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Help, simple math problem.

  1. Dec 12, 2009 #1
    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. jcsd
  3. Dec 12, 2009 #2


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    Well, I'm pretty sure C=1. Can you tell me why? Look at the second and third columns.
    Last edited: Dec 12, 2009
  4. Dec 12, 2009 #3
    Can you further explain why it's 1?
  5. Dec 13, 2009 #4


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    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.
  6. Dec 13, 2009 #5


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

    With that X in: [itex]2A+11B=X+C[/itex] 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 [itex]X\neq 0[/itex]? And finally figure out what the only possible value of X is?
  7. Dec 13, 2009 #6
    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= ? .
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