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

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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