Question Number 14451 by christine last updated on 31/May/17
Answered by mrW1 last updated on 31/May/17
$$\mathrm{4}{x}=\mathrm{9}{n}+\mathrm{5} \\ $$$${x}=\frac{\mathrm{9}{n}+\mathrm{5}}{\mathrm{4}}=\frac{\mathrm{8}{n}+\mathrm{4}+{n}+\mathrm{1}}{\mathrm{4}}=\mathrm{2}{n}+\mathrm{1}+\frac{{n}+\mathrm{1}}{\mathrm{4}} \\ $$$${n}+\mathrm{1}=\mathrm{4}{k} \\ $$$${n}=\mathrm{4}{k}−\mathrm{1} \\ $$$${x}=\frac{\mathrm{9}\left(\mathrm{4}{k}−\mathrm{1}\right)+\mathrm{5}}{\mathrm{4}}=\mathrm{9}{k}−\mathrm{1},\:{k}\epsilon\mathbb{P} \\ $$$${i}.{e}.\:{x}=\mathrm{8},\mathrm{17},\mathrm{26},\mathrm{35}…… \\ $$
Answered by RasheedSindhi last updated on 01/Jun/17
$$\mathrm{4x}\equiv\mathrm{5}\left(\mathrm{mod}\:\mathrm{9}\right) \\ $$$$\mathrm{4x}\equiv\mathrm{5}+\mathrm{27}\left(\mathrm{mod}\:\mathrm{9}\right) \\ $$$$\mathrm{4x}\equiv\mathrm{32}\left(\mathrm{mod}\:\mathrm{9}\right) \\ $$$$\mathrm{As}\:\left(\mathrm{4},\mathrm{9}\right)=\mathrm{1},\mathrm{so}\:\mathrm{by}\:\mathrm{dividing}\:\mathrm{by}\:\mathrm{4} \\ $$$$\mathrm{x}\equiv\mathrm{8}\left(\mathrm{mod}\:\mathrm{9}\right) \\ $$$$\mathrm{x}\equiv\mathrm{8}−\mathrm{9}\left(\mathrm{mod}\:\mathrm{9}\right) \\ $$$$\mathrm{x}\equiv−\mathrm{1}\left(\mathrm{mod}\:\mathrm{9}\right) \\ $$$$\mathrm{x}=\mathrm{9k}−\mathrm{1},\:\forall\:\mathrm{k}\in\mathbb{Z} \\ $$
Commented by mrW1 last updated on 01/Jun/17
$${nice}\:{working}! \\ $$$${i}\:{have}\:{a}\:{question}\:{concerning}\: \\ $$$${negative}\:{numbers}\:{in}\:{connection}\:{with} \\ $$$${mod}\:{operation}.\:{if}\:{i}\:{take}\:{k}=−\mathrm{1}\:{for} \\ $$$${example},\:{i}\:{get}\:{x}=−\mathrm{10},\:{and} \\ $$$$−\mathrm{40}\:{mod}\:\mathrm{9}\:=\mathrm{5} \\ $$$${how}\:{is}\:{this}\:{to}\:{understand}? \\ $$
Commented by Tinkutara last updated on 01/Jun/17
$$\mathrm{Because}\:−\mathrm{40}\:=\:\mathrm{9}\left(−\mathrm{5}\right)\:+\:\mathrm{5} \\ $$$$\therefore\:−\mathrm{40}\:\equiv\:\mathrm{5}\:\left(\mathrm{mod}\:\mathrm{9}\right) \\ $$
Commented by mrW1 last updated on 01/Jun/17
$${thanks}! \\ $$