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Thread: number exponentiation

  1. #1
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    number exponentiation

    "Is there an exponent of number $\displaystyle 3 ( 3^x ) $ that ends with 0001 in decimal system?" Hope you understood.
    It can be written like this :
    $\displaystyle 3^x=10000*k+1$. Find x
    Last edited by teps; Oct 25th 2010 at 11:07 AM.
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  2. #2
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    Quote Originally Posted by teps View Post
    "Is there an exponent of number $\displaystyle 3 ( 3^x ) $ that ends with 0001 in decimal system?" Hope you understood.
    It can be written like this :
    $\displaystyle 3^x=10000*k+1$. Find x
    Hint: Use Euler's theorem. If you want the actual value of $\displaystyle x$, you will need to compute $\displaystyle \varphi(10000)$. 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 $\displaystyle \{x_1,x_2,\ldots,x_{10001}\}$. Since there are 10000 remainders upon division by 10000, two of the numbers $\displaystyle x$ and $\displaystyle y$ in our set must have the property that $\displaystyle 3^x$ and $\displaystyle 3^y$ have the same remainder when divided by 10000. Say $\displaystyle x>y$.

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