Question Number 150746 by naka3546 last updated on 15/Aug/21
$${Find}\:\:{the}\:\:{smallest}\:\:{positive}\:\:{integer}\:\:{n}\:\:{so}\:\:{that}\:\:\left(\mathrm{1}^{\mathrm{2}} \:+\:\mathrm{2}^{\mathrm{2}} \:+\:\mathrm{3}^{\mathrm{2}} \:+\:\ldots\:+\:{n}^{\mathrm{2}} \right)\:\:{is}\:\:{divided}\:\:{by}\:\:{n}\:. \\ $$
Answered by mr W last updated on 15/Aug/21
$${assume}\:{n}>\mathrm{1}. \\ $$$$\Sigma=\frac{{n}\left({n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)}{\mathrm{6}} \\ $$$${this}\:{is}\:{divisible}\:{by}\:{n}\:{for}\:{any}\:{n}\:{if} \\ $$$${n}+\mathrm{1}=\mathrm{6}{k}\:{or}\:{n}=\mathrm{6}{k}−\mathrm{1} \\ $$$${so}\:{the}\:{smallest}\:{n}\:{is}\:{then}\:{n}=\mathrm{5} \\ $$
Answered by naka3546 last updated on 15/Aug/21
$${thank}\:\:{you},\:\:{sir}. \\ $$