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Little mathematical problem

  1. Apr 30, 2005 #1
    This is not for help or anything, I just to see your level of mathematical cleverness.

    Knowing that a, b and c are positive whole numbers;

    [tex]a^{-2} + b^{-2} = c^{-2}[/tex]

    What is the sum of all possible values for a or b from 0 to 100?
     
    Last edited: Apr 30, 2005
  2. jcsd
  3. Apr 30, 2005 #2

    Zurtex

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    Well that's the same as:

    [tex]\frac{1}{a^2} + \frac{1}{b^2} = \frac{1}{c^2}[/tex]

    [tex]\frac{a^2 + b^2}{(ab)^2} = \frac{1}{c^2}[/tex]

    Therefore a2 + b2 must be a divisor of (ab)2 and ab=c.

    The rest is a simple matter of writing a program to work the possibilities out under this and adding them together.
     
    Last edited: Apr 30, 2005
  4. Apr 30, 2005 #3

    arildno

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    Zurtex was the one going into the trap..
     
  5. Apr 30, 2005 #4
    Be careful!
     
  6. Apr 30, 2005 #5

    arildno

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    You are a gentleman, Data!
    I was hoping for a hurried response along the lines:
    "Trap?! What trap?" :biggrin:
     
  7. Apr 30, 2005 #6
    such evil people here! :rofl:


    Now, if the poster were going to be really devious, he'd say that the answer is

    [tex]\sum_{ 0 \leq p_i \leq 100} p_i, \ p_i \ \mbox{prime},[/tex]

    with the claim that [itex]a, \ b, \ c[/itex] don't appear in the equation he gave at all - he meant

    [tex]\hat{a} + \hat{b} - 4= \hat{c} -2,[/tex]

    and he also specified that you choose either [itex]a[/itex] or [itex]b[/itex] to add the values for, so it's just the sum of all primes between 0 and 100.
     
    Last edited: Apr 30, 2005
  8. Apr 30, 2005 #7

    Zurtex

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    Haha, I only opened the door, I didn't go down it :tongue:
     
  9. Apr 30, 2005 #8
    Zurtex you are right on track, but not quite. You can solve it, if you have the proper thinking. One hint; square root the equation.
     
  10. Apr 30, 2005 #9
    [tex]\frac{\sqrt{(a^2+b^2)}} {ab} = \frac{1} {c} [/tex]
     
    Last edited: Apr 30, 2005
  11. Apr 30, 2005 #10
    Not at all.
     
  12. Apr 30, 2005 #11
    Is it, by any chance, zero?
    ab=c, which goes against the definition of a prime.
     
  13. Apr 30, 2005 #12
    I'm terribly sorry I meant whole numbers.
     
  14. Apr 30, 2005 #13
    haha, that changes the whole question :tongue:
     
  15. Apr 30, 2005 #14

    arildno

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    How sad; I thought you had made a really good trap Zurtex fell into, even though he strenuously asserts he only opened the (trap-) door a bit.
    What he forgot, is the influence of gravity:
    If you step upon a trap-door, opening it, however little, gravity pulls you down to the bottom.
     
  16. Apr 30, 2005 #15
    It would have been impossible with prime numbers, so there is no trap :p
     
  17. Apr 30, 2005 #16

    Zurtex

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    :biggrin: pfft, whatever, clearly didn't go as far as Werg, anyway I didn't make a single mistake in my answer :rolleyes:
     
  18. Apr 30, 2005 #17
    Now with the equation try awnsering it!

    Now awnser? I'll give another hint. Inverse the equation

    [tex]\frac{ab} {\sqrt{a^{2} + {b^2}}} = c[/tex]

    If no one can awnser by the end of the day I will awnser it myself :frown:
     
    Last edited: May 1, 2005
  19. May 1, 2005 #18
    Ok here how to solve it.

    First, if a, b and c are whole numbers then [tex]{\sqrt{a^{2}[/tex] has to be a certain number that will devid ab in a whole number. Now the result of [tex]{\sqrt{a^{2}[/tex] is either whole or decimal... not quite. Phytagorus with whole numbers NEVER give decimal numbers that are perfect
     
  20. May 1, 2005 #19
    Ok here how to solve it.

    First, if a, b and c are whole numbers then [tex]{\sqrt{a^{2} + b^{2}}[/tex] has to be a certain number that will divide ab in a whole number. Now the result of [tex]{\sqrt{a^{2} + b^{2}}[/tex] is either whole or decimal... not quite. No product of whole numbers is devidable by their Pythagoaras result into another whole number. So, now you know that [tex]{\sqrt{a^{2} + b^{2}}[/tex] has to be a perfect Pythagoras. But witch numbers when multiplied give a number that is dividable by those two numbers Pythagoras? Only 5 factors. Witch numbers give 5 factors perfect Pythagoras? Factors of 3 and 4. So a and b

    have to be:

    Factors of 5
    And factors of 3 or 4

    [tex]3(5)=15[/tex]

    [tex]4(5)=20[/tex]

    Every perfect Pythagoras with factors of 3 and 4 have common factor witch we will name x.

    [tex]a=15x[/tex]
    [tex]b=20x[/tex]

    [tex]\frac{15x(20x)} {\sqrt{(15x)^{2} + {(20x)^2}}} = c[/tex]

    [tex]\frac{300 x^2} {\sqrt{225 x^2 + 400 x^2}}} = c[/tex]

    [tex]\frac{300 x^2} {\sqrt{x^2 625}}} = c[/tex]

    [tex]\frac{300 x^2} {25x}= c[/tex]

    [tex]12x= c[/tex]

    All the possible values for a and be would be

    [tex]15(x+1+1+...)+20(x+1+1...)[/tex]

    The maximum value for x in the case of 15 is 6 since 15*6=90 and 15*7=105
    The maximum value for x in the case of 20 is 5 since 20*5=100 and 20*6=120

    The sum for all numbers to 0 to 6 is

    [tex]\frac{6^2 + 6} {2}=21[/tex]

    and 5

    [tex]\frac{5^2 + 5} {2}=15[/tex]

    SO

    [tex]15(21) + 20(15)=615[/tex]

    But there is ONE exception. That is 65.

    [tex]\frac{65(156)} {\sqrt{65^{2} + {156^2}}} = 169[/tex]

    [tex]615+65=680[/tex]

    680 is the sum of all possible values for a and b. I must admit this problem disturbed a while and it took me almost a week to solve it! I had to rediscover all of this. I guess a quite clever mathematician would have been able to solve it right away...
     
    Last edited: May 1, 2005
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