Question

Find the square root of 126736 by division method
A. 355
B. 356
C. 366
D. 465

Answer 1

Hint: We use the division method to calculate the square root of the given number after we pair up the elements of the number starting from the right hand side of the number. Create a pair of two and the leftmost left one can be taken as single. Find such a number which when multiplied to the digit in its one’s place gives a number less than or equal to the dividend. Double the quotient and make it as the next divisor and continue the same process.

Complete step-by-step solution:
First we pair the digits in the number in pairs of two each and the remaining one (if any) starting from the right side.
\[126736 = \overline {12} \overline {67} \overline {36} \]
Then we take the highest number whose square will be less than or equal to the first pair i.e. 12
So, we know, \[1 \times 1 = 1,2 \times 2 = 4,3 \times 3 = 9\]
We choose \[3 \times 3 = 9\] because \[9 < 12\]
Now we divide the number by taking this number as a divisor and taking the same number as the quotient.
\[
  3\mathop{\left){\vphantom{1{\overline {12} \overline {67} \overline {36} }}}\right.
\!\!\!\!\overline{\,\,\,\vphantom 1{{\overline {12} \overline {67} \overline {36} }}}}
\limits^{\displaystyle \,\,\, 3} \\
   - 9 \\
  \overline { = 367}
 \]
Now the remainder becomes the next dividend and the new divisor is twice the old divisor followed by a digit which makes a number such that the square of that number will be less than or equal to the new dividend.
So, we have a new dividend as 367 and we can have a divisor as \[2 \times 3\underline {} = 6\underline {} \] where blank is filled by a digit.
Now we try to find a number in the lane of the sixties whose square is less than or equal to our new dividend.
\[
  61 \times 1 = 61 \\
  62 \times 2 = 124 \\
  63 \times 3 = 189 \\
  64 \times 4 = 256 \\
  65 \times 5 = 325 \\
  66 \times 6 = 396
 \]
We can clearly see that \[65 \times 5 = 325\] suits our requirement because \[325 < 367\]
Now we divide the dividend by the number 65 and the quotient 5 comes beside the earlier quotient.
\[
  65\mathop{\left){\vphantom{1{367\overline {36} }}}\right.
\!\!\!\!\overline{\,\,\,\vphantom 1{{367\overline {36} }}}}
\limits^{\displaystyle \,\,\, {35}} \\
   - 325 \\
  \overline { = 4236}
 \]
Now the remainder becomes the next dividend and the new divisor is twice the old divisor followed by a digit which makes a number such that the square of that number will be less than or equal to the new dividend.
So, we have a new dividend as 4212 and we can have a divisor as \[2 \times 35\underline {} = 70\underline {} \] where blank is filled by a digit.
Now we try to find a number in the lane of seven hundreds whose square is less than or equal to our new dividend.
\[
  701 \times 1 = 701 \\
  702 \times 2 = 1404 \\
  703 \times 3 = 2109 \\
  704 \times 4 = 2816 \\
  705 \times 5 = 3525 \\
  706 \times 6 = 4236
 \]
We can clearly see that \[706 \times 6 = 4236\] suits our requirement because \[4236 = 4236\]
Now we divide the dividend by the number 706 and the quotient 6 comes beside the earlier quotient.
\[
  706\mathop{\left){\vphantom{1{4236}}}\right.
\!\!\!\!\overline{\,\,\,\vphantom 1{{4236}}}}
\limits^{\displaystyle \,\,\, {356}} \\
   - 4236 \\
  \overline { = 0}
 \]
Since the remainder comes out to be 0, the square root of 126736 is 356.

\[\therefore \]Option B is correct.

Note: Alternative method:
We can start calculating squares of numbers from \[1\] and check in the list but looking at the number we can use hit and trial method to find near about squares.
Say we know \[300 \times 300 = 90000,400 \times 400 = 160000\]
so we check numbers in between \[300,400\]
Now we know \[350 \times 350 = 12250,360 \times 360 = 129600\]
so we check numbers in between \[350,360\]
\[351 \times 351 = 123201,352 \times 352 = 123904,353 \times 353 = 124609,354 \times 354 = 125316,355 \times 355 = 126025\] and \[356 \times 356 = 126736\]
From the list we see that \[356 \times 356 = 126736\]
So, 356 is square root is 126736
\[\therefore \]Option B is correct.