mọi người help me mình ko biết làm. Ai làm mình vote cho (miễn là ko sai)
\(\overline{abcd}+\overline{abc}+\overline{ab}+a=4433\Rightarrow a\le4\)
Ta có
\(b\le9;c\le9\Rightarrow\overline{abcd}\ge4433-499-49-4=3881\Rightarrow a\ge3\)
\(\Rightarrow3\le a\le4\)
\(10.\overline{abc}+d+\overline{abc}+10.a+b+a=4433\)
\(11.\overline{abc}+11.a+b+d=11.403\)(1)
\(\Rightarrow11.\overline{abc}+11.a< 11.403\)
\(\Rightarrow\overline{abc}+a< 403\)(2)
Nếu \(a=4\Rightarrow\overline{abc}+a=400+\overline{bc}+4=404+\overline{bc}>403\) => (2) không đúng \(\Rightarrow a=3\)
Ta có \(11.403⋮11\Rightarrow11.\overline{abc}+11.a+\left(b+d\right)⋮11\)
Mà \(11.\overline{abc}+11.a⋮11\Rightarrow\left(b+d\right)⋮11\Rightarrow\left(b+d\right)=\left\{0;11\right\}\)
+ Nếu \(b+d=0\)
\(\left(1\right)\Leftrightarrow11.\overline{abc}+11.a=11.403\)
\(\Leftrightarrow\overline{abc}+a=403\)
\(\Rightarrow\overline{3bc}+3=403\)
\(\Rightarrow300+\overline{bc}+3=403\Rightarrow\overline{bc}=100\) (trường hợp này loại)
+ Nếu \(b+d=11\)
\(\left(1\right)\Leftrightarrow11.\overline{abc}+11.a+11=11.403\)
\(\Leftrightarrow11.\overline{abc}+11.a=11.402\Rightarrow\overline{abc}+a=402\)
\(\Rightarrow\overline{3bc}+3=402\)
\(\Rightarrow300+\overline{bc}+3=402\Rightarrow\overline{bc}=99\Rightarrow b=9\Rightarrow d=2\)
Thử
\(3992+399+39+3=4433\)
