Ta có:
\(\overline{abc}:\left(a+b+c\right)=25\)
\(\Rightarrow\overline{abc}=25\left(a+b+c\right)\)
\(\Rightarrow\overline{abc}⋮25\Rightarrow\left[{}\begin{matrix}\overline{abc}=\overline{a00}\\\overline{abc}=\overline{a25}\\\overline{abc}=\overline{a50}\\\overline{abc}=\overline{a75}\end{matrix}\right.\)
TH1:\(\overline{abc}=\overline{a00}\)
\(\Rightarrow\overline{a00}=25.a\)
\(\Rightarrow100a=25.a\)
\(\Rightarrow a=0\), loại.
TH2:\(\overline{abc}=\overline{a25}\)
\(\Rightarrow\overline{a25}=25\left(a+b+c\right)=25\left(a+2+5\right)=25a+175\)
\(\Rightarrow100a+25=25a+175\)
\(\Rightarrow100a-25a=175-25\)
\(\Rightarrow75a=150\Rightarrow a=2\)
\(\Rightarrow a=b=2\), loại.
TH3:\(\overline{abc}=\overline{a50}\)
\(\Rightarrow\overline{a50}=25\left(a+5+0\right)=25\left(a+5\right)=25a+125\)
\(\Rightarrow100a+50=25a+125\)
\(\Rightarrow75a=75\Rightarrow a=1\left(TM\right)\)
TH4:\(\overline{abc}=\overline{a75}\)
\(\Rightarrow\overline{a75}=25\left(a+7+5\right)=25a+300\)
\(\Rightarrow100a+75=25a+300\)\(\Rightarrow75a=225\Rightarrow a=3\left(TM\right)\)
Vậy \(\overline{abc}\in\left\{150;375\right\}\)
