Ta có: \(\overline{ab}\times7=\overline{abc}\)
=>\(7\times\overline{ab}=10\times\overline{ab}+\overline{c}\)
=>\(3\times\overline{ab}=\overline{c}\)
=>c⋮3
=>c∈{3;6;9}
TH1: c=3
=>\(\overline{ab}=\frac33=1\)
=>b=1; a=0
TH2: c=6
=>\(\overline{ab}=\frac63=2\)
=>b=2; a=0
TH3: c=9
=>\(\overline{ab}=\frac93=3\)
=>b=3; a=0