Question Number 112994 by Aina Samuel Temidayo last updated on 10/Sep/20
$$\mathrm{If}\:\mathrm{n}\:\mathrm{is}\:\mathrm{any}\:\mathrm{even}\:\mathrm{number},\:\mathrm{then} \\ $$$$\mathrm{n}\left(\mathrm{n}^{\mathrm{2}} +\mathrm{20}\right)\:\mathrm{is}\:\mathrm{always}\:\mathrm{divisible}\:\mathrm{by}? \\ $$
Answered by mr W last updated on 10/Sep/20
$${n}=\mathrm{2}{k} \\ $$$$\mathrm{2}{k}\left(\mathrm{4}{k}^{\mathrm{2}} +\mathrm{20}\right)=\mathrm{8}{k}\left({k}^{\mathrm{2}} +\mathrm{5}\right) \\ $$$${one}\:{from}\:{k}\:{and}\:{k}^{\mathrm{2}} +\mathrm{5}\:{is}\:{even} \\ $$$$\Rightarrow{n}\left({n}^{\mathrm{2}} +\mathrm{20}\right)\:{is}\:{always}\:{divisible}\:{by}\:\mathrm{16}. \\ $$
Commented by Aina Samuel Temidayo last updated on 10/Sep/20
$$\mathrm{n}=\mathrm{2k}\:\left(\mathrm{k}\in\mathbb{Z}\right) \\ $$$$\mathrm{How}\:\mathrm{is}\:\mathrm{k}^{\mathrm{2}} +\mathrm{5}\:\mathrm{even}? \\ $$
Commented by mr W last updated on 10/Sep/20
$${i}\:{didn}'{t}\:{say}\:{that}\:{k}^{\mathrm{2}} +\mathrm{5}\:{is}\:{even}.\:{i}\:{said} \\ $$$${one}\:{of}\:{k}\:{and}\:{k}^{\mathrm{2}} +\mathrm{5}\:{is}\:{even}. \\ $$$${if}\:{k}\:{is}\:{even}\:{then}\:{k}^{\mathrm{2}} +\mathrm{5}\:{is}\:{odd}. \\ $$$${if}\:{k}\:{is}\:{odd}\:{then}\:{k}^{\mathrm{2}} +\mathrm{5}\:{is}\:{even}. \\ $$$${that}\:{means} \\ $$$${k}\left({k}^{\mathrm{2}} +\mathrm{5}\right)\:{is}\:{always}\:{even}. \\ $$
Commented by Aina Samuel Temidayo last updated on 10/Sep/20
$$\mathrm{Oh}.\:\mathrm{Thanks}. \\ $$