Câu 14
Ta có: a+10a+b+100a+10b+c=100b+10c+b
111a+11b+c=101b+10c
111a=90b+9c
111a=9.10b+9c
111a=9.(10b+c)
Vì 111a=9.(10b+c) nên 111a⋮9
a=3,9
TH1:a=3
111.3=9.(10b+c)
333=9.(10b+c)
37=10b+c
37=10.3+7
⇒b=3,c=7
TH2:a=9
111.9=9.(10b+c)
999=9.(10b+c)
111=10b+c
111=10.11+1(loại)
Vậy a,b=3;c=7