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Math Help - Find all positive integers such that...?

  1. #1
    Super Member fardeen_gen's Avatar
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    Find all positive integers such that...?

    Find all the positive integers (x,y,z) such that xyz = 5(x + y + z)
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  2. #2
    Member Henderson's Avatar
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    (3,4,5)
    (2,5,7)
    (2,4,10)
    (2,3,25)

    I believe these are the only solutions. My method of solving was nothing to speak of, aside from starting with the recognition that at least one of the variables had to be a multiple of five.
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  3. #3
    Super Member fardeen_gen's Avatar
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    The solution itself does not carry any importance. Does anybody know the method of solving this problem?
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  4. #4
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    Quote Originally Posted by fardeen_gen View Post
    Find all the positive integers (x,y,z) such that xyz = 5(x + y + z)
    If x,y\geq 4 then 4xy > 5(x+y+4).

    We will show that if x,y,z\geq 4 then xyz > 5(x+y+z). We will use induction starting at z=4. At z=4 this statement is true by above. Say z is true i.e. xyz > 5(x+y+z) then xy(z+1) = xyz + xy > 5(x+y+z) + 5 = 5(x+y+(z+1)). This completes induction.

    This tells us that if z\geq 4 then it must mean that x=1,2,3 and y=1,2,3 for there to be a possible solution. And if z=1,2,3 then there is a chance to be a solution. Since this is a symmetric equation it means if (a,b,c) is a solution then any permutation is a solution. Therefore, it is sufficient to check whether this equation has integer solution for z=1,2,3.

    If z=1 then the equation is xy = 5(x+y+1). We can write xy-5x-5y=5. Then add and subtract 25 to get, xy-5x-5y+25 = 30, finally factor, (x-5)(y-5)=30. The possible solutions are x-5=\pm 1,\pm 2,\pm 3,\pm 5,\pm 6,\pm 10,\pm 15,\pm 30 and y-5 = \pm 30,\pm 15,\pm 10, \pm 6, \pm 5,\pm 3,\pm 2,\pm 1. But because x,y>0 we cannot choose the negative solutions. Thus, in fact the solutions are x=6,7,8,10,11,15,,20,35 and y=35,20,15,11,10,8,7,6 (in that order). Because of permuting solutions we can ignore some and write: (6,35,1),(7,20,1),(8,15,1),(10,11,1).

    If z=2 then the equation is 2xy = 5(x+y+2). Rewrite as (2x-5)(2y-5) = 45. The solutions are (no negatives) 2x-5=1,3,5,9,15,45 and 2y-5=45,15,9,5,3,1. Thus, x=3,4,5,7,10,25 and y=25,10,7,5,4,3. This gives us (3,25,2),(4,10,2),(5,7,2).

    If z=3 then equation is 3xy = 5(x+y+3). Rewrite as (3x-5)(3y-5)=70. Thus, 3x-5=1,2,5,7,10,14,35,70 and 3x-5=70,35,14,10,7,5,2,1. This gives x=2,4,5,25 and y=25,5,4,2. This gives us (2,25,3),(4,5,3).

    And if z\geq 4 then as explained above there are no other solutions if we think of permutations.
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