the last digit of the cube root = 7
Hence the right most digit of the cube root  = 7

Step 3:
Take the next group which is 185.
From point 3, we see that 185 lies between 125 and 216 which are cubes of 5 and 6 respectively. So we will take the cube root of the smaller number i.e. 125 which is 5.
So 5 is the tens digit of the answer.

So, the answer = 57.

Try some of the perfect cubes like 287496, 658503, 46656.

Isn't that interesting and easy technique! Try some examples, enjoy this interesting technique. We shall discuss how to calculate cube root of other numbers in coming articles.


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Vedic Math - Square Root-2

In this article, we shall continue discussing the remaining part of discussion 'how to find the Square Root using Vedic Math'. In last article , we discussed the technique for 4-digit numbers, In this article, we shall discuss the another technique which is useful for bigger numbers. In this method, we shall use "Duplex" (mentioned in General Squaring).

So, First observation:

• if number is 69563217 then n=8, Digits in the square root is 8/2=4, pairing is 69'56'32'17 and the first digit will be 8(8²=64)
• if number is 764613731 then n=9, Digits in the square root is (9+1)/2=5, pairing is 7'64'61'37'31 and the first digit will be 2 (2²=4)

Recall "Duplex"
• for a single digit 'a', D = a². e.g. D(4) = 16
• for a 2-digit number of the form 'ab', D = 2( a x b ). e.g. D(23) = 2(2x3) = 12
• for a 3-digit number like 'abc', D = 2( a x c ) + b². e.g. D(231) = 2(2x1) + 3² = 13
• for a 4-digit number 'abcd', D = 2( a x d ) + 2( b x c ) e.g. D(2314) = 2(2x4) + 2(3x1) = 22
• for a 5-digit number 'abcde', D = 2( a x e ) + 2( b x d ) + c² e.g. D(14235) = 2(1x5) + 2(4x3) + 2² = 38  and so on.

As we know how to calculate the duplex of a number, now we learn how to use it in calculating the square root of a number?
We will explain using an example.

Example:  √734449

Step1: n=6, Digits in the square root is 6/2=3, pairing is 73'44'49. Rearrange the numbers of two-digit groups from right to left as follows:
         | 73 :  4  4  4  9
.|    :
-----------------
.|    :
     As you see, in above representation, we provide spaces in front of the numbers to perform straight division, if required.

Step2: Now, find the perfect square less than the first group 73 i.e 64 and its square root is 8. Write down this 8 and the reminder 9 (73-64=9) as shown below:
         | 73 :   4   4   4   9
     16| 64 :9
     ------------------
         | 8  :

     We also calculate twice of number '8' (i.e. 8 x 2 = 16), and put that number to the left of the "|" on the second line as shown above. Here, number '16' is the divisor and which is always double of the quotient (here, quotient is 8).

Step3: Next is the gross dividend, the number which we have written after the colon on the second line appended in front of the next digit of the square. Thus, our gross dividend is 94.

 
     Since there are no digits to the right of the " : " on the answer line, we will not subtract anything here. If there are any digits on the answer line to the right of the " : ", then we calculate duplexes for that digit and subtract it from dividend. But here, without subtracting anything from the gross dividend, we divide 94 by the divisor 16 and put down the second Quotient digit 5 and the second reminder 14 in their proper place.

  

Step4: Third gross dividend-unit is 144. From 144 subtract 25 [ Duplex value of the second quotient digit (number to the right of the ":" on the answer line) D(5) = 25 ] ,get 119 as the actual dividend. Now, divide it by 16 and set the Quotient 7 and reminder 7 in their proper places.  

Step5: Fourth gross dividend-unit is 74. From 74 subtract Duplex D(57) [because D(57) = 2(5 X 7) = 70 ] obtain 4 , divide this 4 by 16 and put down Quotient as 0 and reminder 4 in their proper places

     We put a decimal point after the third digit since we know that the square root of a 6-digit number has to have 3 digits before the decimal point (mentioned in Step1).

Step6: Fifth gross dividend-unit is 49. From 49 subtract Duplex(570) = 49 and get 0.

  

 This means the work has been completed, the given expression is a prefect square and 857 is its square root.

Now, let us discuss some of the complicated cases.

