Question Number 199708 by tri26112004 last updated on 08/Nov/23
$${n}^{\mathrm{20}} +\mathrm{11}^{{n}} =\mathrm{2023}\:\left({n}\in{N}\right) \\ $$$${n}=¿ \\ $$
Commented by Rasheed.Sindhi last updated on 08/Nov/23
$${f}\left({n}\right)={n}^{\mathrm{20}} +\mathrm{11}^{{n}} \left({increasing}\:{function}\right) \\ $$$${f}\left(\mathrm{1}\right)=\mathrm{12} \\ $$$${f}\left({k}\right)=\mathrm{2023} \\ $$$${f}\left(\mathrm{2}\right)=\mathrm{1048697} \\ $$$${f}\left(\mathrm{1}\right)<{f}\left({k}\right)<{f}\left(\mathrm{2}\right) \\ $$$$\mathrm{1}<{k}<\mathrm{2}\: \\ $$$$\mathcal{T}{here}'{s}\:{no}\:{natural}\:{between}\:\mathrm{1\&}\:\mathrm{2}. \\ $$