Consider the following statement!

$\overline{ABCD}+\overline{EFG}=8768$, and $\overline{ABC}+\overline{DEFG}=6005$.

If $A,\,B,\,C,\,D,\,E,\,F,$ and $G$ are different numbers, Then, what is $\overline{ABCD}$?

My idea is :

$$\begin{aligned} &\overline{ABCD}+\overline{EFG}=8768\\ &1000(A)+100(B+E)+10(C+F)+(D+G)=8768\\ &\begin{cases} &A=8\\ &B+E=7\\ &C+F=6\\ &D+G=8 \end{cases} \end{aligned} $$

But i don't think that this is the best idea. Cz when i tried to solve this, i got the wrong numbers that doesn't satisfied the equation.

Please help.

  • $\begingroup$ It's solvable... the answer is $8625$ (from the multiple choice and it's fits with that equation), but i don't know how to get this. $\endgroup$ – user516076 Oct 11 at 10:19

Let $\,\alpha = \overline{ABC}\,$ and $\,\beta = \overline{EFG}.\,$ Then the system becomes

$$ \color{white}{text}\\ \left\{ \begin{array}{ll} (10\alpha + D) + \beta \;&=\; 8768 \\ \alpha + (1000D + \beta) \;&=\; 6005 \\ \end{array} \right. \color{white}{text}\\ $$

Subtracting and simplifying yields $\,\alpha - 111D = 307.\,$

From the second given equation it is clear that $\,D = 5;\,$ this leads to $\,\alpha = 862\,$ and $\,\beta = 143.\,$

Thus $\,A=8, B=6, C=2, D=5, E=1, F=4\,$ and $\,G=3.\,$ The required solution then is

$$\overline{ABCD} = \boxed{8625}$$

  • 2
    $\begingroup$ Well A very good answer $\endgroup$ – Max Chan Oct 11 at 10:56

The key is rendering the digits in the right order.

First: the second sum is $6005$ and $\overline{ABC}$ must be at least $102$. So we must have $D=5$ with the sum getting over $6000$ as the result of a "carry". Then:

The first sum forces $G=3$ to get $8$ in the units digit.

The second sum forces $C=2$ to get $5$ in the units digit.

The first sum forces $F=4$ to get $6$ in the tens digit.

The second sum forces $B=6$ to get $0$ in the tens digit, and we see there will be a "carry" into the hundreds digit of this sum.

Keep going in this manner to get the rest.


When you have $X+Y\le 9+9$ the most you can carry is $1$.

And if you have $X+Y+1\le 9+9+1$ the most you can carry is $1$.

So you can never carry more than one. Now if you have $X+Y\to 0$ that means $X+Y$ or $X+Y + 1=10$ and you do carry $1$ to the next column.


$\overline{ABC} + \overline{DEFG} = 6005$

has $A+E\to 0$ so we carry and $D+1 = 6$ and $D=5$.


$\overline{ABCD}+\overline{EFG}= \overline{ABC}5+\overline{EFG}=8768$

$5 + G = 8$ or $5+G =18$ but $5+G \le 14$ so $G= 3$.

$\overline{ABC} + 5\overline{EF}3 = 6005$

$C+3 = 15$ or $C+3 = 5$ but $C+3 \le 12$ so $C+3 =5$ so $C=2$.

$\overline{ABCD}+\overline{EFG}= \overline{AB}25+\overline{EF}3=8768$

So $F+2=6$ (obviously not $16$) so $F = 4$

$\overline{AB}2 + 5\overline{E}43 = 6005$

So $B+4=10$ and $B=6$


$6+E = 7$ so $E=1$

And $\overline A625 +143=8768$ and $\overline{A}62 + 5143 = 6005$ so

$A=8$ and $A+1+1 =10$. So $A=8$.

So $\overline{ABCD} = 8625$.

  • $\begingroup$ Thanks for the detail answer. $\endgroup$ – user516076 Oct 11 at 11:22

Well I think it may be solvable, Consider the result you get from the first equation, Then I expand the second equation: $$1000(D)+100(C+E)+10(B+F)+(C+G)=6005$$ Then try to calucate by yourself!

  • $\begingroup$ How do you find 6 unknowns from just 1 equation? Usually you'd need 6 equations for that. You may be able to notice and use some things in the equation, like that $C$ and $G$ are the only variables contributing to the resulting $5$, but that also applies to the original equations and solving it like this or in some other manner would be non-trivial, to say the least. It would greatly improve the answer to expand on the steps required after getting this equation. $\endgroup$ – Dukeling Oct 11 at 17:31

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