Written as the product of its prime factors $1800={{2}^{3}}\times {{3}^{2}}\times {{5}^{2}}$. Expr

Question

Written as the product of its prime factors $1800={{2}^{3}}\times {{3}^{2}}\times {{5}^{2}}$. Express $588$ as the product of its prime factors. Hence, write down the smallest positive integer, $k$, such that $1800k$ is a perfect cube.

Accepted Answer

(a) $2^2\times3\times7^2$ (b) $15$

Prime Factorisation and Perfect Cube Conditions

Prime Factorisation and Perfect Cube Conditions

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