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Question Number 137269 by mohssinee last updated on 31/Mar/21

Commented by mohssinee last updated on 31/Mar/21

n∈N^∗

$${n}\in\mathbb{N}^{\ast} \\ $$

Answered by aleks041103 last updated on 31/Mar/21

Check for n=1 40^1 1!=40∣120=5!=(5.1)! Suppose its true for some n>1. Then: (5(n+1))!=(5n+5)!= =(5n+1)(5n+2)(5n+3)(5n+4)(5n+5)(5n)!= =5(5n+1)(5n+2)(5n+3)(5n+4)(n+1)(5n)! Since 40^n n!∣(5n)! ⇒(5n)!=m.40^n n! ⇒(5(n+1))!=5(5n+1)(5n+2)(5n+3)(5n+4)(n+1)!40^n m Let′s look at two cases 1) n≡1(mod 2) and n≡1 or 3(mod 4) 5n≡1(mod 2) and 5n≡1 or 3(mod 4) ⇒5n+1≡2 or 0(mod 4) 5n+3≡0 or 2(mod 4) Therefore both are devisible by 2 and one is devisible by 4. ⇒ 5.2.4=40∣5(5n+1)(5n+2)(5n+3)(5n+4) ⇒5(5n+1)(5n+2)(5n+3)(5n+4)=40k 2) n≡0 or 2(mod 4) Analogously 5(5n+1)(5n+2)(5n+3)(5n+4)=40k Then (5(n+1))!=40.k.40^n .(n+1)!.m ⇒(mk)40^(n+1) (n+1)!=(5(n+1))! ⇒40^(n+1) (n+1)!∣(5(n+1))! By induction, the statement is true!

$${Check}\:{for}\:{n}=\mathrm{1} \\ $$$$\mathrm{40}^{\mathrm{1}} \mathrm{1}!=\mathrm{40}\mid\mathrm{120}=\mathrm{5}!=\left(\mathrm{5}.\mathrm{1}\right)! \\ $$$$ \\ $$$${Suppose}\:{its}\:{true}\:{for}\:{some}\:{n}>\mathrm{1}. \\ $$$${Then}: \\ $$$$\left(\mathrm{5}\left({n}+\mathrm{1}\right)\right)!=\left(\mathrm{5}{n}+\mathrm{5}\right)!= \\ $$$$=\left(\mathrm{5}{n}+\mathrm{1}\right)\left(\mathrm{5}{n}+\mathrm{2}\right)\left(\mathrm{5}{n}+\mathrm{3}\right)\left(\mathrm{5}{n}+\mathrm{4}\right)\left(\mathrm{5}{n}+\mathrm{5}\right)\left(\mathrm{5}{n}\right)!= \\ $$$$=\mathrm{5}\left(\mathrm{5}{n}+\mathrm{1}\right)\left(\mathrm{5}{n}+\mathrm{2}\right)\left(\mathrm{5}{n}+\mathrm{3}\right)\left(\mathrm{5}{n}+\mathrm{4}\right)\left({n}+\mathrm{1}\right)\left(\mathrm{5}{n}\right)! \\ $$$${Since}\:\mathrm{40}^{{n}} {n}!\mid\left(\mathrm{5}{n}\right)!\:\Rightarrow\left(\mathrm{5}{n}\right)!={m}.\mathrm{40}^{{n}} {n}! \\ $$$$\Rightarrow\left(\mathrm{5}\left({n}+\mathrm{1}\right)\right)!=\mathrm{5}\left(\mathrm{5}{n}+\mathrm{1}\right)\left(\mathrm{5}{n}+\mathrm{2}\right)\left(\mathrm{5}{n}+\mathrm{3}\right)\left(\mathrm{5}{n}+\mathrm{4}\right)\left({n}+\mathrm{1}\right)!\mathrm{40}^{{n}} {m} \\ $$$${Let}'{s}\:{look}\:{at}\:{two}\:{cases} \\ $$$$\left.\mathrm{1}\right)\:{n}\equiv\mathrm{1}\left({mod}\:\mathrm{2}\right)\:{and}\:{n}\equiv\mathrm{1}\:{or}\:\mathrm{3}\left({mod}\:\mathrm{4}\right) \\ $$$$\mathrm{5}{n}\equiv\mathrm{1}\left({mod}\:\mathrm{2}\right)\:{and}\:\mathrm{5}{n}\equiv\mathrm{1}\:{or}\:\mathrm{3}\left({mod}\:\mathrm{4}\right) \\ $$$$\Rightarrow\mathrm{5}{n}+\mathrm{1}\equiv\mathrm{2}\:{or}\:\mathrm{0}\left({mod}\:\mathrm{4}\right) \\ $$$$\:\:\:\:\:\mathrm{5}{n}+\mathrm{3}\equiv\mathrm{0}\:{or}\:\mathrm{2}\left({mod}\:\mathrm{4}\right) \\ $$$${Therefore}\:{both}\:{are}\:{devisible}\:{by}\:\mathrm{2} \\ $$$${and}\:{one}\:{is}\:{devisible}\:{by}\:\mathrm{4}. \\ $$$$\Rightarrow\:\mathrm{5}.\mathrm{2}.\mathrm{4}=\mathrm{40}\mid\mathrm{5}\left(\mathrm{5}{n}+\mathrm{1}\right)\left(\mathrm{5}{n}+\mathrm{2}\right)\left(\mathrm{5}{n}+\mathrm{3}\right)\left(\mathrm{5}{n}+\mathrm{4}\right) \\ $$$$\Rightarrow\mathrm{5}\left(\mathrm{5}{n}+\mathrm{1}\right)\left(\mathrm{5}{n}+\mathrm{2}\right)\left(\mathrm{5}{n}+\mathrm{3}\right)\left(\mathrm{5}{n}+\mathrm{4}\right)=\mathrm{40}{k} \\ $$$$\left.\mathrm{2}\right)\:{n}\equiv\mathrm{0}\:{or}\:\mathrm{2}\left({mod}\:\mathrm{4}\right) \\ $$$${Analogously} \\ $$$$\mathrm{5}\left(\mathrm{5}{n}+\mathrm{1}\right)\left(\mathrm{5}{n}+\mathrm{2}\right)\left(\mathrm{5}{n}+\mathrm{3}\right)\left(\mathrm{5}{n}+\mathrm{4}\right)=\mathrm{40}{k} \\ $$$$ \\ $$$${Then} \\ $$$$\left(\mathrm{5}\left({n}+\mathrm{1}\right)\right)!=\mathrm{40}.{k}.\mathrm{40}^{{n}} .\left({n}+\mathrm{1}\right)!.{m} \\ $$$$\Rightarrow\left({mk}\right)\mathrm{40}^{{n}+\mathrm{1}} \left({n}+\mathrm{1}\right)!=\left(\mathrm{5}\left({n}+\mathrm{1}\right)\right)! \\ $$$$\Rightarrow\mathrm{40}^{{n}+\mathrm{1}} \left({n}+\mathrm{1}\right)!\mid\left(\mathrm{5}\left({n}+\mathrm{1}\right)\right)! \\ $$$$ \\ $$$${By}\:{induction},\:{the}\:{statement}\:{is}\:{true}! \\ $$

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