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Homework Help: Simple Math question

  1. Sep 16, 2005 #1
    Hi,
    This is a math problem.

    3AB + BA = 4AC

    In this problem i have to find BA which has to be a [SIZE=3]two digit number[/SIZE].This is what i did this

    3AB + AB = 4AC
    4AB = 4AC
    B=C

    It's not working. Can anyone please give me a hint, thanks for looking at this problem. Have a nice day.
     
  2. jcsd
  3. Sep 16, 2005 #2
    Can BA, a two digit number, always equal AB?
     
  4. Sep 17, 2005 #3

    VietDao29

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    Homework Helper

    I assume AB is a form of a 2 digit number and not A x B.
    And because AB is not A x B, you can't say that AB = BA.
    So AB = 10A + B, BA = 10B + A, AC = 10A + C. 0 < A, B <= 9, and 0 <= C <= 9.
    3AB + BA = (30A + 3B) + (10B + A) = 31A + 13B.
    4AC = 40A + 4C.
    Since you have 3AB + BA = 4AC <=> 31A + 13B = 40A + 4C <=> 9A + 4C = 13B (1). So you have to find A, C such that 9A + 4C is divisible by 13 and 0 < 9A + 4C <= 13 * 9 = 117.
    Since you notice that 9 + 4 = 13, gcd(9, 13) = 1, and gcd(4, 13) = 1. You try to prove that there exists no C such that C <> A, and 9A + 4C is divisible by 13.
    So let C = A + a (a <> 0, and -9 <= a <= 8). So 9A + 4C = 13A + 4a. 13A is already divisible by 13, gcd(4, 13) = 1, but because 9A + 4C needs to be divisible by 13, 4a must be also divisible by 13, so a must be divisible by 13 (because gcd(4, 13) = 1). And there exist no number a such that a <> 0, and -9 <= a <= 8, and a is divisible by 13, so hence
    there exists no C such that C <> A, and 9A + 4C is divisible by 13.
    So what can you say about A, C? What can you say about A, B, C?
    Viet Dao,
     
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