# Homework Help: Help, simple math problem.

1. Dec 12, 2009

### johnnyies

1. The problem statement, all variables and given/known data
http://img51.imageshack.us/img51/7789/imagesj.png [Broken]
a, b, c are different digits. Figure out A.

2. Relevant equations
~

3. 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.

Last edited by a moderator: May 4, 2017
2. Dec 12, 2009

### Dick

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

Last edited: Dec 12, 2009
3. Dec 12, 2009

### johnnyies

Can you further explain why it's 1?

4. Dec 13, 2009

### 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.

5. Dec 13, 2009

### 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?

6. Dec 13, 2009

### 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= ? .