Đáp án:
$\overline{abc} \in \{ 100;200;300;400;500;600;700;800;900\}$
Giải thích các bước giải:
Ta có:
$A=\dfrac{\overline{abc}}{a+b+c}+1918=\dfrac{100.a+10.b+c}{a+b+c} \leq \dfrac{100a+100b+100c}{a+b+c}+1918$
$\Rightarrow A \leq \dfrac{100(a+b+c)}{a+b+c}+1918=100+1918=2018$
Dấu "=" xảy ra khi $10b+c=100b+100c \Rightarrow -90b=99c$
$\Rightarrow b,c $ trái dấu
Mà $b,c \geq 0$
$\Rightarrow b=c=0$
$\Rightarrow a\in \{1;2;3;4;5;6;7;8;9\} , b=c=0$