Question Number 180938 by Shrinava last updated on 19/Nov/22
$$\overline {\mathrm{ab}}\:+\:\overline {\mathrm{ba}}\:=\:\overline {\mathrm{c3}} \\ $$$$\mathrm{find}:\:\:\:\mathrm{a}+\mathrm{b}+\mathrm{2c}=? \\ $$
Answered by Rasheed.Sindhi last updated on 19/Nov/22
$$\overline {\mathrm{ab}}\:+\:\overline {\mathrm{ba}}\:=\:\overline {\mathrm{c3}}\:;\:\mathrm{a}+\mathrm{b}+\mathrm{2c}=? \\ $$$$\mathrm{10a}+\mathrm{b}+\mathrm{10b}+\mathrm{a}=\mathrm{10c}+\mathrm{3} \\ $$$$\mathrm{11a}+\mathrm{11b}=\mathrm{10c}+\mathrm{3} \\ $$$$\mathrm{a}+\mathrm{b}=\frac{\mathrm{10c}+\mathrm{3}}{\mathrm{11}}\:\in\mathbb{N}\:\wedge\:\mathrm{0}\leqslant\mathrm{c}\leqslant\mathrm{9} \\ $$$$\Rightarrow\:\:\:\mathrm{c}=\mathrm{3} \\ $$$$\mathrm{a}+\mathrm{b}+\mathrm{2c}=\frac{\mathrm{10}\left(\mathrm{3}\right)+\mathrm{3}}{\mathrm{11}}+\mathrm{2}\left(\mathrm{3}\right)=\mathrm{9} \\ $$
Answered by manxsol last updated on 19/Nov/22
$${ab}\:+ \\ $$$$\frac{{ba}}{{c}\mathrm{3}} \\ $$$${a},{b}\neq\mathrm{0}\:\:{a}+{b}\langle\mathrm{10} \\ $$$${posibilitaes} \\ $$$$\begin{array}{|c|c|c|}{{a}}&\hline{{b}}&\hline{{c}}\\{\mathrm{1}}&\hline{\mathrm{2}}&\hline{\mathrm{3}}\\{\mathrm{2}}&\hline{\mathrm{1}}&\hline{\mathrm{3}}\\\hline\end{array} \\ $$$$\left({a},{b},{c}\right)=\left(\mathrm{1},\mathrm{2},\mathrm{3}\right\} \\ $$$$\left({a},{b},{c}\right)=\left(\mathrm{2},\mathrm{1},\mathrm{3}\right) \\ $$$${a}+{b}+\mathrm{2}{c}=\mathrm{9} \\ $$$$ \\ $$