If $999\times \mathrm{abc}=\mathrm{def123}$ in decimal system, then find $\mathrm{a,b,c,d,e\ and\ f}.$

Question

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}$).

Answer 1

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 $0\leq a,b,c,d,e,f\leq9$

Answer 2

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?

Answer 3

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

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

Answer 4

\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}

Answer 5

Since $\gcd (123,1000)=1$ we have $$(1000-1)(1000-x)\equiv 123 \pmod {1000}\iff x\equiv 123 \pmod {1000}\iff$$ $$\iff abc=1000-123=877.$$