Case 1: Take an example, which is complicated.
Example: √36481  (n=5, Digits in square root is (5+1)/2=3)
Step1:
       | 3 :  6  4  8  1
    2 |    :2
    --------------------
       | 1 :

Step2: Divisor 2, can fully divide 26 with quotient 13 and no remainder. But in the duplex method, we always restrict our quotients to be single digits. In other words, we add numbers to the answer row one digit at a time. Because of this, we put down 9 on the answer row as the quotient, and put down 8 as our next remainder (remember that 9*2 + 8 = 26).

       | 3 :  6   4   8   1
     2|    :2   8
 ---------------------------
       | 1 : 9

Step3: Next gross dividend is 84. From 84 subtract Duplex D(9),get 3. Divide this 3 by 2 and put down Quotient as 1 and reminder 1 in their proper places

        | 3 :  6   4    8    1
     2 |   :2    8   1
 ---------------------------
        | 1 : 9  1

Step4: Gross dividend is 18. From 18 subtract Duplex D(91), get 0. Divide this 0 by 2 and put down Quotient as 0 and reminder 0 in their proper places
        | 3 :  6    4    8    1
     2 |   : 2    8    1    0
 ---------------------------
       | 1 :  9     1     0

Step5: Gross dividend is 01. From 01 subtract Duplex D(910), get 0.
        | 3 : 6   4    8    1
     2 |   :2    8    1    0
 ---------------------------
        | 1 : 9   1 .  0    0
This completes the procedure. The final answer is 191.

Case 2: Now, we move to the next complication.
Example: √16384   (n=5, Digits in square root is (5+1)/2=3)
Step1:
        | 1 :  6   3   8   4
     2 |    :0   0
     -----------------
        | 1 :  3
We see that divisor 2, can fully divide 06 with quotient 3 and no remainder. This would then lead to a new gross dividend to 3, and a net dividend to -6 because the duplex of 3 is 9.
This type of complication occurs many times. To solve this problem, we reduce the second quotient to 2 and carry over a remainder of 2 to the next step.  As shown below:
| 1 :  6   3   8   4
 2 |    :0   2
    --------------------
    | 1 :  2

Step2: Next gross dividend is 23, and a net dividend is 19 (23 - the duplex of 2, which is 4).Divide this 19 by 2 and put down Quotient as 1 and reminder 1 in their proper places
        | 1 :  6   3   8   4
     2 |    :0   2   1
     -------------------
        | 1 :  2   9
Again, the same case arises. Divisor 2 divides 18 with quoitent 9 and reminder 0. And then the new gross dividend is 4 and net dividend is -32 (4 - D(29)= -32). So, we reduce the third quotient to 8 and carry over a reminder of 3 to the next step.
 | 1 :  6   3   8   4
      2 |    :0   2   3
     --------------------
        | 1 :  2   8

Step3: Gross dividend is 38 and net dividend is 6 (38 - D(28) = 6). Divide this 6 by 2 and put down Quotient as 3 and reminder 0 in their proper places
 | 1 :  6   3   8   4
      2 |    :0   2   3   0
     ------------------
         | 1 :  2   8 . 3
  But again net dividend comes negative. so, we reduce the quotient to 2, and we get
 | 1 :  6   3   8   4
      2 |    :0   2   3   2
     ------------------
        | 1 :  2   8 . 2
Here gross dividend is 24 and net dividend which is again negative. Again we reduce the quotient to 1, but again the net dividend comes negative. So, now the new quotient is 0.
 | 1 :  6   3   8   4
      2 |    :0   2   3   6
     ------------------
        | 1 :  2   8 . 0

Step4: Gross dividend is 64. From 64 subtract Duplex D(280), gets 0.
         | 1 :  6   3   8   4
      2 |    :0   2   3   6
     ---------------------
         | 1 :  2    8 . 0 0
The final answer is 128.


Following are few of the examples:

(1)   √552049   (n=6, Digits in square root is 6/2=3)
         |55 :  2    0    4    9
     14|     :6    6    2    0
--------------------
 |  7 :  4   3 .   0   0  (A perfect Square)

 
(2)   √14047504  (n=8, Digits in square root is 8/2=4)
       |14 :  0    4     7     5    0    4
    6 |     :5     8   11   13    1
---------------------------
        | 3 :  7   4    8   .   1 ....       ( Not a perfect square. As number of digits in square root is 4 and it didn't terminate after 4 digits )

(3) √119716  (n=6, Digits in square root is 6/2=3)
       |11 :  9    7    1    6
    6 |     :2    5     5    3
     ---------------------
       | 3 :  4    6 .  0    0   (A perfect Square)


Hopefully, this lesson will be helpful to handle the computation of square roots.

