addition problem

I have tried, but can't figure this problem out. It looks easy ,but I can't figure it out. Any help would be appreciated. It is extra credit.
The letters A, B, and C are 3 different digits. What is ABC if ABC
+ACB
-------
BCA

I'll start you off. Since A and C are different, we know that we're going to have to carry twice. So we get the following system of equations:

C+B=A+10
B+C+1=C+10
A+A+1=B

See if you can take it from here.

Hello, irene12!

The letters A, B, and C are 3 different digits. What is ABC if:

. . . . \(\displaystyle \begin{array}{cccc} & _1 & _2 & _3 \\ & A&B&C \\ +&A&C&B \\ \hline & B & C & A \end{array}\)

Click to expand...

In column-2, we have: .\(\displaystyle B + C \,\text{ ends in }\,C\)

\(\displaystyle \text{There are two choices for }B:\;\;\begin{Bmatrix} B \:=\: 0 \\ B \:=\: 9 & \text{with a "carry" from column 3.}\end{Bmatrix}\)


\(\displaystyle \text{The three-digit sum would not begin with }B = 0.\)
. . \(\displaystyle \text{Hence: }\:B = 9.\)

. . . . \(\displaystyle \begin{array}{cccc} & _1 & _2 & _3 \\ & A&9&C \\ +&A&C&9 \\ \hline & 9 & C & A \end{array}\)


\(\displaystyle \text{In column-1, we have: }\:A \,=\,4\)

. . . . \(\displaystyle \begin{array}{cccc} & _1 & _2 & _3 \\ & 4&9&C \\ +&4&C&9 \\ \hline & 9 & C & 4 \end{array}\)


\(\displaystyle \text{In column-3, we have: }\:C \,=\, 5\)

. . . . \(\displaystyle \begin{array}{cccc} & _1 & _2 & _3 \\ & 4&9&5 \\ +&4&5&9 \\ \hline & 9 & 5 & 4 \end{array}\)


\(\displaystyle \text{Therefore: }\:ABC \:=\:495\)

Thank you for the help. I see how you did this, but I'm not sure how one could solve C+B=A+10, and I'm not sure how one knows that B is 9. Very Good. Thank you again.

Irene, in my partial solution you are not solving a single equation. You need to solve the system of equations simultaneously. That is, you are looking for digits A, B and C that satisfy all 3 equations.

There are many ways to solve a system of equations: elimination, substitution and Gauss Jordon Reduction would be the best in this example.

I think substitution is actually quickest here. Let me do it that way as an example for you:

C+B=A+10
B+C+1=C+10
A+A+1=B

From the second equation, we immediately get B=9.
Substituting this into the third equation we get

2A+1=9
2A=8
A=4

Now substituting A=4 and B=9 into the first equation gives the following:

C+9=4+10
C+9=14
C=5

Thank you for breaking it down for me. I definitely understand it now. Thanks again