Question Number 104038 by bramlex last updated on 19/Jul/20
$${what}\:{is}\:{the}\:{largest}\:{positive} \\ $$$${integer}\:{n}\:{such}\:{that}\:{n}^{\mathrm{3}} +\mathrm{100}\:{is} \\ $$$${divisible}\:{by}\:{n}+\mathrm{10}\:?\: \\ $$
Answered by john santu last updated on 19/Jul/20
$${Let}\:{n}^{\mathrm{3}} +\mathrm{100}\:=\:\left({n}+\mathrm{10}\right)\left({n}^{\mathrm{2}} +{an}+{b}\right)+{c} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:{n}^{\mathrm{3}} +{n}^{\mathrm{2}} \left(\mathrm{10}+{a}\right)+{n}\left({b}+\mathrm{10}{a}\right)+\mathrm{10}{b}+{c} \\ $$$${equating}\:{coefficients}\:{yields} \\ $$$$\begin{cases}{\mathrm{100}+{a}\:=\:\mathrm{0}}\\{\mathrm{10}{a}+{b}\:=\:\mathrm{0}}\\{\mathrm{10}{b}\:+\:{c}\:=\mathrm{100}}\end{cases} \\ $$$${solving}\:{this}\:{system}\:{yields} \\ $$$${a}=−\mathrm{10},\:{b}\:=\:\mathrm{100}\:\&\:{c}\:=\:−\mathrm{900} \\ $$$${therefore}\:{by}\:{the}\:{Euclidean} \\ $$$${Algorithm}\:{we}\:{get}\:{n}+\mathrm{10}\:= \\ $$$${gcd}\left({n}^{\mathrm{3}} +\mathrm{100},\:{n}+\mathrm{10}\right)\:= \\ $$$${gcd}\left(−\mathrm{900},{n}+\mathrm{10}\right)\:= \\ $$$${gcd}\left(\mathrm{900},{n}+\mathrm{10}\right)\:.\:{The}\:{maximum} \\ $$$${value}\:{for}\:{n}\:{is}\:{hence}\:{n}\:=\:\mathrm{890}. \\ $$$$\left({JS}\:\circledast\right) \\ $$
Commented by bramlex last updated on 19/Jul/20
$${nice}\:!\bullet\smile\bullet \\ $$