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Vedic Math - Square roots

Earlier we discussed "Squaring numbers near base" and "General Squaring through Duplex Process" and now we will find out how to calculate the square root of numbers. To understand this, let us first learn basic rules for finding the square root.

(1). The given number is first arranged in two-digit groups from right to left. If on left hand side, a single digit is left, that will also be counted as a group.
(2). The number of digits in the square root will be the same as the number of groups derived from the number. Examples are:

• 25 will be having one group as '25', hence square root should be of one digit.
• 144 will be having two groups as '44' and '1', hence the square root should be of two digits.
• 1024 will be having two groups as '24' and '10', hence the square root should be of two digits.

(3). If the given number has 'n' digits then the square root will have n/2 or (n+1)/2 digits
(4). The squares of the first nine natural numbers are 1,4,9,16,25,36,49,64, and 81. All of these squares end with 1, 4, 5, 6, 9, 0. This means

• An exact square never ends in 2, 3, 7 or 8
• If a number ends in 2, 3, 7 or 8, its square root will always be an irrational number
• If an exact square ends in 1, its square root ends in 1 or 9
• If an exact square ends in 4, its square root ends in 2 or 8
• If an exact square ends in 5, its square root ends in 5
• If an exact square ends in 6, its square root ends in 4 or 6
• If an exact square ends in 9, its square root ends in 3 or 7

(5). If a perfect square is an odd number, the square root is also an odd number
(6). If a perfect square is an even number, the square root is also an even number
(7). A whole number, which ends with an odd numbers of 0's, can never be the square of a whole number
(8). An exact square never ends in a 6 if the penultimate digit(digit that is next to the last digit) is even (eg. exact squares can not end in 26, 46, 86, etc.)
(9).An exact square never has an odd penultimate digit unless the final digit is a 6 (thus, exact squares can not end in 39,71, etc.)
(10).An exact square never ends with an even number when the last two digits taken together are not divisible by 4 (thus, no exact square can end in 22, 34 and other non-multiples of 4 if the last digit is even)

Firstly, we use "The First by the First and the Last by the Last" technique to solve the square root.

(1). √6889
     There are two groups of figures, '68' and '89'. So we expect 2-digit answer.
     Now see since 68 is greater than 64(8²) and less than 81(9²), the first figure must be 8.

     So, 6889 is between 6400 and 8100, that means, between 80² and 90².
     Now look at the last figure of 6889, which is 9.
     Squaring of numbers 3 and 7 ends with 9.
     So, either the answer is 83 or 87.
     There are two easy ways of deciding. One is to use the digit sums.
     If 87² = 6889
     Then converting to digit sums
     (L.H.S. is 8+7 = 15 -> 1+5 -> 6 and R.H.S. is 6+8+8+9 -> 31 -> 3+1 -> 4)
     We get 6² -> 4, which is not correct.
     But 83² = 6889 becomes 2² -> 4, so the answer must be 83.
     The other method is to recall that since 85² = 7225 and 6889 is below this. 6889 must be below 85. So it must be 83.

Note: To find the square root of a perfect 4-digit square number we find the first figure by looking at the first figures and we find two possible last figures by looking at the last figure. We then decide which is correct either by considering the digit sums or by considering the square of their mean.

(2). √5776
     The first 2-digit(i.e. 57) at the beginning is between 49 and 64, so the first figure must be 7.
     The last digit (i.e. 6) at the end tells us the square root ends in 4 or 6.
     So the answer is 74 or 76.
     74² = 5776 becomes 2² -> 7 which is not true in terms of digit sums, so 74 is not the answer.
     76² = 5776 becomes 4² > 16 -> 7, which is true, so 76 is the answer.
     Alternatively to choose between 74 and 76 we note that 75² = 5625 and 5776 is greater than this so the square root must be greater than 75. So it must be 76.

Second technique is useful for bigger numbers and in this method, we use "Duplex". In the next article, we shall continue to discuss this second technique. Until then, good luck and happy computing!!


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Vedic Math - Fourth Power of 2 Digit Numbers

We discussed the cube of 2-digit number in previous article. In this article, we shall describe the fourth power of 2-digit numbers using the same formula.

