Question

Find the square root of following numbers with the division method 15217501

Answer 1

$\large\underline{\sf{Solution-}}$

$\rm \: \sqrt{15217501}$

So, using long division method, we have

$\begin{gathered}\begin{gathered} \:\: \begin{array}{c|c} {\underline{\sf{}}}&{\underline{\sf{\:\:3900.96 \:\:}}}\\ {\underline{\sf{3}}}& {\sf{\:\:15217501.0000 \:\:}} \\{\sf{}}& \underline{\sf{\:9 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:  }} \\ {\underline{\sf{69}}}& {\sf{\:\: 621 \: \: \: \: \: \: \: \: \: \: \: \:   \:\:}} \\{\sf{}}& \underline{\sf{\:\:621 \: \: \: \: \: \: \: \: \: \: \: \: \:  \:}} \\ {\underline{\sf{78009}}}& {\sf{\: 750100 }} \\{\sf{}}& \underline{\sf{\:702081}} \\ {\underline{\sf{780186}}}& {\sf{ \: \: \: \: 4801900\:}} \\{\sf{}}& \underline{\sf{\: \: \: \: 4681116\:}}\\ {\underline{\sf{}}}& {\sf{ \: \: \: \: \: \: 120784\:}} \end{array}\end{gathered}\end{gathered}$

Hence,

$\rm\implies \: \: \boxed{\tt{ \: \rm \: \sqrt{15217501} = 3900.96 \: \: }}$

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ADDITIONAL INFORMATION :-

1. The number of digits in the square root of consisting of 2n + 1 digits is n + 1.

2. The number of digits in the square root of consisting of 2n digits is n.

3. Between squares of any two consecutive natural numbers n and n + 1, there always exist 2n natural numbers.

4. The square of any natural number can never ends with the digit 2, 3, 7, 8 or odd number of zeroes.

5. The sum of the first n odd natural numbers is n^2.

6. The sum of the first n even natural numbers is n(n + 1).

Answer 2

$</p><p>\begin{gathered}\begin{gathered}\begin{gathered} \:\: \begin{array}{c|c} {\underline{\rm{}}}&{\underline{\rm{\:\:3900.96 \:\:}}}\\ {\underline{\rm{3}}}& {\rm{\:\:15217501.0000 \:\:}} \\{\rm{}}& \underline{\rm{\:9 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:  }} \\ {\underline{\rm{}}}& {\rm{\:\: 621 \: \: \: \: \: \: \: \: \: \: \: \:   \:\:}} \\ \underline{\rm{69}}& \underline{\rm{\:\:621 \: \: \: \: \: \: \: \: \: \: \: \: \:  \:}} \\ {\underline{\rm{78009}}}& {\rm{\: 750100 }} \\{\rm{}}& \underline{\rm{\:702081}} \\ {\underline{\rm{780186}}}& {\rm{ \: \: \: \: 4801900\:}} \\{\rm{}}& \underline{\rm{\: \: \: \: 4681116\:}}\\ {\underline{\rm{}}}& {\rm{ \: \: \: \: \: \: 120784\:}} \end{array}\end{gathered}\end{gathered} \\ \end{gathered}$

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∴ The Value of √15217501 = 3900.96