Question

Find the smallest numberby which 23625 must be multiplied to obtain a perfect cube

Answer 1

$<b><u>Hello Friend!!!!</u></b>$

$<u><b>Here Is Your Answer</b></u>$

$<u>In order to solve it we need to first find the prime factors of 23625</u>$

$<b>And Here It Is :-</b>$

$\begin{array}{l|lc} \sf{3} &\sf{23625} \\ \cline{1-2} \sf3 & \sf7875 \\ \sf3 & 2625 \\ \sf5 & \sf875 \\ \sf5 & 175 \\ \sf5 & \sf35 \\ \sf7 & \sf7 \\ & \sf1 \end{array}$

$<b>As you can see above there the following pairs in the prime-factorization. </b>$

$23625={\underbrace{\sf 3x3x3}_{\sf 3x}}\sf{x}{\underbrace{\sf 5x5x5}_{\sf3x}}\sf{x7}.$

$<b>As you know we need pairs of three numbers to form a cube. </b>$

$<u>And there are a triplet of 3 and 5 but 7 is left alone so we need to multiply 7 x 7(which is 49) to 23625 to make it a perfect cube.</u>$

Which Is :-

$\implies 23625 \sf{x}49$

$\implies 1157625$

$<b><u>Hope The Above Answer Helped.</b></u>$

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