Question

Find the smallest number by which 6336 must be divided to obtain a perfect square. Also, find the square root of the perfect square so obtained.​

Answer 1

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

Let first find the prime factorization of 6336

$\red{\bf :\longmapsto\:Prime \: factorization \: of \: 6336}$

$\begin{gathered}\begin{gathered}\begin{gathered} \:\: \begin{array}{c|c} {\underline{\sf{2}}}&{\underline{\sf{\:\:6336 \:\:}}}\\ {\underline{\sf{2}}}& \underline{\sf{\:\:3168 \:\:}} \\\underline{\sf{2}}&\underline{\sf{\:\:1584\:\:}} \\ {\underline{\sf{2}}}& \underline{\sf{\:\:792 \:\:}} \\ {\underline{\sf{2}}}& \underline{\sf{\:\:396 \:\:}} \\ {\underline{\sf{2}}}& \underline{\sf{\:\:198 \:\:}} \\ {\underline{\sf{3}}}& \underline{\sf{\:\:99 \:\:}} \\ {\underline{\sf{3}}}& \underline{\sf{\:\:33 \:\:}} \\ {\underline{\sf{11}}}& \underline{\sf{\:\:11 \:\:}}\\\underline{\sf{}}&{\sf{\:\:1 \:\:}} \end{array}\end{gathered}\end{gathered}\end{gathered}$

$\rm :\longmapsto\:6336 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 11$

$\rm :\longmapsto\:6336 = {2}^{2} \times {2}^{2} \times {2}^{2} \times {3}^{2} \times 11$

So, in order to obtain a perfect square, 6336 should be divided by 11.

So,

Smallest number is 11, by which 6336 must be divided to obtain a perfect square.

So, required number = 576

So,

$\rm :\longmapsto\:576 = {2}^{2} \times {2}^{2} \times {2}^{2} \times {3}^{2}$

$\bf\implies \: \sqrt{576}$

$\rm \:= \: \:\sqrt{ {2}^{2} \times {2}^{2} \times {2}^{2} \times {3}^{2} }$

$\rm \:= \: \: 2 \times 2 \times 2 \times 3$

$\rm \:= \: \:24$

$\bf\implies \: \sqrt{576} = 24$

Answer 2

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Step-by-step explanation:

Answer. So, in order to obtain a perfect square, 6336 should be divided by 11. So, Smallest number is 11, by which 6336 must be divided to obtain a perfect square.