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Very tough question

  1. Nov 13, 2013 #1
    1. The problem statement, all variables and given/known data
    Not sure if I'm posting in the right area. Stuck on this question.

    1/a + 1/ab + 1/abc = 25/84
    Find the value of a+b+c

    2. Relevant equations



    3. The attempt at a solution
    I tried making abc = 84 then bc+c +1 = 25

    but still couldn't solve it.
     
  2. jcsd
  3. Nov 13, 2013 #2

    rcgldr

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    I think you're supposed to assume that a, b, c are integers, which limits the number of possibilities to integer factors of 84, since as you mentioned, abc = 84.
     
  4. Nov 13, 2013 #3
    yes I am going to assume a, b, c are integers... but I cannot make the assumption that abc=84 because abc = 84x where x is an integer.
     
  5. Nov 13, 2013 #4

    phyzguy

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    I think there are multiple solutions. I found 10 solutions if negative integers are allowed, 3 if you restrict to positive integers.
     
  6. Nov 13, 2013 #5

    FeDeX_LaTeX

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    Assuming ##a,b,c \in \mathbb{Z}^{+}##, that gets you some of the solutions, since ##bc + c + 1 = 25 \implies c(b + 1) = 24## giving you some case-checking to do.

    You may also wish to consider making the substitution ##bc = 24 - c##, or see how changing ##a## affects ##bc##.
     
  7. Nov 15, 2013 #6
    yes I got bc+c+1=25⟹c(b+1)=24
    but is there some way to do it from here without guess and check?

    where should i sub in bc=24-c?
     
  8. Nov 15, 2013 #7

    phyzguy

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    I don't think there is any way to find the solutions except to restrict the range of possible integers, then exhaustively search through them for possible solutions. As you say, "guess and check".

    As the original problem is posed, even restricted to positive integers, I find three solutions.
     
  9. Nov 15, 2013 #8

    rcgldr

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    not reallly, but as mentioned earlier, assuming that a, b, c are integers and that abc = 84, then a, b, c are each one of the numbers or the product of a pair of the numbers selected from the factors of 84, which are 2,2,3,7.
     
  10. Nov 16, 2013 #9
    Yes, a, b, c, are positive integers and there's actually another condition... a>= b>= c
     
  11. Nov 17, 2013 #10
    c = 84 / (25ab-84b-84) ; a,b,c <>0

    if a, b, c are real number there are many solutions and a , b , c must not be zero.

    if c is integer then 25ab-84b-84 should be the member of {1,-1, 2, -2, 3, -3, 4, -4, 6, -6, 7, -7, 14, -14, 21, -21, 42,-42,84,-84}

    so you have 20 case to test

    case 1 : 25ab-84b-84 = -84 , 25ab -84b = 0 then a =0 that prohibit .
    ...
     
  12. Nov 17, 2013 #11

    FeDeX_LaTeX

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    Are you sure? I don't think there are any solutions with those constraints.
     
  13. Nov 17, 2013 #12

    phyzguy

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    I agree. There are no solutions with the conditions you've given. You'd better check the problem again.
     
  14. Nov 18, 2013 #13
    Oh I'm very sorry - definitely need more sleep. it's a<= b<= c
     
  15. Nov 19, 2013 #14

    phyzguy

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    So with those conditions there is exactly one solution. Have you found it yet?
     
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