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Need desparate help on this question concerning finding a positive integer

  1. Mar 14, 2005 #1
    fi have no idea what to do and i tried posting it on another forum and nobody replied so please help me! thank you so much!



    find a positive integer n so that 40n is a fifth power (of an integer) 500n is a sixth power, and 200n is a seventh power, or explain why it is impossible to do so (hint: let x = e2(n) and y = e5(n) use the given assumptions about n to find systems of linear congruences that x and y must satisfy. you may leave n in terms of its canonical prime factorization. show as much work as possible
     
  2. jcsd
  3. Mar 14, 2005 #2
    40 = 2^3 * 5, 500 = 2^2 * 5^3, 200 = 2^3 * 5^2.

    For simplicity, assume n = 2^x * 5^y for some natural numbers x, y.

    Then 40n = 2^(x + 3) * 5^(y + 1). If this is to be a fifth power, both exponents must be divisible by 5, i.e.

    x + 3 == 0 (mod 5),
    y + 1 == 0 (mod 5).

    You can find similar requirements for 500n and 200n. Then you solve this system of equations.
     
    Last edited: Mar 14, 2005
  4. Mar 14, 2005 #3
    Thank you for answering my question. I just need some clarification though. When you have the
    x + 3 == 0 (mod 6),
    y + 1 == 0 (mod 6).

    do you solve the congruence x + 3 == 0(mod6) and then you solve the second y+1 == 0(mod6) and once you do you subsitute the remainder into the n = 2^x * 5^y

    For the 500n = 2^(x + 2) * 5^(y + 3).
    x+2 == 0(mod6)
    y+3 ==0(mod6)

    what happens when you solve all the equations how do you know what n would be if you have different numbers?
     
  5. Mar 14, 2005 #4
    I made a mistake in my original post, 40n was supposed to be a fifth power. The equations need to be changed accordingly.

    No. You must find x, y such that all six equations are satisfied /simultaneously/:

    x + 3 == 0 (mod 5),
    y + 1 == 0 (mod 5),
    x + 2 == 0 (mod 6),
    y + 3 == 0 (mod 6),
    x + 3 == 0 (mod 7),
    y + 2 == 0 (mod 7).

    Have you heard about the Chinese remainder theorem? (This system can be solved without explicitely using the CRT though).

    After you have found a solution, you plug it in to n = 2^x * 5^y. That's how you find out what n "is" (but n is not uniquely determined, i.e. there are infinitely many n that will work. The original problem was to find /an/ n that worked, not all n).
     
  6. Mar 14, 2005 #5
    Question again

    how can you use the chinese remainder theorem if you have an x and y and you want to do it simulatenously?
     
  7. Mar 14, 2005 #6
    Um, by first solving

    x + 3 == 0 (mod 5)
    x + 2 == 0 (mod 6)
    x + 3 == 0 (mod 7)

    using the CRT, and then doing the same to

    y + 1 == 0 (mod 5)
    y + 3 == 0 (mod 6)
    y + 2 == 0 (mod 7).
     
  8. Mar 14, 2005 #7
    to use the chinese remainder theorem wouldn't you have to rewrite the equation s for the x's as
    x==-3(mod5)
    x==-2(mod6)
    x==-3(mod7)

    so then we can rewrite it using the q's. and once we figure out what the general form of it the x would be the remainder of it
     
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