Riddle: A special $6$-digit number

You already know that the number has to be of the form $$ \underline{1}\, \underline{ }\, \underline{ }\, \underline{ }\, \underline{ }\, \underline{\color{green}{7}} $$

Now since $$\color{green}7\times 2=1\color{red}{4},\quad \color{green}7\times 4=2\color{red}{8}, \quad \color{green}7\times 5=3\color{red}{5},\quad \color{green}7\times 6=4\color{red}{2}$$ we see that the other digits have to be $4,8,5,2$.

The last two digits have to be $57$ since $$27\times 4=1\color{red}{0}8,\quad 87\times 3=2\color{red}{6}1,\quad 47\times 2=\color{red}{9}4$$

The last three digits have to be $857$ since $$457\times 2=\color{red}{9}14,\quad 257\times 4=1\color{red}{0}28$$

The last four digits have to be $2857$ since $$4857\times 2=\color{red}{9}714$$