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Homework Help: A simple problem on linear equation in two variables

  1. Jul 22, 2010 #1
    1. The problem statement, all variables and given/known data
    if a+11b is completely divisible by 13 and a+13b is completely divisible by 11 then find the the smallest possible value of a+b. a and b are positive integers.
    the answer is 28


    2. Relevant equations

    a+11b= 13x
    a+13b= 11y

    3. The attempt at a solution
    i have equated them many times but am still getting 3 variables. hope anyone knows how to solve this.
     
  2. jcsd
  3. Jul 22, 2010 #2
    With the system of equations you've posted,

    You won't be able to solve for all a, b, x and y since you only have 2 equations.
     
  4. Jul 22, 2010 #3
    it is not necessary that you take x and y. you may also want to try to make expressions which are always divisible with 13 or 11 as per the condition. just refer to the question. not necessary that you consider my equations. they are not mentioned anywhere in the book. just calculate a+b in any manner. if you want i can also tell you the answer.
     
  5. Jul 23, 2010 #4
    The point of these forums is for us to help you.

    If you make more of an attempt maybe we'll be able to better assist you. As it stands with your current attempt(and current system of equations), solving for all those variables is impossible.

    Maybe you need to rework your equations.
     
  6. Jul 24, 2010 #5

    HallsofIvy

    User Avatar
    Science Advisor

    This is a problem in diophantine equations- a and b must be integer so that, while a single equation in two variables has an infinite number of solutions, it is possible to write a formula for them.

    Here, you are saying that a+ 11b= 13x for some integers a, b, and x, and that a+ 13b= 11y for some integer y.

    We can subtract one equation from another to bet 2b= 11y- 13x. Now, 11 divides into 13 once with remainder 2: 13- 11= 2. So one solution, for b= 1, is x=-1, y= -1. But if we take x=-1+ 11k, y= -1+ 13k, then 11y- 13k= 11(-1+ 13k)- 13(-1+ 11k)= -11+ 11(13)k+ 13- 13(11)k= 2 for all k. And since x= -1+ 11k, y= -1+ 13k is the "general solution" for b= 1, x= -b+ 11k, y= -b+ 13k (technically, it should be "-b+ 11kb" but I has absorbed the b into the integer k) is the general solution for any b.

    Now go back to a= 11y- 13b= 11(-b+ 13k)- 13b= -24b+ 143k. We have a+ b= -23b+ 143k. Since a and b are both postive, a+ b must be postive:-23b+ 143k> 0 or 143k> 23b.
     
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