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Question Number 124130 by soumyasaha last updated on 01/Dec/20

$$$$$$\mathrm{Let}\mathrm{P}={(3\sqrt{6}+7)}^{89}\mathrm{and}\mathrm{F}\mathrm{is}\mathrm{fractional}$$$$\mathrm{part}\mathrm{of}\mathrm{P}.$$$$\mathrm{Then}\mathrm{find}\mathrm{the}\mathrm{remainder}\mathrm{when}$$$$\left(\mathrm{PF}\right)+{\left(\mathrm{PF}\right)}^2+{\left(\mathrm{PF}\right)}^3\mathrm{is}\mathrm{divided}\mathrm{by}31.$$

Answered by MJS_new last updated on 01/Dec/20

$$3\sqrt{6}\approx 7.34847\Rightarrow \mathrm{the}\mathrm{fractional}\mathrm{part}\mathrm{of}7+3\sqrt{6}$$$$\mathrm{is}-7+3\sqrt{6}$$$$\mathrm{the}\mathrm{fractional}\mathrm{part}\mathrm{of}{(7+3\sqrt{6})}^{2n+1}\mathrm{is}{(-7+3\sqrt{6})}^{2n+1}$$$$(\mathrm{fractional}\mathrm{part}\mathrm{of}{(7+3\sqrt{6})}^{2n}\mathrm{is}1-{(-7+3\sqrt{6})}^{2n})$$$$\Rightarrow$$$$PF={(7+3\sqrt{6})}^{89}{(-7+3\sqrt{6})}^{89}=5^{89}$$$$\Rightarrow$$$$\left(PF\right)+{\left(PF\right)}^2+{\left(PF\right)}^3=5^{267}+5^{178}+5^{89}$$$$\mathrm{the}\mathrm{remainders}\mathrm{of}{5}^n/31\mathrm{are}\{\begin{array}{l}1;n=3k\\ 5;n=3k+1\\ 25;n=3k+2\end{array}$$$$89=3\times 29+2$$$$178=3\times 59+1$$$$267=3\times 89$$$$\Rightarrow \mathrm{remainder}\mathrm{is}1+5+25=31\equiv 0$$

Commented by soumyasaha last updated on 01/Dec/20

$$\mathrm{Thanks}\mathrm{Sir}$$

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