If $999\times \mathrm{abc}=\mathrm{def123}$ in decimal system, then find $\mathrm{a,b,c,d,e\ and\ f}.$ Consider the following multiplication in decimal system:
$$999\times \mathrm{abc}=\mathrm{def123}$$
then find the value of digits $\mathrm{a,b,c,d,e\ and\ f}.$
Here $\mathrm{abc}$ means not $(a\times b\times c$), $\mathrm{abc}$ is a number of 3 digits decimal system (e.g. if $\mathrm{abc}=123$, then $
\mathrm{a=1,\ b=2,\ c=3}$).
 A: The question is basically
$$(1000-1)\times abc=abc000-abc=def123$$
$c=7$,
 $b=7$,
 $a=8$,
which gives
$d=8$, $e=7$, $f=6$
Here's how to do it:
$$a\ b \ c \ 0 \ 0 \ 0 $$
$$-\ \ \ \  \ a\ b\ c $$
$$---------$$
$$\ d \ e\ f\ \ 1 \  2 \ 3$$
Note that $0\leq a,b,c,d,e,f\leq9$
A: Hint:
$$(c+10b+100a)(9+9\cdot10+9\cdot100)=9c+10(9b+9c)+100(9a+9b+9c)+\cdots$$
$\implies9c\equiv3\pmod{10}\iff c\equiv9^{-1}\cdot3\equiv9\cdot3\equiv7\pmod{10}$ as $9\cdot9\equiv1\pmod{10}$
As $0\le c\le9,c=7$
Again, $23\equiv9\cdot7+10(9b+9\cdot7)\pmod{100}$
$\iff-4\equiv9(b+7)\pmod{10}\iff b+7\equiv-36\iff b\equiv-43\equiv7$
Now $0\le b\le9$
Finally $123\equiv9\cdot7+10(9\cdot7+9\cdot7)+100\cdot(7+7+a)\pmod{1000}$
Can you take it from here?
A: Using the expansion of a natural number into powers of $10$ your equation becomes
\begin{equation*}
999\times 10^2a+999\times 10b+999c=10^5d+10^4e+10^3f+123.\tag{1}
\end{equation*}
This implies that $c=7$ because the final digit of the right-hand side of $(1)$ is $3 $, so should be the left-hand side's; and since we have $9(7)=63$, the unit's digit of $999c $ shall equal $3 $. Plugging this value into $(1)$,  simplifying and dividing both sides by $10$, we obtain
\begin{equation*}
999\times 10a+999b+687=10^4d+10^3e+10^2f,\tag{2}
\end{equation*}
which yields $b=7$ because now the final digit of both sides of $(2)$ must equal $0$, and since $9(7)+7=70$, the final digit of $999b+687 $ shall equal $0 $. Proceeding in a similar way to the above, we find that the single solution of $(1)$ is 

$a=8$, $b=7$, $c=7$, $d=8$, $e=7$, $f=6$.

A: \begin{align}
999\times \overline{abc}&=\overline{def123} \\
1000\times \overline{abc} &= \overline{def123} + \overline{abc} \\
\overline{abc000} &= \overline{def000} + 123 + \overline{abc} \\
\text{so we must have }\quad 123 + \overline{abc} &= 1000\\
\therefore \overline{abc} &= 877\\
\text{and also }\quad \overline{abc000} &= \overline{def000} + 1000\\
\overline{abc} &= \overline{def} + 1\\
\therefore \overline{def} &= 876\\
\end{align}
A: Since $\gcd (123,1000)=1$ we have $$(1000-1)(1000-x)\equiv 123 \pmod {1000}\iff  x\equiv 123 \pmod {1000}\iff$$ $$\iff  abc=1000-123=877.$$ 
