Question

find the smallest number by which 604 must be divided to obtain a perfect cube.

Answer 1

$$\begin{gathered} \mathrm{The}\mathrm{prime}\mathrm{factorisation}\mathrm{of}604\mathrm{is}, \\ 604=2\times 2\times 151 \\ \mathrm{Now},\mathrm{prime}\mathrm{factor}2\mathrm{and}151\mathrm{do}\mathrm{not}\mathrm{occur}\mathrm{in}\mathrm{the}\mathrm{group}\mathrm{of}3,\mathrm{so}\mathrm{it}\mathrm{is}\mathrm{not}\mathrm{a}\mathrm{perfect}\mathrm{cube}. \\ \mathrm{In}\mathrm{order}\mathrm{to}\mathrm{make}\mathrm{it}\mathrm{a}\mathrm{perfect}\mathrm{cube},604\mathrm{must}\mathrm{be}\mathrm{divided}\mathrm{by}604\left(2\times 2\times 151\right) \\ 604÷604=\left(2\times 2\times 151\right)÷\left(2\times 2\times 151\right)=1\left(\mathrm{a}\mathrm{perfect}\mathrm{cube}\mathrm{number}\right) \end{gathered}$$