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Question Number 126400 by F_Nongue last updated on 20/Dec/20

$$If{a}_n=6^n+8^{n}find\frac{a_{1991}}{49}.$$

Commented by talminator2856791 last updated on 20/Dec/20

$$\mathrm{are}\mathrm{you}\mathrm{wanting}\mathrm{remainder}?$$

Commented by MJS_new last updated on 20/Dec/20

$$a_{1991}>{10}^{1798}\Rightarrow \frac{a_{1991}}{49}>{10}^{1796}$$

Commented by MJS_new last updated on 20/Dec/20

$$\mathrm{the}\mathrm{remainder}\mathrm{is}42$$

Commented by F_Nongue last updated on 20/Dec/20

$$yezIwanttheremainder$$

Commented by MJS_new last updated on 20/Dec/20

$$\mathrm{the}\mathrm{remainder}\mathrm{you}\mathrm{get}\mathrm{by}\mathrm{finding}\mathrm{the}\mathrm{periodical}$$$$\mathrm{remainders}\mathrm{of}\frac{a_n}{49}.\mathrm{start}\mathrm{with}n=1,2,3,...$$$$\mathrm{until}\mathrm{you}\mathrm{have}\mathrm{one}\mathrm{full}\mathrm{period}.\mathrm{then}\mathrm{find}\mathrm{the}$$$$\mathrm{position}\mathrm{of}1991\mathrm{within}\mathrm{this}\mathrm{period}$$

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

$$6^{1991}+8^{1991}\left(\mathrm{mod}49\right)$$$$\varphi \left(49\right)=7^2-7=42$$$$6^{1991}={\left(6^{47}\right)}^{42+17}\equiv 6^{17}$$$$8^{1991}\equiv 8^{17}$$$${\mathrm{a}}_{1991}\equiv 6^{17}+8^{17}\left(\mathrm{mod}49\right)$$$$6^2=36\equiv -13\left(\mathrm{mod}49\right)$$$$6^{17}={\left(6^2\right)}^8.6\equiv {13}^8.6={\left({13}^2\right)}^4.6\equiv {22}^4.6$$$$\equiv {\left({22}^2\right)}^2.6\equiv {(-6)}^2.6\equiv 6^2.6\equiv -13.6\equiv -78\equiv 20\left(\mathrm{mod}49\right)$$$$8^2=64\equiv 15\left(\mathrm{mod}49\right)$$$$8^{17}\equiv {\left(8^2\right)}^8.8\equiv {15}^8.8={\left({15}^2\right)}^4.8\equiv {(-20)}^4.8={20}^4.8$$$$={\left({20}^2\right)}^2.8\equiv 8^2.8\equiv 15.8\equiv 22\left(\mathrm{mod}49\right)$$$${\mathrm{a}}_{1991}\equiv 20+22\equiv 42\left(\mathrm{mod}49\right)$$$$$$

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