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Analytical Ability

  1. Jan 8, 2004 #1
    How one can find the solution to [tex]x^y+y^x=100[/tex] Where x,y belong to integers?
     
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  3. Jan 8, 2004 #2

    HallsofIvy

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    I can't answer how one would solve such an equation in general!

    However, as soon as I saw that "100" on the right side of the equation, I thought "64+ 36= 82+ 62= 100".
    (Think 3-4-5 right triangle.)

    Hmmm, 8= 23 so 82= (23)2= 26.

    Sure enough, 26+ 62= 100.

    x= 2, y= 6 is a solution.
     
  4. Jan 8, 2004 #3
    I got another solution apart from(2,6)
    It can also be (x=1,y=99)

    I believe there is no other way than churning the combination of numbers out of the mind

    Also, we got 1 equation and two variables
     
  5. Jan 8, 2004 #4

    NateTG

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    Well, let's take a quick look:

    Clearly [tex]x[/tex] and [tex]y[/tex] must be greater than or equal to zero. If one is less than zero, there is a fractional part, if both are less than zero, the sum is on the interval [tex](-2,0)[/tex].

    So, we have:
    [tex]x=0[/tex] no solution.
    [tex]x = 1 \rightarrow y=99[/tex]
    [tex]x = 2 \rightarrow y=6[/tex] (Thanks Halls)
    [tex]x = 3[/tex] no solution.
    Since [tex]x=3[/tex] we have [tex]x^y \in {1,3,9,27,81}[/tex] but none of those work since the complements mod 100 are not powers of the appropriate exponents.
    [tex]x = 4[tex] no solution.
    [tex]x = 5[tex] no solution.
    [tex]x = 6 \rightarrow y=2[/tex]
    Now, since [tex]y[/tex] is monotone decreasing in the next solution, [tex]y\leq1[/tex] so it's
    [tex]x=99 \rightarrpw y=1[/tex]
    since the solutions are symetric.

    Which gives you a complete list of solutions in the integers:
    (1,99),(2,6),(6,2),(99,1)
     
  6. Jan 8, 2004 #5
    Okay from reply of Halls and NateTG? I believe there is no general way

    Anyway thanks Guys for using the symmetry and Mind

    I do found the two solutions i was rather looking for framing the answers, NateTG's Explanation is good
     
  7. Jan 9, 2004 #6

    NateTG

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    Considering that you really only have to check about [tex]log n[/tex] values for [tex]x^y+y^x=n[/tex] it's really not so bad.

    Consider that if [tex]x=y[/tex] then you have
    [tex]2x^x=n[tex]
    so
    [tex]x ln x = ln \frac{n}{2}[/tex]

    So, you'd really only have to check up to [tex]x=8[/tex] or so for [tex]n=1000000[/tex]

    Unless n is really big (so big that it's not practical to store it on a computer) that approach will give you all possible solutions fairily quickly.

    Unless you've got something very specific in mind, that's already a pretty good solution.

    I suppose it would be good if you want to ask questions like:
    Are there n with arbitrarily many solutions?
    or something similar.
     
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