Question Number 110519 by bobhans last updated on 29/Aug/20
$$\:\:\:\:\:\:\mathrm{17x}\:\equiv\:\mathrm{3}\:\left(\mathrm{mod}\:\mathrm{29}\right) \\ $$
Commented by kaivan.ahmadi last updated on 29/Aug/20
$$\mathrm{17}{x}\overset{\mathrm{29}} {\equiv}\mathrm{3}\Rightarrow\mathrm{34}{x}\overset{\mathrm{29}} {\equiv}\mathrm{6}\Rightarrow\mathrm{5}{x}\overset{\mathrm{29}} {\equiv}\mathrm{6}\Rightarrow\mathrm{30}{x}\overset{\mathrm{29}} {\equiv}\mathrm{36}\Rightarrow \\ $$$${x}\overset{\mathrm{29}} {\equiv}\mathrm{7}\Rightarrow{x}=\mathrm{29}{k}+\mathrm{7};\:{k}\in\mathbb{Z} \\ $$
Commented by Rasheed.Sindhi last updated on 29/Aug/20
$$\sqcap\overset{\bullet} {\mid}\sqsubset\exists\:\underset{\bullet} {\mid} \\ $$
Answered by bemath last updated on 29/Aug/20
$$\mathrm{we}\:\mathrm{want}\:\mathrm{to}\:\mathrm{work}\:\mathrm{out}\:\mathrm{a}\:\mathrm{multiplicative} \\ $$$$\mathrm{inverse}\:\mathrm{of}\:\mathrm{17}\:\left(\mathrm{mod}\:\mathrm{29}\right) \\ $$$$\rightarrow\mathrm{17}×\mathrm{12}=\mathrm{1}\:\left(\mathrm{mod}\:\mathrm{29}\right) \\ $$$$\mathrm{so}\:\mathrm{we}\:\mathrm{have}\:\mathrm{12}×\mathrm{17}{x}\:\equiv\:\mathrm{12}×\mathrm{3}\:\left(\mathrm{mod}\:\mathrm{29}\right) \\ $$$$\Rightarrow\:{x}\:\equiv\:\mathrm{36}\:\left({mod}\:\mathrm{29}\right) \\ $$$$\Rightarrow{x}\:\equiv\:\mathrm{7}\:\left(\mathrm{mod}\:\mathrm{29}\right)\:\mathrm{or}\:\mathrm{we}\:\mathrm{got}\: \\ $$$$\Rightarrow{x}\:=\:\mathrm{7}\:+\:\mathrm{29}{k}.\: \\ $$