# Thread: Very embarrassing question

1. ## 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 $x$
$1 and $2^{32767}-1$ is divisible by $x$.

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

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

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

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

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

2. Continuing your idea, for any natural $x>1$, $(x^n-1)/(x-1)=1+x+\dots+x^{n-1}$. This is checked directly. (This is also a way to find the sum of a geometric progression.)

3. Originally Posted by novice
I got this from the exercise in the introductory chapter of Velleman's "How to Prove It."

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

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

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

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

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

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

Hi, I don't understand what you problem is: you found $x=2^{1057}-1$ (or $x=2^{31}-1$), this answers the problem. What more do you want? The number of digits of $x$?

4. 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.

5. Originally Posted by Laurent
Hi, I don't understand what you problem is: you found $x=2^{1057}-1$ (or $x=2^{31}-1$), this answers the problem. What more do you want? The number of digits of $x$?
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.