Question Number 122358 by mathocean1 last updated on 16/Nov/20
![Given a, b ∈[0;4] 309a+15c=226b 1) show that 2b≡0[3] and 3a≡1[5]](https://www.tinkutara.com/question/Q122358.png)
$${Given}\:{a},\:{b}\:\in\left[\mathrm{0};\mathrm{4}\right] \\ $$$$\mathrm{309}{a}+\mathrm{15}{c}=\mathrm{226}{b} \\ $$$$\left.\mathrm{1}\right)\:{show}\:{that}\:\mathrm{2}{b}\equiv\mathrm{0}\left[\mathrm{3}\right]\:{and}\:\mathrm{3}{a}\equiv\mathrm{1}\left[\mathrm{5}\right] \\ $$$$ \\ $$
Answered by mindispower last updated on 16/Nov/20
![309a+15c=226b=225b+b 309a+15c=3(103a+5c)≡0[3] ⇒3(75b)+b≡0[3]⇔b≡0[3]⇒2b=0.2[3]≡0[3] 309a=226b−15c≡b[5] 309=4+5.61 ⇔4a≡b[5]⇒−a=b[5]⇒a=−b[5] ⇒3a=2b[5] 2b=3k first result,⇒2b∈{0,6} b=0⇒309.0+15.0=0,{0,0,0} solution 3a≠1[5] b≠0⇒2b=6≡1[5] 3a=2b[5]⇔3a=1[5],we muste b∈[1,4]](https://www.tinkutara.com/question/Q122374.png)
$$\mathrm{309}{a}+\mathrm{15}{c}=\mathrm{226}{b}=\mathrm{225}{b}+{b} \\ $$$$\mathrm{309}{a}+\mathrm{15}{c}=\mathrm{3}\left(\mathrm{103}{a}+\mathrm{5}{c}\right)\equiv\mathrm{0}\left[\mathrm{3}\right] \\ $$$$\Rightarrow\mathrm{3}\left(\mathrm{75}{b}\right)+{b}\equiv\mathrm{0}\left[\mathrm{3}\right]\Leftrightarrow{b}\equiv\mathrm{0}\left[\mathrm{3}\right]\Rightarrow\mathrm{2}{b}=\mathrm{0}.\mathrm{2}\left[\mathrm{3}\right]\equiv\mathrm{0}\left[\mathrm{3}\right] \\ $$$$\mathrm{309}{a}=\mathrm{226}{b}−\mathrm{15}{c}\equiv{b}\left[\mathrm{5}\right] \\ $$$$\mathrm{309}=\mathrm{4}+\mathrm{5}.\mathrm{61} \\ $$$$\Leftrightarrow\mathrm{4}{a}\equiv{b}\left[\mathrm{5}\right]\Rightarrow−{a}={b}\left[\mathrm{5}\right]\Rightarrow{a}=−{b}\left[\mathrm{5}\right] \\ $$$$\Rightarrow\mathrm{3}{a}=\mathrm{2}{b}\left[\mathrm{5}\right] \\ $$$$\mathrm{2}{b}=\mathrm{3}{k}\:{first}\:{result},\Rightarrow\mathrm{2}{b}\in\left\{\mathrm{0},\mathrm{6}\right\} \\ $$$${b}=\mathrm{0}\Rightarrow\mathrm{309}.\mathrm{0}+\mathrm{15}.\mathrm{0}=\mathrm{0},\left\{\mathrm{0},\mathrm{0},\mathrm{0}\right\}\:{solution} \\ $$$$\mathrm{3}{a}\neq\mathrm{1}\left[\mathrm{5}\right] \\ $$$${b}\neq\mathrm{0}\Rightarrow\mathrm{2}{b}=\mathrm{6}\equiv\mathrm{1}\left[\mathrm{5}\right] \\ $$$$\mathrm{3}{a}=\mathrm{2}{b}\left[\mathrm{5}\right]\Leftrightarrow\mathrm{3}{a}=\mathrm{1}\left[\mathrm{5}\right],{we}\:{muste}\:{b}\in\left[\mathrm{1},\mathrm{4}\right] \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Commented by mathocean1 last updated on 18/Nov/20
$${thanks}\:{sir} \\ $$