Question Number 146408 by tabata last updated on 13/Jul/21
$${find}\:\mathrm{3}{x}\approxeq\mathrm{4}\left(\:{mod}\:\mathrm{5}\right) \\ $$
Commented by tabata last updated on 13/Jul/21
$${help}\:{me}\:{sir} \\ $$
Commented by tabata last updated on 13/Jul/21
$$????????? \\ $$
Commented by otchereabdullai@gmail.com last updated on 16/Jul/21
$$\mathrm{nice}\:\mathrm{question} \\ $$
Answered by gsk2684 last updated on 13/Jul/21
$${to}\:{find}\:{x}\:{such}\:{that}\:\mathrm{5}\:{divides}\:\mathrm{3}{x}−\mathrm{4} \\ $$$$\mathrm{3}{x}−\mathrm{4}=\mathrm{5}{n} \\ $$$${x}=\frac{\mathrm{5}{n}+\mathrm{4}}{\mathrm{3}}\:,{n}\in{Z} \\ $$$${x}\in\left\{…,\frac{\mathrm{4}}{\mathrm{3}},\frac{\mathrm{9}}{\mathrm{3}},\frac{\mathrm{14}}{\mathrm{3}},…\right\} \\ $$
Commented by tabata last updated on 13/Jul/21
$${sir}\:{can}\:{you}\:{give}\:{me}\:{steb}\:{by}\:{steb}\:{please} \\ $$$$ \\ $$$${how}\:{x}\:\in\left\{…..,\frac{\mathrm{4}}{\mathrm{3}},\frac{\mathrm{9}}{\mathrm{3}},\frac{\mathrm{14}}{\mathrm{3}},….\right\} \\ $$
Commented by gsk2684 last updated on 13/Jul/21
$${just}\:{by}\:{substiting}\:{the}\:{value}\:{of}\:{n}\: \\ $$$${if}\:{n}={o}\:{then}\:{x}=\frac{\mathrm{5}\left(\mathrm{0}\right)+\mathrm{4}}{\mathrm{3}}=\frac{\mathrm{4}}{\mathrm{3}} \\ $$$${if}\:{n}=\mathrm{1}\:{then}\:{x}=\frac{\mathrm{5}\left(\mathrm{1}\right)+\mathrm{4}}{\mathrm{3}}=\frac{\mathrm{9}}{\mathrm{3}} \\ $$$${if}\:{n}=\mathrm{2}\:{then}\:{x}=\frac{\mathrm{5}\left(\mathrm{2}\right)+\mathrm{4}}{\mathrm{3}}=\frac{\mathrm{14}}{\mathrm{3}} \\ $$$${and}\:{so}\:{on}. \\ $$
Answered by Rasheed.Sindhi last updated on 13/Jul/21
$$\:\:\:\:\:\:\:\:\:\:\:\mathrm{3}{x}\equiv\mathrm{4}\left({mod}\:\mathrm{5}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\mathrm{3}{x}\equiv\mathrm{4}+\mathrm{5}\left({mod}\:\mathrm{5}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\mathrm{3}{x}\equiv\mathrm{9}\left({mod}\:\mathrm{5}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:{x}\equiv\mathrm{3}\left({mod}\:\mathrm{5}\right) \\ $$$${x}=\mathrm{3}+\mathrm{5}{k};\:\forall{k}\in\mathbb{Z} \\ $$
Answered by physicstutes last updated on 13/Jul/21
![3x ≡ 4 (mod 5) ⇒ 3x = 5y + 4 where y ∈Z so we solve the linear diophantine equation 3x−5y = 4 5 = 1(3) + 2 3 = 1(2) + 1 2 = 2(1)+0 Reversing the steps in euclids algorithm just done above, 1 = 3−1(2) = 3−1[5−1(3)] = 3−1(5)+1(3) = 2(3)−1(5) = 3[5−1(3)]−1(5) = −3(3)−2(5) ∵ x = 3 is a good solution other solutions are such that x ≡ 3 (mod 5)](https://www.tinkutara.com/question/Q146424.png)
$$\mathrm{3}{x}\:\equiv\:\mathrm{4}\:\left(\mathrm{mod}\:\mathrm{5}\right)\:\Rightarrow\:\mathrm{3}{x}\:=\:\mathrm{5}{y}\:+\:\mathrm{4}\:\mathrm{where}\:{y}\:\in\mathbb{Z} \\ $$$$\mathrm{so}\:\mathrm{we}\:\mathrm{solve}\:\mathrm{the}\:\mathrm{linear}\:\mathrm{diophantine}\:\mathrm{equation} \\ $$$$\:\mathrm{3}{x}−\mathrm{5}{y}\:=\:\mathrm{4} \\ $$$$\mathrm{5}\:=\:\mathrm{1}\left(\mathrm{3}\right)\:+\:\mathrm{2} \\ $$$$\:\mathrm{3}\:=\:\mathrm{1}\left(\mathrm{2}\right)\:+\:\mathrm{1} \\ $$$$\mathrm{2}\:=\:\mathrm{2}\left(\mathrm{1}\right)+\mathrm{0} \\ $$$$\mathrm{Reversing}\:\mathrm{the}\:\mathrm{steps}\:\mathrm{in}\:\mathrm{euclids}\:\mathrm{algorithm}\:\mathrm{just}\:\mathrm{done}\:\mathrm{above}, \\ $$$$\:\mathrm{1}\:=\:\mathrm{3}−\mathrm{1}\left(\mathrm{2}\right) \\ $$$$\:\:\:\:=\:\mathrm{3}−\mathrm{1}\left[\mathrm{5}−\mathrm{1}\left(\mathrm{3}\right)\right] \\ $$$$\:\:\:\:=\:\mathrm{3}−\mathrm{1}\left(\mathrm{5}\right)+\mathrm{1}\left(\mathrm{3}\right) \\ $$$$\:\:\:\:=\:\mathrm{2}\left(\mathrm{3}\right)−\mathrm{1}\left(\mathrm{5}\right) \\ $$$$\:\:\:\:=\:\mathrm{3}\left[\mathrm{5}−\mathrm{1}\left(\mathrm{3}\right)\right]−\mathrm{1}\left(\mathrm{5}\right) \\ $$$$\:\:\:\:\:=\:−\mathrm{3}\left(\mathrm{3}\right)−\mathrm{2}\left(\mathrm{5}\right) \\ $$$$\because\:{x}\:=\:\mathrm{3}\:\mathrm{is}\:\mathrm{a}\:\mathrm{good}\:\mathrm{solution}\:\mathrm{other}\:\mathrm{solutions}\:\mathrm{are}\:\mathrm{such}\:\mathrm{that} \\ $$$$\:{x}\:\equiv\:\mathrm{3}\:\left(\mathrm{mod}\:\mathrm{5}\right) \\ $$