Find the digits of a number

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.

• 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. – 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

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

• Well A very good answer – 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.

So

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

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

So

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

$$5 + G = 8$$ 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 $$B=6$$

$$\overline{A}625+\overline{E}43=8768$$

$$6+E = 7$$ 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 $$\overline{ABCD} = 8625$$.

• Thanks for the detail answer. – 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!

• 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. – Dukeling Oct 11 at 17:31