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solve-in-N-n-4-42-n-2-




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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