Question Number 85224 by mathocean1 last updated on 20/Mar/20
$${solve}\:{in}\:\mathbb{N} \\ $$$$\left({n}−\mathrm{4}\right)!=\mathrm{42}\left({n}−\mathrm{2}\right)! \\ $$
Commented by jagoll last updated on 20/Mar/20
$$\left(\mathrm{n}−\mathrm{4}\right)!\:=\:\mathrm{42}\left(\mathrm{n}−\mathrm{2}\right)\left(\mathrm{n}−\mathrm{3}\right)\left(\mathrm{n}−\mathrm{4}\right)! \\ $$$$\mathrm{1}\:=\:\mathrm{42}\left(\mathrm{n}−\mathrm{2}\right)\left(\mathrm{n}−\mathrm{3}\right) \\ $$$$\mathrm{n}\:=\:\varnothing \\ $$
Commented by mathocean1 last updated on 20/Mar/20
$${thank}\:{you}\:{sir}! \\ $$$${And}\:{this}\:{equation},\:{what}\:{about}\:{it}\: \\ $$$${A}_{{n}} ^{\mathrm{4}} =\mathrm{42}{A}_{{n}} ^{\mathrm{2}} \:\:{in}\:\mathbb{N}\:{can}\:{help}\:{me}\:{to}\:{solve}\:{it}? \\ $$
Commented by jagoll last updated on 20/Mar/20
$$\mathrm{A}_{\mathrm{n}} ^{\mathrm{4}} −\mathrm{42A}_{\mathrm{n}} ^{\mathrm{2}} \:=\:\mathrm{0} \\ $$$$\mathrm{A}_{\mathrm{n}} ^{\mathrm{2}} \:\left(\mathrm{A}_{\mathrm{n}} ^{\mathrm{2}} −\mathrm{42}\right)\:=\:\mathrm{0} \\ $$$$\begin{cases}{\mathrm{A}_{\mathrm{n}} ^{\mathrm{2}} \:=\:\mathrm{0}\:}\\{\mathrm{A}_{\mathrm{n}} ^{\mathrm{2}} =\mathrm{42}}\end{cases}\:.\:\mathrm{what}\:\mathrm{is}\:\mathrm{A}_{\mathrm{n}} \:\mathrm{is}\:\mathrm{this}\:\mathrm{case}? \\ $$
Commented by mathocean1 last updated on 20/Mar/20
$${A}_{{n}} ^{\mathrm{2}} =\frac{{n}!}{\left({n}−\mathrm{2}\right)!}\:\:\:? \\ $$
Commented by mathocean1 last updated on 20/Mar/20
$${So}\:{i}\:{can}\:{have}\:{n}!=\mathrm{42}\left({n}−\mathrm{2}\right)! \\ $$
Commented by mathocean1 last updated on 20/Mar/20
$$\left({n}−\mathrm{1}\right)!×\left({n}−\mathrm{2}\right)!=\mathrm{42}\left({n}−\mathrm{2}\right)! \\ $$$${so}\:\left({n}−\mathrm{1}\right)!=\mathrm{42} \\ $$$${but}\:{how}\:{can}\:{i}\:{have}\:{the}\:{value}\left[{of}\:{n}?\right. \\ $$
Commented by jagoll last updated on 20/Mar/20
$$\mathrm{n}!\:=\:\mathrm{42}\:\left(\mathrm{n}−\mathrm{2}\right)! \\ $$$$\mathrm{n}\:\left(\mathrm{n}−\mathrm{1}\right)\left(\mathrm{n}−\mathrm{2}\right)!\:=\:\mathrm{42}\:\left(\mathrm{n}−\mathrm{2}\right)! \\ $$$$\mathrm{n}^{\mathrm{2}} −\mathrm{n}−\mathrm{42}=\mathrm{0} \\ $$$$\left(\mathrm{n}−\mathrm{7}\right)\left(\mathrm{n}+\mathrm{6}\right)=\mathrm{0} \\ $$$$\mathrm{n}\:=\:\mathrm{7}\:\leftarrow\mathrm{this}\:\mathrm{solution} \\ $$