What is the smallest number by which 625 must | vedclass

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(B) First,find the prime factorization of $625$:$\begin{array}{l|l} 5 & 625 \ \hline 5 & 125 \ \hline 5 & 25 \ \hline & 5 \end{array}$Thus,$625 = 5

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$5$

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(B) First,find the prime factorization of $625$:$\begin{array}{l|l} 5 & 625 \ \hline 5 & 125 \ \hline 5 & 25 \ \hline & 5 \end{array}$Thus,$625 = 5 	imes 5 	imes 5 	imes 5 = 5^3 	imes 5$.To make the quotient a perfect cube,we must divide $625$ by the factor that is not part of the triplet,which is $5$.Therefore,$625 \div 5 = 125$,and $125 = 5^3$,which is a perfect cube.The smallest number is $5$.

What is the smallest number by which $625$ must be divided so that the quotient is a perfect cube?

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