x-Z-15-48a-1-x-mod-17-find-x-

Question Number 144961 by mathdanisur last updated on 30/Jun/21

$${x}\in\mathbb{Z}^{+} \\ $$$$\mathrm{15}^{\mathrm{48}\boldsymbol{{a}}+\mathrm{1}} \:\equiv\:{x}\:\left({mod}\:\mathrm{17}\right)\:\:{find}\:\:{x}=?\: \\ $$

Answered by Rasheed.Sindhi last updated on 01/Jul/21

15^8 ≡x(mod17) 15^8 ≡1(mod17) (15^8 )^(6a) ≡(1)^(6a) (mod17) 15^(48a) ≡1 15^(48a) .15≡1.15(mod17) 15^(48a+1) ≡15(mod17) x≡15(mod17) x=15+17k; k∈{0,1,2,....}

$$\mathrm{15}^{\mathrm{8}} \equiv{x}\left({mod}\mathrm{17}\right) \\ $$$$\mathrm{15}^{\mathrm{8}} \equiv\mathrm{1}\left({mod}\mathrm{17}\right) \\ $$$$\left(\mathrm{15}^{\mathrm{8}} \right)^{\mathrm{6}{a}} \equiv\left(\mathrm{1}\right)^{\mathrm{6}{a}} \left({mod}\mathrm{17}\right) \\ $$$$\mathrm{15}^{\mathrm{48}{a}} \equiv\mathrm{1} \\ $$$$\mathrm{15}^{\mathrm{48}{a}} .\mathrm{15}\equiv\mathrm{1}.\mathrm{15}\left({mod}\mathrm{17}\right) \\ $$$$\mathrm{15}^{\mathrm{48}{a}+\mathrm{1}} \equiv\mathrm{15}\left({mod}\mathrm{17}\right) \\ $$$${x}\equiv\mathrm{15}\left({mod}\mathrm{17}\right) \\ $$$${x}=\mathrm{15}+\mathrm{17}{k};\:{k}\in\left\{\mathrm{0},\mathrm{1},\mathrm{2},….\right\} \\ $$

Commented by mathdanisur last updated on 01/Jul/21