a) Vì \(15=3.5\Rightarrow\overline{2a13b}⋮3;5\)
Xét số \(\overline{2a13b}⋮5\Rightarrow b=0;5.\)
*Trường hợp 1: \(b=5\Rightarrow\overline{2a13b}=\overline{2a135}\)
Xét số \(\overline{2a135}⋮3\)khi \(\left(2+a+1+3+5\right)⋮3\Leftrightarrow\left(11+a\right)⋮3\)
\(\Rightarrow a=1;4;7.\)
*Trường hợp 2: \(b=0\Rightarrow\overline{2a31b}=\overline{2a310}\)
Xét số \(\overline{2a310}⋮3\)khi \(\left(2+a+1+3+0\right)⋮3\Leftrightarrow\left(6+a\right)⋮3\)
\(\Rightarrow a=0;3;6;9.\)
Vậy \(\left(a;b\right)\in\left\{\left(1;5\right);\left(4;5\right);\left(7;5\right);\left(0;0\right);\left(3;0\right);\left(6;0\right);\left(9;0\right)\right\}.\)
b) Vì \(45=5.9\Rightarrow\overline{51ab}⋮5;9\)
Xét số \(\overline{51ab}⋮5\Rightarrow b=0;5\)
*Trường hợp 1: \(b=0\Rightarrow\overline{51ab}=\overline{51a0}\)
Xét số \(\overline{51a0}⋮9\)khi \(\left(5+1+a+0\right)⋮9\Leftrightarrow\left(6+a\right)⋮9\)
\(\Rightarrow a=3.\)
*Trường hợp 2: \(b=5\Rightarrow\overline{51ab}=\overline{51a5}\)
Xét số \(\overline{51a5}⋮9\)khi \(\left(5+1+a+5\right)⋮9\Leftrightarrow\left(11+a\right)⋮9\)
\(\Rightarrow a=7.\)
Vậy \(\left(a;b\right)\in\left\{\left(3;0\right);\left(7;5\right)\right\}.\)
