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Question Number 85224 by mathocean1 last updated on 20/Mar/20

solve in N (n−4)!=42(n−2)!

$${solve}\:{in}\:\mathbb{N} \\ $$$$\left({n}−\mathrm{4}\right)!=\mathrm{42}\left({n}−\mathrm{2}\right)! \\ $$

Commented by jagoll last updated on 20/Mar/20

(n−4)! = 42(n−2)(n−3)(n−4)! 1 = 42(n−2)(n−3) n = ∅

$$\left(\mathrm{n}−\mathrm{4}\right)!\:=\:\mathrm{42}\left(\mathrm{n}−\mathrm{2}\right)\left(\mathrm{n}−\mathrm{3}\right)\left(\mathrm{n}−\mathrm{4}\right)! \\ $$$$\mathrm{1}\:=\:\mathrm{42}\left(\mathrm{n}−\mathrm{2}\right)\left(\mathrm{n}−\mathrm{3}\right) \\ $$$$\mathrm{n}\:=\:\varnothing \\ $$

Commented by mathocean1 last updated on 20/Mar/20

thank you sir! And this equation, what about it A_n ^4 =42A_n ^2 in N can help me to solve it?

$${thank}\:{you}\:{sir}! \\ $$$${And}\:{this}\:{equation},\:{what}\:{about}\:{it}\: \\ $$$${A}_{{n}} ^{\mathrm{4}} =\mathrm{42}{A}_{{n}} ^{\mathrm{2}} \:\:{in}\:\mathbb{N}\:{can}\:{help}\:{me}\:{to}\:{solve}\:{it}? \\ $$

Commented by jagoll last updated on 20/Mar/20

A_n ^4 −42A_n ^2 = 0 A_n ^2 (A_n ^2 −42) = 0 { ((A_n ^2 = 0 )),((A_n ^2 =42)) :} . what is A_n is this case?

$$\mathrm{A}_{\mathrm{n}} ^{\mathrm{4}} −\mathrm{42A}_{\mathrm{n}} ^{\mathrm{2}} \:=\:\mathrm{0} \\ $$$$\mathrm{A}_{\mathrm{n}} ^{\mathrm{2}} \:\left(\mathrm{A}_{\mathrm{n}} ^{\mathrm{2}} −\mathrm{42}\right)\:=\:\mathrm{0} \\ $$$$\begin{cases}{\mathrm{A}_{\mathrm{n}} ^{\mathrm{2}} \:=\:\mathrm{0}\:}\\{\mathrm{A}_{\mathrm{n}} ^{\mathrm{2}} =\mathrm{42}}\end{cases}\:.\:\mathrm{what}\:\mathrm{is}\:\mathrm{A}_{\mathrm{n}} \:\mathrm{is}\:\mathrm{this}\:\mathrm{case}? \\ $$

Commented by mathocean1 last updated on 20/Mar/20

A_n ^2 =((n!)/((n−2)!)) ?

$${A}_{{n}} ^{\mathrm{2}} =\frac{{n}!}{\left({n}−\mathrm{2}\right)!}\:\:\:? \\ $$

Commented by mathocean1 last updated on 20/Mar/20

So i can have n!=42(n−2)!

$${So}\:{i}\:{can}\:{have}\:{n}!=\mathrm{42}\left({n}−\mathrm{2}\right)! \\ $$

Commented by mathocean1 last updated on 20/Mar/20

(n−1)!×(n−2)!=42(n−2)! so (n−1)!=42 but how can i have the value[of n?

$$\left({n}−\mathrm{1}\right)!×\left({n}−\mathrm{2}\right)!=\mathrm{42}\left({n}−\mathrm{2}\right)! \\ $$$${so}\:\left({n}−\mathrm{1}\right)!=\mathrm{42} \\ $$$${but}\:{how}\:{can}\:{i}\:{have}\:{the}\:{value}\left[{of}\:{n}?\right. \\ $$

Commented by jagoll last updated on 20/Mar/20

n! = 42 (n−2)! n (n−1)(n−2)! = 42 (n−2)! n^2 −n−42=0 (n−7)(n+6)=0 n = 7 ←this solution

$$\mathrm{n}!\:=\:\mathrm{42}\:\left(\mathrm{n}−\mathrm{2}\right)! \\ $$$$\mathrm{n}\:\left(\mathrm{n}−\mathrm{1}\right)\left(\mathrm{n}−\mathrm{2}\right)!\:=\:\mathrm{42}\:\left(\mathrm{n}−\mathrm{2}\right)! \\ $$$$\mathrm{n}^{\mathrm{2}} −\mathrm{n}−\mathrm{42}=\mathrm{0} \\ $$$$\left(\mathrm{n}−\mathrm{7}\right)\left(\mathrm{n}+\mathrm{6}\right)=\mathrm{0} \\ $$$$\mathrm{n}\:=\:\mathrm{7}\:\leftarrow\mathrm{this}\:\mathrm{solution} \\ $$

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