The Algebraic Expression of (a + b)⁴

 (a + b)⁴ = a⁴ + 4a³b + 6a²b² + 4ab³+ b⁴

 We can rewrite the above equation as:
           a⁴       a³b          a²b²           ab³        b⁴

                   3a³b        5a²b²          3ab³
So, apply the same rule which we applied in previous article, while finding cubic of the number. Consider the first term as a⁴ and the remaining terms get multiplied by b/a with the previous term.
The Difference comes in second row, in fourth power, we multiply 2nd and 4th term by 3 and 3rd term by 5.

Example: 11⁴

            1    1    1    1    1
                 3    5    3
          -------------------------
            1    4    6    4    1
          -------------------------

Example: 32⁴
         
            81     54      36     24     16
                   162    180     72
          -------------------------------------
          104      8       5        7       6
          -------------------------------------

The "Binomial Theorem" is thus capable of practical application more comprehensively in Vedic Math. Here it is been utilised for splendid purpose as described above, with Vedic Sutras.

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Vedic Math - Cube of 2 Digit Numbers

Cube of numbers plays an important role in mathematics calculations, like in finding cube root of the numbers. So it is useful if we can do the cube of numbers quickly. We are trying for the same here using 'Anurupya Sutra' of Vedic Math. Let us learn how we can do it.

Cubes of the single digits i.e. from (1 to 9) are given below:
 1³  = 1,              2³  = 8,              3³  = 27,            4³  = 64,            5³  = 125,
6³  = 216,          7³  = 343,          8³  = 512,           9³  = 729,          10³  = 1000

If we observe closely, the last digit of every cubic number is unique i.e. numbers from (1 to 9) does not repeat. This observation will be very helpful while calculating cube roots of any 2-digit numbers.

Let us first see the Algebraic Expression for Cube root:

(a + b)³ = a³ + 3a²b + 3ab²+ b³

Above expression for cube root of (a + b) contain 4 terms in total.

• 1st term is  a³
• 2nd term is a²b =  a³  x (b/a)  = 1st term x (b/a)
• 3rd term is ab² = a²b x (b/a) = 2nd term x (b/a)
• 4th term is b³   = ab² x (b/a) = 3rd term x (b/a)

Here (b/a) is the common ratio

Also, as the whole, 2nd term is 3a²b = a²b + 2a²b          {split as sum of two terms}
                      and, 3rd term is 3ab² = ab² + 2ab²           {split as sum of two terms}

So to find the cube, we have to compute a³ and b/a.

In Vedic Math, same formula can be used in a different way to find the cube of 2-digit numbers i.e. ab. Apply formula on 'ab' like (a+b)³ as stated above, and add the results of different rows in vertical columns. You will be able to do the cube of any two digit numbers quickly.

We shall use 'Anurupya Sutra' of Vedic Math for this cube calculation, which states:
"If you start with the cube of first digit and take the next three numbers (in the top row) in a Geometrical Proportion (in the ratio of original digits themselves), you will find that the fourth figure on the right hand will be just the cube of second digit".

Following is the step by step description of finding the cube of 2-digit number:

• Step 1: In the first row, start with a^3 as 1st term and multiplying 1st term by (b/a) to get 2nd term.
• Step 2: Repeat the multiplication till 4th term.
• Step 3: In the second row, double the two middle terms (i.e. 2nd term and 3rd term) and write just below 2nd term and 3rd term.
• Step 4: Add them vertically in columns. Carry forward the 10th place digit to next column.

The example given below will describe this method well.

Example: 11³
Here a = 1 , b = 1 ,  a³ = 1 ,  b/a = 1/1 = 1          (Here common ratio is equal to 1)

Now see the formation of the table:
First Row             1     1     1     1
Second Row               2     2
                    -------------------------
Add                     1     3     3     1
11³ = 1331

Example: 13³
Here a = 1 , b = 3 ,  a³ = 1 ,  b/a = 3/1 = 3          (Here common ratio is greater than 1)

Now see the formation of the table:
First Row              1     3     9     27       (Note: 4th term is just  b³  as shown above in algebraic  
Second Row                6     18               expression)
                          -------------------------
Add                      1     9     27    27
                          -------------------------
                         =   1     9      7     7       (Apply carry over rule)
                                    2     2
 =   1      9      9     7
                    2
                 =   1      1     9     7
                                    1
 =   2      1     9     7
13³= 2197

Example: 52³
Here a = 5 , b = 2 ,  a³ = 125 ,  b/a = 2/5            (Here common ratio is less than 1)

