<s>[INST] HCF LCM Question $540= 2^2 \times 3^3 \times 5$
 Find the smallest positive integer $K$ such that $\frac{540}{k}$ is a cube number . 
 [/INST] $540=2^2\cdot3^3\cdot 5$ . We have to divide by a number $k=2^a3^b5^c$ with $a\equiv 2\bmod 3,b\equiv0\bmod 3$ and $c\equiv1\bmod 3$. Clearly the smallest choices are $a=2,b=0,c=1$. So $k=2^25^1$.
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