Question

what is the least number to subtracted from each of the following numbers to get a perfect square number also found the square root of the perfect square numbers thus obtained (i) 402 (ii) 1989 (iii) 3250 (iv) 825 (v) 4000


please answer the question
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Answer 1

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

$\rm :\longmapsto\: \sqrt{402}$

Using Long Division we have

$\begin{gathered} \:\: \begin{array}{c|c} {\underline{\sf{}}}&{\underline{\sf{\:\:20 \:\:}}}\\ {\underline{\sf{2}}}& {\sf{\:\:402 \:\:}} \\{\sf{}}& \underline{\sf{\:\: \: \: \: \: \: \: \: \: 4  \:  \:  \:  \:  \:  \: \:\:}} \\ {\underline{\sf{40}}}& {\sf{\:\: \: \: \: \: 002 \: \:  \:\:}} \end{array}\end{gathered}$

So, 2 must be subtracted from 402 to make it a perfect square.

So, Required number is 402 - 2 = 400

$\rm :\longmapsto\: \sqrt{400} = 20$

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

$\rm :\longmapsto\: \sqrt{1989}$

Using Long Division

$\begin{gathered} \:\: \begin{array}{c|c} {\underline{\sf{}}}&{\underline{\sf{\:\:44 \:\:}}}\\ {\underline{\sf{4}}}& {\sf{\:\:1989 \:\:}} \\{\sf{}}& \underline{\sf{\:\: \: \: \: \: \:\: \: \: 16 \: \:  \:  \:  \:  \:  \: \:\:}} \\ {\underline{\sf{84}}}& {\sf{\:\: \: \: \: \:389 \:  \:\:}} \\{\sf{}}& \underline{\sf{\:\: \: 336\:}} \\ {\underline{\sf{}}}& {\sf{\:\: \: \: \: \: \: 53  \:\:}} \end{array}\end{gathered}$

So, 53 must be subtracted from 1989 to make it a perfect square.

So, Required number is 1989 - 53 = 1936

$\rm :\longmapsto\: \sqrt{1936} = 44$

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

$\rm :\longmapsto\: \sqrt{3250}$

$\begin{gathered} \:\: \begin{array}{c|c} {\underline{\sf{}}}&{\underline{\sf{\:\:57 \:\:}}}\\ {\underline{\sf{5}}}& {\sf{\:\:3250 \:\:}} \\{\sf{}}& \underline{\sf{\:\: \: \: \: \: \:\: \: \: 25 \: \:  \:  \:  \:  \:  \: \:\:}} \\ {\underline{\sf{104}}}& {\sf{\:\: \: \: \: \:750 \:  \:\:}} \\{\sf{}}& \underline{\sf{\:\: \: 749\:}} \\ {\underline{\sf{}}}& {\sf{\:\: \: \: \: \: \: 1  \:\:}} \end{array}\end{gathered}$

So, 1 must be subtracted from 3250 to make it a perfect square.

So, Required number is 3250 - 1 = 3249

$\rm :\longmapsto\: \sqrt{3249} = 57$

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

$\rm :\longmapsto\: \sqrt{825}$

$\begin{gathered} \:\: \begin{array}{c|c} {\underline{\sf{}}}&{\underline{\sf{\:\:28 \:\:}}}\\ {\underline{\sf{2}}}& {\sf{\:\:825 \:\:}} \\{\sf{}}& \underline{\sf{\:\: \: \: \: \: \:\: \: \: 4\: \:  \:  \:  \:  \:  \: \:\:}} \\ {\underline{\sf{48}}}& {\sf{\:\: \: \: \: 425\:  \:\:}} \\{\sf{}}& \underline{\sf{\:\: \: 384\:}} \\ {\underline{\sf{}}}& {\sf{\:\: \: \: \: \: \: 41  \:\:}} \end{array}\end{gathered}$

So, 41 must be subtracted from 825 to make it a perfect square.

So, Required number is 825 - 41 = 784

$\rm :\longmapsto\: \sqrt{784} = 28$

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

$\rm :\longmapsto\: \sqrt{4000}$

Using Long Division, we have

$\begin{gathered} \:\: \begin{array}{c|c} {\underline{\sf{}}}&{\underline{\sf{\:\:63 \:\:}}}\\ {\underline{\sf{6}}}& {\sf{\:\:4000 \:\:}} \\{\sf{}}& \underline{\sf{\:\: \: \: \: \: \:\: \: \: 36 \: \:  \:  \:  \:  \:  \: \:\:}} \\ {\underline{\sf{123}}}& {\sf{\:\: \: \: \: \:400 \:  \:\:}} \\{\sf{}}& \underline{\sf{\:\: \: 369\:}} \\ {\underline{\sf{}}}& {\sf{\:\: \: \: \: \: \: 31  \:\:}} \end{array}\end{gathered}$

So, 31 must be subtracted from 4000 to make it a perfect square.

So, Required number is 4000 - 31 = 3969

$\rm :\longmapsto\: \sqrt{3969} = 63$

Answer 2

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