Question Number 124512 by mathocean1 last updated on 03/Dec/20
$${determinate}\:{the}\:{smallest}\:{integer} \\ $$$${which}\:{has}\:\mathrm{28}\:{divisors}. \\ $$
Commented by mr W last updated on 04/Dec/20
$$\mathrm{960}? \\ $$
Commented by mathocean1 last updated on 04/Dec/20
$${please}\:{sir}\:{can}\:{you}\:{detail} \\ $$
Commented by mr W last updated on 04/Dec/20
$${n}={p}^{{a}} {q}^{{b}} {r}^{{c}} ... \\ $$$${with}\:{p},{q},{r},...={prime}=\mathrm{2},\mathrm{3},\mathrm{5},... \\ $$$$\left({a}+\mathrm{1}\right)\left({b}+\mathrm{1}\right)\left({c}+\mathrm{1}\right)...=\mathrm{28}=\mathrm{7}×\mathrm{4}=\mathrm{7}×\mathrm{2}×\mathrm{2} \\ $$$${case}\:\mathrm{1}:\:{a}=\mathrm{6},\:{b}=\mathrm{3} \\ $$$${n}_{{min}} =\mathrm{2}^{\mathrm{6}} \mathrm{3}^{\mathrm{3}} =\mathrm{1728} \\ $$$${case}\:\mathrm{2}:\:{a}=\mathrm{6},\:{b}=\mathrm{1},\:{c}=\mathrm{1} \\ $$$${n}_{{min}} =\mathrm{2}^{\mathrm{6}} \mathrm{3}^{\mathrm{1}} \mathrm{5}^{\mathrm{1}} =\mathrm{960} \\ $$$$ \\ $$$$\Rightarrow{n}_{{min}} =\mathrm{960} \\ $$