Solve Simple Math Problem with Different Digits - Help Needed!

[johnnyies]

Homework Statement

http://img51.imageshack.us/img51/7789/imagesj.png
a, b, c are different digits. Figure out A.

Homework Equations

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The Attempt at a Solution

As far as I know, 2A+ 11B = 111C

A + A + B = X + C but the leading digit is added to B in the next column to get C

Not sure though. I'm probably overlooking the answer to this one.

 

[Dick]

Well, I'm pretty sure C=1. Can you tell me why? Look at the second and third columns.

 

[johnnyies]

Can you further explain why it's 1?

 

[HallsofIvy]

What else could it be? The largest A and B could be is 9 and 8. 9+ 9+ 8= 26 so the largest number that could be "carried" to the second column is 2. Whether B is 8 or 9, adding a 2 carries only a 1 into the third column.

 

[Mentallic]

In other words, you can't get 222 out of adding 2A+11B with A,B between 0-9.

With that X in: $2A+11B=X+C$ I reckon it's best to change it to 10X and obviously (by the same idea that hallsofivy has given) X can be reduced down to being either 0,1 or 2.

Now, can you explain why $X\neq 0$? And finally figure out what the only possible value of X is?

 

[icystrike]

Step By step:
1) What is the maximum number can 2A+B gives?
2)What could be in the tenth place of 2A+B?
3)What is in the tenth place could add B to give 10C+C , given C is a digit , it explains the statement made by Dick:"C=1" & B= 9
4)Equating to first equation that 2A+B=10C+C , A= ? .