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Math Help - number exponentiation

  1. #1
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    number exponentiation

    "Is there an exponent of number 3   ( 3^x ) that ends with 0001 in decimal system?" Hope you understood.
    It can be written like this :
    3^x=10000*k+1. Find x
    Last edited by teps; October 25th 2010 at 12:07 PM.
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  2. #2
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    Quote Originally Posted by teps View Post
    "Is there an exponent of number 3   ( 3^x ) that ends with 0001 in decimal system?" Hope you understood.
    It can be written like this :
    3^x=10000*k+1. Find x
    Hint: Use Euler's theorem. If you want the actual value of x, you will need to compute \varphi(10000). See here for how to do that.
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  3. #3
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    This can be done by a pigeon hole argument.

    Choose 10001 different integers \{x_1,x_2,\ldots,x_{10001}\}. Since there are 10000 remainders upon division by 10000, two of the numbers x and y in our set must have the property that 3^x and 3^y have the same remainder when divided by 10000. Say x>y.

    Hence 3^x-3^y = 10000k for some integer k, whereas 3^y(3^{x-y}-1) = 10000k. But 10000 and 3^y are relatively prime, hence 3^{x-y}-1 = 10000l for some integer l, like you asked for.
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  4. #4
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    Thank you. Got it
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