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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:
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.
If you like the article, you may contribute by:
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.
Step6: Fifth gross dividend-unit is 49. From 49 subtract Duplex(570) = 49 and get 0.
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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6 comments:
plz solve this problem 25281....
14047504... is a perfect square root of 3748
best complicated case is 114921
It is solved easily
In a six digit square number the hundred and unit place can be calculated easily. But how to say the tenth place or middle value instantly. Any clue?
Can you find the square root of 95824521 using this method? Please help.
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