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Question Number 112465 by Aina Samuel Temidayo last updated on 08/Sep/20

Answered by floor(10²Eta[1]) last updated on 08/Sep/20

3∣3.4^n +51=3(4^n +17) (clearly a multiple of 3) 9∣3.4^n +51⇔3∣4^n +17⇔3∣(4^n +17)−18=4^n −1 4^n −1=(4−1)(4^(n−1) +4^(n−2) +...+4+1) =3k⇒3∣4^n −1⇒3∣4^n +17⇒9∣3.4^n +51.

$$\mathrm{3}\mid\mathrm{3}.\mathrm{4}^{\mathrm{n}} +\mathrm{51}=\mathrm{3}\left(\mathrm{4}^{\mathrm{n}} +\mathrm{17}\right)\:\left(\mathrm{clearly}\:\mathrm{a}\:\mathrm{multiple}\:\mathrm{of}\:\mathrm{3}\right) \\ $$$$\mathrm{9}\mid\mathrm{3}.\mathrm{4}^{\mathrm{n}} +\mathrm{51}\Leftrightarrow\mathrm{3}\mid\mathrm{4}^{\mathrm{n}} +\mathrm{17}\Leftrightarrow\mathrm{3}\mid\left(\mathrm{4}^{\mathrm{n}} +\mathrm{17}\right)−\mathrm{18}=\mathrm{4}^{\mathrm{n}} −\mathrm{1} \\ $$$$\mathrm{4}^{\mathrm{n}} −\mathrm{1}=\left(\mathrm{4}−\mathrm{1}\right)\left(\mathrm{4}^{\mathrm{n}−\mathrm{1}} +\mathrm{4}^{\mathrm{n}−\mathrm{2}} +...+\mathrm{4}+\mathrm{1}\right) \\ $$$$=\mathrm{3k}\Rightarrow\mathrm{3}\mid\mathrm{4}^{\mathrm{n}} −\mathrm{1}\Rightarrow\mathrm{3}\mid\mathrm{4}^{\mathrm{n}} +\mathrm{17}\Rightarrow\mathrm{9}\mid\mathrm{3}.\mathrm{4}^{\mathrm{n}} +\mathrm{51}. \\ $$$$ \\ $$

Commented by Aina Samuel Temidayo last updated on 08/Sep/20

Thanks. It′s been a long time. Where have you been?

$$\mathrm{Thanks}.\:\mathrm{It}'\mathrm{s}\:\mathrm{been}\:\mathrm{a}\:\mathrm{long}\:\mathrm{time}.\:\mathrm{Where} \\ $$$$\mathrm{have}\:\mathrm{you}\:\mathrm{been}? \\ $$

Commented by floor(10²Eta[1]) last updated on 08/Sep/20

i just not use this app so often

$$\mathrm{i}\:\mathrm{just}\:\mathrm{not}\:\mathrm{use}\:\mathrm{this}\:\mathrm{app}\:\mathrm{so}\:\mathrm{often} \\ $$

Commented by Aina Samuel Temidayo last updated on 08/Sep/20

Ok.

$$\mathrm{Ok}. \\ $$

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