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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

$$3\mid 3.4^{\mathrm{n}}+51=3(4^{\mathrm{n}}+17)\left(\mathrm{clearly}\mathrm{a}\mathrm{multiple}\mathrm{of}3\right)$$$$9\mid 3.4^{\mathrm{n}}+51\iff 3\mid 4^{\mathrm{n}}+17\iff 3\mid (4^{\mathrm{n}}+17)-18=4^{\mathrm{n}}-1$$$$4^{\mathrm{n}}-1=(4-1)(4^{\mathrm{n}-1}+4^{\mathrm{n}-2}+...+4+1)$$$$=3\mathrm{k}\Rightarrow 3\mid 4^{\mathrm{n}}-1\Rightarrow 3\mid 4^{\mathrm{n}}+17\Rightarrow 9\mid 3.4^{\mathrm{n}}+51.$$$$$$

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

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

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

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

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

$$\mathrm{Ok}.$$

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