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What is the smallest number by which $6912$ must be divided so that the number formed is a perfect cube?
Given: A number $6912$.
To do: To find the smallest number by which $6912$ must be divided so that the number formed is a perfect cube.
Solution:
The factors of:
$6912=\underline{2\times2\times2}\times\underline{2\times2\times2}\times2\times2\times\underline{3\times3\times3}=2^2\times2^6\times3^3$
Therefore, we should divide $6912$ by $2^2=4$, the smallest number to get $1728$ which is a cube of $12$.
Updated on: 10-Oct-2022
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