abc= bc×5
=> abc-bc×5= 0
=> bc(a-5)=0
=> bc=0 =>b=0, c=0
Hoặc a-5=0 => a=5
\(abc=bc.5\Rightarrow c⋮5\Rightarrow c\in\left\{0;5\right\}\)
\(+,c=0\Rightarrow ab0=b0.5\Leftrightarrow ab=b.5\Leftrightarrow b⋮5\Rightarrow b\in\left\{0;5\right\}\)
\(b=0\Rightarrow a00=0\left(loại\right)\)
\(b=5\Rightarrow a50=50.5=250\Rightarrow a=2\Rightarrow abc=250\left(thoaman\right)\)
\(+,c=5\Rightarrow ab5=b5.5\Leftrightarrow ab.10+5=b.50+25\Leftrightarrow ab.2+1=b.10+5\Leftrightarrow ab.2=b.10+4\Leftrightarrow ab=b.5+2\Rightarrow b\in\left\{2;7\right\}\)
\(b=2\Rightarrow a25=25.5=125\Rightarrow a=1\Rightarrow abc=125\left(thoaman\right)\)
\(b=7\Rightarrow a75=75.5=375\Rightarrow a=3\Rightarrow abc=375\left(thoaman\right)\)
\(Vậy:abc\in\left\{250;125;375\right\}\)
