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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 $12$.

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