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-

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

Let P = {(3 \sqrt{6} + 7)}^{89} and F is fractional

part of P .

Then find the remainder when

(PF) + {(PF)}^{2} + {(PF)}^{3} is divided by 31 .

Answered by MJS_new last updated on 01/Dec/20

3 \sqrt{6} \approx 7.34847 \Rightarrow the fractional part of 7 + 3 \sqrt{6}

is - 7 + 3 \sqrt{6}

the fractional part of {(7 + 3 \sqrt{6})}^{2 n + 1} is {(- 7 + 3 \sqrt{6})}^{2 n + 1}

(fractional part of {(7 + 3 \sqrt{6})}^{2 n} is 1 - {(- 7 + 3 \sqrt{6})}^{2 n})

\Rightarrow

P F = {(7 + 3 \sqrt{6})}^{89} {(- 7 + 3 \sqrt{6})}^{89} = 5^{89}

\Rightarrow

(P F) + {(P F)}^{2} + {(P F)}^{3} = 5^{267} + 5^{178} + 5^{89}

the remainders of 5^{n} / 31 are {\begin{cases} 1; n = 3 k \\ 5; n = 3 k + 1 \\ 25; n = 3 k + 2 \end{cases}

89 = 3 \times 29 + 2

178 = 3 \times 59 + 1

267 = 3 \times 89

\Rightarrow remainder is 1 + 5 + 25 = 31 \equiv 0

Commented by soumyasaha last updated on 01/Dec/20

Thanks Sir