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Let-P-3-6-7-89-and-F-is-fractional-part-of-P-Then-find-the-remainder-when-PF-PF-2-PF-3-is-divided-by-31-

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Question Number 124130 by soumyasaha last updated on 01/Dec/20
Let P = ( 3(√6) + 7 )^(89) and F is fractional part of P. Then find the remainder when (PF) + (PF)^2 + (PF)^3 is divided by 31.
$$ \\ $$$$\:\:\mathrm{Let}\:\:\mathrm{P}\:=\:\left(\:\mathrm{3}\sqrt{\mathrm{6}}\:+\:\mathrm{7}\:\right)^{\mathrm{89}} \:\mathrm{and}\:\:\:\mathrm{F}\:\mathrm{is}\:\mathrm{fractional} \\ $$$$\:\:\mathrm{part}\:\mathrm{of}\:\:\mathrm{P}. \\ $$$$\:\:\mathrm{Then}\:\mathrm{find}\:\mathrm{the}\:\mathrm{remainder}\:\mathrm{when} \\ $$$$\:\:\:\left(\mathrm{PF}\right)\:+\:\left(\mathrm{PF}\right)^{\mathrm{2}} \:+\:\left(\mathrm{PF}\right)^{\mathrm{3}} \:\mathrm{is}\:\mathrm{divided}\:\mathrm{by}\:\mathrm{31}. \\ $$
Answered by MJS_new last updated on 01/Dec/20
3(√6)≈7.34847 ⇒ the fractional part of 7+3(√6) is −7+3(√6) the fractional part of (7+3(√6))^(2n+1) is (−7+3(√6))^(2n+1) (fractional part of (7+3(√6))^(2n) is 1−(−7+3(√6))^(2n) ) ⇒ PF=(7+3(√6))^(89) (−7+3(√6))^(89) =5^(89) ⇒ (PF)+(PF)^2 +(PF)^3 =5^(267) +5^(178) +5^(89) the remainders of 5^n /31 are { ((1; n=3k)),((5; n=3k+1)),((25; n=3k+2)) :} 89=3×29+2 178=3×59+1 267=3×89 ⇒ remainder is 1+5+25=31≡0
$$\mathrm{3}\sqrt{\mathrm{6}}\approx\mathrm{7}.\mathrm{34847}\:\Rightarrow\:\mathrm{the}\:\mathrm{fractional}\:\mathrm{part}\:\mathrm{of}\:\mathrm{7}+\mathrm{3}\sqrt{\mathrm{6}} \\ $$$$\mathrm{is}\:−\mathrm{7}+\mathrm{3}\sqrt{\mathrm{6}} \\ $$$$\mathrm{the}\:\mathrm{fractional}\:\mathrm{part}\:\mathrm{of}\:\left(\mathrm{7}+\mathrm{3}\sqrt{\mathrm{6}}\right)^{\mathrm{2}{n}+\mathrm{1}} \:\mathrm{is}\:\left(−\mathrm{7}+\mathrm{3}\sqrt{\mathrm{6}}\right)^{\mathrm{2}{n}+\mathrm{1}} \\ $$$$\left(\mathrm{fractional}\:\mathrm{part}\:\mathrm{of}\:\left(\mathrm{7}+\mathrm{3}\sqrt{\mathrm{6}}\right)^{\mathrm{2}{n}} \:\mathrm{is}\:\mathrm{1}−\left(−\mathrm{7}+\mathrm{3}\sqrt{\mathrm{6}}\right)^{\mathrm{2}{n}} \right) \\ $$$$\Rightarrow \\ $$$${PF}=\left(\mathrm{7}+\mathrm{3}\sqrt{\mathrm{6}}\right)^{\mathrm{89}} \left(−\mathrm{7}+\mathrm{3}\sqrt{\mathrm{6}}\right)^{\mathrm{89}} =\mathrm{5}^{\mathrm{89}} \\ $$$$\Rightarrow \\ $$$$\left({PF}\right)+\left({PF}\right)^{\mathrm{2}} +\left({PF}\right)^{\mathrm{3}} =\mathrm{5}^{\mathrm{267}} +\mathrm{5}^{\mathrm{178}} +\mathrm{5}^{\mathrm{89}} \\ $$$$\mathrm{the}\:\mathrm{remainders}\:\mathrm{of}\:\mathrm{5}^{{n}} /\mathrm{31}\:\mathrm{are}\:\begin{cases}{\mathrm{1};\:{n}=\mathrm{3}{k}}\\{\mathrm{5};\:{n}=\mathrm{3}{k}+\mathrm{1}}\\{\mathrm{25};\:{n}=\mathrm{3}{k}+\mathrm{2}}\end{cases} \\ $$$$\mathrm{89}=\mathrm{3}×\mathrm{29}+\mathrm{2} \\ $$$$\mathrm{178}=\mathrm{3}×\mathrm{59}+\mathrm{1} \\ $$$$\mathrm{267}=\mathrm{3}×\mathrm{89} \\ $$$$\Rightarrow\:\mathrm{remainder}\:\mathrm{is}\:\mathrm{1}+\mathrm{5}+\mathrm{25}=\mathrm{31}\equiv\mathrm{0} \\ $$
Commented by soumyasaha last updated on 01/Dec/20
Thanks Sir
$$\mathrm{Thanks}\:\mathrm{Sir} \\ $$

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