Question

find the smallest number by which 23265 must divide to obtain a perfect cube​

Answer 1

prime factorization of 23265:

$\Large{ \begin{array}{c|c} \tt & \\ \cline \: \tt{3} \: & \sf { 23265 } \\ \cline \: \tt {5 }\: & \sf{ 7755} \\ \cline \: \tt{3} & \: \sf{ 1551} \\ \cline \: \tt {11} & \: \sf{ 517 }\\ \cline \tt {47} & \sf{ 47}\\ & \: \sf{ 1 } \end{array}}$

23265 = 3×3×5×11×47

Grouping the factors in triplets of equal factors we get,

23265 = (3×3×?)(?×5×?)(?×11×?)(?×?×47)

so we need , 3 × 5 × 5 × 11 × 11 × 47 = 426525

therefore , 426525 is the smallest number to be multiplied to get a perfect square.

Answer 2

Step-by-step explanation:

we know that we have to find the smallest perfect cube number that can be divisible by 23265.

so we have to do is

23265 = 3×3×5×11×47

Grouping the factors in triplets, so that we get the answer.

23265 = (3×3×?)(?×5×?)(?×11×?)(?×?×47)

so we need , 3 × 5 × 5 × 11 × 11 × 47 = 426525

there 426525 is the smallest perfect cube number that can be divisible by 23265.

I hope you got your answer and understood that.

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