Question Number 137269 by mohssinee last updated on 31/Mar/21
Commented by mohssinee last updated on 31/Mar/21
$${n}\in\mathbb{N}^{\ast} \\ $$
Answered by aleks041103 last updated on 31/Mar/21
$${Check}\:{for}\:{n}=\mathrm{1} \\ $$$$\mathrm{40}^{\mathrm{1}} \mathrm{1}!=\mathrm{40}\mid\mathrm{120}=\mathrm{5}!=\left(\mathrm{5}.\mathrm{1}\right)! \\ $$$$ \\ $$$${Suppose}\:{its}\:{true}\:{for}\:{some}\:{n}>\mathrm{1}. \\ $$$${Then}: \\ $$$$\left(\mathrm{5}\left({n}+\mathrm{1}\right)\right)!=\left(\mathrm{5}{n}+\mathrm{5}\right)!= \\ $$$$=\left(\mathrm{5}{n}+\mathrm{1}\right)\left(\mathrm{5}{n}+\mathrm{2}\right)\left(\mathrm{5}{n}+\mathrm{3}\right)\left(\mathrm{5}{n}+\mathrm{4}\right)\left(\mathrm{5}{n}+\mathrm{5}\right)\left(\mathrm{5}{n}\right)!= \\ $$$$=\mathrm{5}\left(\mathrm{5}{n}+\mathrm{1}\right)\left(\mathrm{5}{n}+\mathrm{2}\right)\left(\mathrm{5}{n}+\mathrm{3}\right)\left(\mathrm{5}{n}+\mathrm{4}\right)\left({n}+\mathrm{1}\right)\left(\mathrm{5}{n}\right)! \\ $$$${Since}\:\mathrm{40}^{{n}} {n}!\mid\left(\mathrm{5}{n}\right)!\:\Rightarrow\left(\mathrm{5}{n}\right)!={m}.\mathrm{40}^{{n}} {n}! \\ $$$$\Rightarrow\left(\mathrm{5}\left({n}+\mathrm{1}\right)\right)!=\mathrm{5}\left(\mathrm{5}{n}+\mathrm{1}\right)\left(\mathrm{5}{n}+\mathrm{2}\right)\left(\mathrm{5}{n}+\mathrm{3}\right)\left(\mathrm{5}{n}+\mathrm{4}\right)\left({n}+\mathrm{1}\right)!\mathrm{40}^{{n}} {m} \\ $$$${Let}'{s}\:{look}\:{at}\:{two}\:{cases} \\ $$$$\left.\mathrm{1}\right)\:{n}\equiv\mathrm{1}\left({mod}\:\mathrm{2}\right)\:{and}\:{n}\equiv\mathrm{1}\:{or}\:\mathrm{3}\left({mod}\:\mathrm{4}\right) \\ $$$$\mathrm{5}{n}\equiv\mathrm{1}\left({mod}\:\mathrm{2}\right)\:{and}\:\mathrm{5}{n}\equiv\mathrm{1}\:{or}\:\mathrm{3}\left({mod}\:\mathrm{4}\right) \\ $$$$\Rightarrow\mathrm{5}{n}+\mathrm{1}\equiv\mathrm{2}\:{or}\:\mathrm{0}\left({mod}\:\mathrm{4}\right) \\ $$$$\:\:\:\:\:\mathrm{5}{n}+\mathrm{3}\equiv\mathrm{0}\:{or}\:\mathrm{2}\left({mod}\:\mathrm{4}\right) \\ $$$${Therefore}\:{both}\:{are}\:{devisible}\:{by}\:\mathrm{2} \\ $$$${and}\:{one}\:{is}\:{devisible}\:{by}\:\mathrm{4}. \\ $$$$\Rightarrow\:\mathrm{5}.\mathrm{2}.\mathrm{4}=\mathrm{40}\mid\mathrm{5}\left(\mathrm{5}{n}+\mathrm{1}\right)\left(\mathrm{5}{n}+\mathrm{2}\right)\left(\mathrm{5}{n}+\mathrm{3}\right)\left(\mathrm{5}{n}+\mathrm{4}\right) \\ $$$$\Rightarrow\mathrm{5}\left(\mathrm{5}{n}+\mathrm{1}\right)\left(\mathrm{5}{n}+\mathrm{2}\right)\left(\mathrm{5}{n}+\mathrm{3}\right)\left(\mathrm{5}{n}+\mathrm{4}\right)=\mathrm{40}{k} \\ $$$$\left.\mathrm{2}\right)\:{n}\equiv\mathrm{0}\:{or}\:\mathrm{2}\left({mod}\:\mathrm{4}\right) \\ $$$${Analogously} \\ $$$$\mathrm{5}\left(\mathrm{5}{n}+\mathrm{1}\right)\left(\mathrm{5}{n}+\mathrm{2}\right)\left(\mathrm{5}{n}+\mathrm{3}\right)\left(\mathrm{5}{n}+\mathrm{4}\right)=\mathrm{40}{k} \\ $$$$ \\ $$$${Then} \\ $$$$\left(\mathrm{5}\left({n}+\mathrm{1}\right)\right)!=\mathrm{40}.{k}.\mathrm{40}^{{n}} .\left({n}+\mathrm{1}\right)!.{m} \\ $$$$\Rightarrow\left({mk}\right)\mathrm{40}^{{n}+\mathrm{1}} \left({n}+\mathrm{1}\right)!=\left(\mathrm{5}\left({n}+\mathrm{1}\right)\right)! \\ $$$$\Rightarrow\mathrm{40}^{{n}+\mathrm{1}} \left({n}+\mathrm{1}\right)!\mid\left(\mathrm{5}\left({n}+\mathrm{1}\right)\right)! \\ $$$$ \\ $$$${By}\:{induction},\:{the}\:{statement}\:{is}\:{true}! \\ $$