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Math Help - Equation problem help

  1. #1
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    Equation problem help

    \frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}=1

    Note that a,b,c and d are natural numbers and you have to find all the possibilities. I only found that a=b=c=d=4 or a=b=c=6 and d=2.
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  2. #2
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    Quote Originally Posted by brainmixer View Post
    \frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}=1

    Note that a,b,c and d are natural numbers and you have to find all the possibilities. I only found that a=b=c=d=4 or a=b=c=6 and d=2.

    \frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}=1

    first, a symetry exists among a, b, c, d
    so, suppose a=<b=<c=<d
    in this way, 4/a>=1
    a>=1
    well, you can suppose a=1 or a=2 or a=3 or a=4

    * if a=1
    there is no solutions
    * if a=2
    then, \frac{1}{b}+\frac{1}{c}+\frac{1}{d}=\frac{1}{2}
    or b=<c=<d
    so 3/b>=1/2
    b=<6
    in this way, b=1 or b=2 or b=3 or b=4 or b=5 or b=6
    **if b=1 : no solutions
    **if b=2 : no solutions
    **if b=3 : \frac{1}{c}+\frac{1}{d}=\frac{1}{6}
    or c=<d
    so 2/c=>1/6
    c<=12
    in this way, c=1 or c=2 or c=3 or c=4 or c=5 or c=6 or c=7 or c=8 or c=9 or c=10 or c=11 or c=12
    ***if c=1 or c=2 or c=3 or c=4 or c=5 or c=6 : no solutions
    ***if c=7: \frac{1}{7}+\frac{1}{d}=\frac{1}{6}
    \frac{1}{d}=\frac{1}{42}
    so, d=42
    solution 1 : a=2, b=3, c=7, d=42
    ***if c=8 : \frac{1}{8}+\frac{1}{d}=\frac{1}{6}
    \frac{1}{d}=\frac{1}{24}
    so, d=24
    solution 2 : a=2, b=3, c=8, d=24
    ***if c=9 : \frac{1}{9}+\frac{1}{d}=\frac{1}{6}
    \frac{1}{d}=\frac{1}{18}
    so, d=18
    solution 3 : a=2, b=3, c=9, d=18
    ***if c=10 : \frac{1}{10}+\frac{1}{d}=\frac{1}{6}
    \frac{1}{d}=\frac{1}{15}
    so, d=15
    solution 4 : a=2, b=3, c=10, d=15
    ***if c=11 : no solution
    ***if c=12 : \frac{1}{12}+\frac{1}{d}=\frac{1}{6}
    \frac{1}{d}=\frac{1}{12}
    so, d=12
    solution 5 : a=2, b=3, c=12, d=12
    **if b=4:
    .....

    it seems too long....but for sure it's an exhaustive way to solve this problem.
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