Now see the formation of the table:
First Row                 125     50     20     8
Second Row                     100     40
                           ---------------------------
Add                         125   150     60     8
                           ---------------------------
                           =  125       0        0     8        (Apply carry over rule)
                                        15        6
  =  140      6    0    8
52³= 140608

Some more examples are as follows:
(1) 16³ =  1    6    36    216
                       12    72
               -----------------------
                   4    0     9      6
               -----------------------

(2) 32³ =  27    18    12     8
                           36    24
                -----------------------
                   32     7      6      8
                -----------------------

(3) 97³ =  729    567      441      343
                           1113      882
                     -----------------------------
                    912       6          7          3
                     ----------------------------- 
or better way for number near base, 97³ = (100-3)³    where a=100, b= -3 and b/a= -3/100
                                                                       = 1000000    -30000      900    -27
                                                                       =                   -60000    1800                                                                                                                                                                      
                                                                           ------------------------------------
  1000000    -90000    2700    -27
   ------------------------------------
                                                                     = 912673

Hope it will help!!                        
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Vedic Math - General Squaring

Today, the topic which we are going to discuss is the 'General Procedure to Square any number'. Earlier we discussed about the squaring of numbers near base, however, general procedure is another nice formula to do the squaring and is applicable universally. The method or sutra is "Vertically and Crosswise", but here it is used in a different sense; based on a procedure known as 'Dwandwa Yoga' or 'Duplex Combination Process' or 'Duplex'; denoted as (D).

'Duplex' term is used in two different sense; for squaring and for multiplication. And for current formula, it will be used in both the senses. If we are having a single or central digit, then 'Duplex' means squaring that digit (a² ). Secondly it can be used for even digits number or on numbers having equidistant digits, then 'Duplex' means to double of cross multiplication of the equidistant numbers (2ab). This concept is very important to understand the current formula and will be used in future articles also. Let us see few example to understand it more:

For 1 digit  – D(a) = single digit = a²
                     e.g. D(5) = 5²  = 25
For 2 digits – D(ab) = even digits number = twice the product of the digits (2ab)
                     e.g. D(26) = 2(2)(6) = 24
For 3 digits – D(abc) = product of equidistant digits from center and square of center digits
     = twice the product of the outer digits (2ac) + the square of the middle digit (b² )
                     e.g. D(734) = 2(7)(4) + 3²
                        = 56 + 9 = 65
For 4 digits – D(abcd) = product of equidistant numbers 
                    = twice the product of the outer digits (2ad) + twice the product of the inner digits (2bc)
                     e.g. D(1034) = 2(1)(4) + 2(0)(3)
                           = 8 + 0 = 8
For 5 digits – e.g. D(10345) = product of equidistant digits and square of center digits
= 2(1)(5) + 2(0)(4) + 3² 
                        = 10 + 0 + 9 = 19

and so on. This is called Duplex.


Now, let us come to original question i.e. how to square a number. And the square of any number is just the total of its Duplexes.

For Example,
34² = 1156
= D(3) = 9, D(34) = 24, D(4) = 16

Combining these three results in the usual way, we get
= 9 | 24 | 16

Now add these results as
= 9 |   4 |   6
        2     1    
= 9 |   5 | 6
        2
= 11 | 5 | 6
= 1156

56²  = 3136
     D(5) = 25, D(56) = 60, D(6) = 36
by combining, we get 25 / 60 / 36 = 3136

Equivalent Algebraic Expression is: (10a + b)²  = 100(a² ) + 10(2ab) + b² .

This method can also be explained by multiplying a number by itself using the general multiplication method.

Note :- If a number consists of n digits, its square must have 2n or 2n-1 digits.

Following are some more examples:

263²  =
      D(2) = 4, D(26) = 24, D(263) = 48, D(63) = 36, D(3) = 9
      4 / 24 / 48 / 36 / 9 = 69169
4332²  =
      D(4) = 16, D(43) = 24, D(433) = 33, D(4332) = 34, D(332) = 21, D(32) = 12, D(2) = 4
      16 / 24 / 33 / 34 / 21 / 12 / 4 = 18766224

3247²  = 9 / 12 / 28 / 58 / 46 / 56 / 49 = 10543009

46325²  = 16 / 48 / 60 / 52 / 73 / 72 / 34 / 20 / 25 = 2146005625


We hope this method will help you in squaring of any number quickly. If you find this difficult, you may use another method which we have discussed earlier( Squaring numbers near base ). Every method will become easy with practice. In our next article, we shall discuss about the cubing of the number.


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