Đáp án: $6224427$
Giải thích các bước giải:
Ta có: $99=9\cdot 11$
Vì $(9,11)=1$
$\to A=\overline{62xy427}\quad\vdots\quad 99$
$\to A\quad\vdots\quad 9,11$
Để $A\quad\vdots\quad9$
$\to 6+2+x+y+4+2+7\quad\vdots\quad9$
$\to x+y+21\quad\vdots\quad9$
$\to x+y+3\quad\vdots\quad9$
Mà $0+0+3\le x+y+3\le 9+9+3\to 3\le x+y+3\le 21$
$\to x+y+3\in\{9,18\}$
Để $A\quad\vdots\quad 11$
$\to (6+x+4+7)-(2+y+2)\quad\vdots\quad 11$
$\to x-y+13\quad\vdots\quad 11$
$\to x-y+2\quad\vdots\quad 11$
Mà $0-9+2\le x-y+2\le 9-0+2\to -7\le x-y+2\le 11$
$\to x-y+2\in\{0,11\}$
Nếu $x+y+3=9\to x-y+2=8-2y$ chẵn
$\to x-y+2=0\to 8-2y=0\to y=4$
$\to x=2$
$\to A=6224427$
Nếu $x+y+3=18\to x-y+2=17-2y$ lẻ
$\to x-y+2=11\to 17-2y=11\to y=3$
$\to x=12$ loại
