Question Number 209316 by klipto last updated on 06/Jul/24
$$\boldsymbol{\mathrm{if}}\:\:\mathrm{2}\boldsymbol{\mathrm{n}}^{\mathrm{2}} +\mathrm{3}\boldsymbol{\mathrm{n}}^{\mathrm{3}} =\boldsymbol{\mathrm{n}}! \\ $$$$\boldsymbol{\mathrm{find}}\:\boldsymbol{\mathrm{n}} \\ $$
Answered by A5T last updated on 07/Jul/24
$${n}\left(\mathrm{2}+\mathrm{3}{n}\right)=\left({n}−\mathrm{1}\right)! \\ $$$$\left({n}−\mathrm{1}\right)!\leqslant{n}\left(\mathrm{2}+\mathrm{3}{n}\right) \\ $$$${n}\left(\mathrm{2}+\mathrm{3}{n}\right)\geqslant{n}\left({n}−\mathrm{1}\right)\left({n}−\mathrm{2}\right) \\ $$$$\Rightarrow\mathrm{3}{n}\geqslant{n}^{\mathrm{2}} −\mathrm{3}{n}\Rightarrow{n}^{\mathrm{2}} \leqslant\mathrm{6}{n}\Rightarrow{n}\leqslant\mathrm{6} \\ $$$${Checking}\Rightarrow{n}=\mathrm{6} \\ $$