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Find-the-smallest-positive-integer-n-so-that-1-2-2-2-3-2-n-2-is-divided-by-n-




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

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