câu hỏi khó

S

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Gọi số đó là $\overline{abc}$, ta có:

$\overline{abc}=\overline{bc}.3+8$

$\rightarrow 100a+10b+c=30b+3c+8$

$\rightarrow 100a=20b+2c+8$

$\rightarrow 50a=10b+c+4$

$10b+c+4 < 103$

$\rightarrow a<3$ mà a khác 0 nên $a=1$ hoặc $2$

Xét $a=1$

$\rightarrow 46=10b+c$

$\rightarrow \overline{bc}=46$

$\rightarrow \overline{abc}=146$

Xét $a=2$

$\rightarrow 96=10b+c$

$\rightarrow \overline{bc}=96$

$\rightarrow \overline{abc}=296$

Vậy $\overline{abc}=146$ hoặc $296$



 
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