HCF LCM Question

Question

$540= 2^2 \times 3^3 \times 5$ Find the smallest positive integer $K$ such that $\frac{540}{k}$ is a cube number .

Answer 1

$540=2^2\cdot3^3\cdot 5$ . 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 $k=2^25^1$.

Answer 2

To be a cube, a numbers prime factors must be a multiple of 3 power. $540 = 2^2*3^3*5$ so the only powers that are 3 or higher is $3^3$. So the only possible answers are $540/20 = 27$ or $540/540 = 1$.

From your comment it sounds like you meant to ask what is the least $k$ so that $540*k$ is a cube. In which case we must bump up the power of 2 and 5.

$(2^2*3^3*5)*2*5^2 = 2^3*3^3*5^3 \iff 540*50 = 27000 = 30^3$