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Any one can find this answer

  1. Jul 24, 2006 #1
    answer = "a" is constant
    a / 9 = x + 8
    a / 8 = x + 7
    a / 7 = x + 6
    a / 6 = x + 5
    a / 5 = x + 4
    a / 4 = x + 3
    a / 3 = x + 2
    a / 2 = x + 1


    a = ?
     
  2. jcsd
  3. Jul 24, 2006 #2
    Why do you think there's a solution? There isn't one.

    Pick any two of the equations and solve them simultaneously for [itex]a, x[/itex]. You will have a unique solution. Plug that solution into any other equation and see if it works (it won't).
     
  4. Jul 24, 2006 #3
    is this a modulus question?
     
  5. Jul 24, 2006 #4
    i got it by microsoft excel if you cant fint it answer is 254+254+1245+766
     
  6. Jul 24, 2006 #5
    I got [tex] a=-n(n+1) [/tex] perhaps he missed something...

    [tex] \frac{a}{n}=x+(n-1) [/tex]
     
  7. Jul 25, 2006 #6
    yes that is wrong
    answer = "a" is constant
    a / 9 = x9 + 8
    a / 8 = x8 + 7
    a / 7 = x7 + 6
    a / 6 = x6 + 5
    a / 5 = x5 + 4
    a / 4 = x4 + 3
    a / 3 = x3 + 2
    a / 2 = x2 + 1

    a = ? a = 2519
    i dont know how to explain
     
  8. Jul 25, 2006 #7
    2519 strikes a bell, it is close to 1/2 7!, which should help you find the answer.
     
  9. Jul 25, 2006 #8

    HallsofIvy

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    Science Advisor

    You've still written that completely wrong. You must mean:
    a= 9x1+ 8
    a= 8x2+ 7
    a= 7x3 + 6
    a= 6x4 + 5
    a= 5x5 + 4
    a= 4x6 + 3
    a= 3x7 + 2
    a= 2x8 + 1
    where the "x"s are all (possibly different) integers.

    That's a "Chinese remainder theorem" problem and can also be stated as
    a= 8 (mod 9)
    a= 7 (mod 8)
    a= 6 (mod 7)
    a= 5 (mod 6)
    a= 4 (mod 5)
    a= 3 (mod 4)
    a= 2 (mod 3)
    a= 1 (mod 2)

    The smallest number divisible by 2, 3, 4, 5, 6, 7, 8, 9 is 9*8*7*5= 2520 (using the highest power of each prime).
    2519 is one less than that: 2519= 2520-1 so 2519= 9(280)- 1= 9(279)+ 9- 1= 9(279)+ 8. 2519= 8(315)- 1= 8(314)+ 8- 1= 8(314)+ 7, etc.
     
  10. Jul 26, 2006 #9
    Ah.."Hallsoftivy" but the problem with "mod" isn't the same posted above..i will try this last one that seems to be clearer...good luck.
     
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