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Thread: Very embarrassing question

  1. #1
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    Very embarrassing question

    I got this from the exercise in the introductory chapter of Velleman's "How to Prove It."

    Exercise 1b: Find an integer $\displaystyle x$
    $\displaystyle 1<x<2^{32767}-1$ and $\displaystyle 2^{32767}-1$ is divisible by $\displaystyle x$.

    I have got $\displaystyle 32767 = 2^{(31)(1057)}$, such that $\displaystyle x= 2^{1057}-1$ and that

    $\displaystyle \frac{2^{32767}-1}{2^{1057}-1}$

    $\displaystyle log_{10} (y) = log_{10} (2^{32767}-1) - log_{10}(2^{1057}-1)$ is too large to handle.

    $\displaystyle log_2 (y) = log_2 (2^{32767}-1) - log_2(2^{1057}-1)$ is unmanageable.

    $\displaystyle log_2$ is a foreign language to me. I don't know how to proceed from here.

    Someone please help.
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  2. #2
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    Continuing your idea, for any natural $\displaystyle x>1$, $\displaystyle (x^n-1)/(x-1)=1+x+\dots+x^{n-1}$. This is checked directly. (This is also a way to find the sum of a geometric progression.)
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  3. #3
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    Quote Originally Posted by novice View Post
    I got this from the exercise in the introductory chapter of Velleman's "How to Prove It."

    Exercise 1b: Find an integer $\displaystyle x$
    $\displaystyle 1<x<2^{32767}-1$ and $\displaystyle 2^{32767}-1$ is divisible by $\displaystyle x$.

    I have got $\displaystyle 32767 = 2^{(31)(1057)}$, such that $\displaystyle x= 2^{1057}-1$ and that

    $\displaystyle \frac{2^{32767}-1}{2^{1057}-1}$

    $\displaystyle log_{10} (y) = log_{10} (2^{32767}-1) - log_{10}(2^{1057}-1)$ is too large to handle.

    $\displaystyle log_2 (y) = log_2 (2^{32767}-1) - log_2(2^{1057}-1)$ is unmanageable.

    $\displaystyle log_2$ is a foreign language to me. I don't know how to proceed from here.

    Someone please help.
    Hi, I don't understand what you problem is: you found $\displaystyle x=2^{1057}-1$ (or $\displaystyle x=2^{31}-1$), this answers the problem. What more do you want? The number of digits of $\displaystyle x$?
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  4. #4
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    emakarov,
    Thank you for coming to my rescue. I partially understood your suggestion. I will spend the rest of my day contemplating it. If I still don't get it, you will see this popping up again.
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  5. #5
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    Quote Originally Posted by Laurent View Post
    Hi, I don't understand what you problem is: you found $\displaystyle x=2^{1057}-1$ (or $\displaystyle x=2^{31}-1$), this answers the problem. What more do you want? The number of digits of $\displaystyle x$?
    Laurent,
    You are absolutely right. I wasn't wrong at all for thinking that I had posted an embarrassing question.

    Oh, I am a brain dead!

    Thank you for helping me.
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