# Thread: Howmany digits in this prime number?

1. ## Howmany digits in this prime number?

Hey Guys,

This is my first post and I imagine there will be a few more judging by how many questions I'm stumped on. Here the first question:

some of you many recognize this value...

2^6972593 - 1

A) Without multiplying it out, determine how many digits are needed to write out this number using base-10

B) Use the factor theorem to show that if 2^p - 1, where p does not equal 3, is a prime number , then p is neither divisible by 4 or divisible by 3. Alternatively , prove that if p is divisible by 4 or 3, then 2^p - 1 is divisible by some number other than +/- itself or +/- .

Thanks for the help

2. Originally Posted by cndman
Hey Guys,

This is my first post and I imagine there will be a few more judging by how many questions I'm stumped on. Here the first question:

some of you many recognize this value...

2^6972593 - 1

A) Without multiplying it out, determine how many digits are needed to write out this number using base-10

2=10^{0.301029996}

should get you there

CB

3. Originally Posted by cndman
Hey Guys,
This is my first post and I imagine there will be a few more judging by how many questions I'm stumped on. Here the first question:
some of you many recognize this value...
2^6972593 - 1

A) Without multiplying it out, determine how many digits are needed to write out this number using base-10
Just adding a little explanation to CB's answer.
$2^{6972593} = 10^x$

$6972593 \log(2) = x \log(10)$

$\dfrac{ 6972593 \log(2)}{\log(10)} = x$
You will need to round x UP to the next integer to get the number of DIGITS in the decimal value.

.

4. Originally Posted by aidan
Just adding a little explanation to CB's answer.
$2^{6972593} = 10^x$

$6972593 \log(2) = x \log(10)$

$\dfrac{ 6972593 \log(2)}{\log(10)} = x$
You will need to round x UP to the next integer to get the number of DIGITS in the decimal value.

.
You will note that in my post I gave a hint on how to approach this, you should not add explanation to the hint untill there has been some response from the OP or at least something like 48 hours has ellapsed.

CB

5. Hey Guys,

This is what I originally figured out:

log(2)^1000000*6.972593, know that low log(2) ~ 0.3010

therefore 1000000*.3010 ~ 300000 * 6.97 = 2.091*10^6

However I think that would be considered multiplying it out, I just couldn't figure out any other way.

Aidan I appreciate the explanation, just one question log(10) = 1 correct? assuming its base 10 and logb(b) = 1.

6. Originally Posted by cndman
Hey Guys,

This is what I originally figured out:

log(2)^1000000*6.972593, know that low log(2) ~ 0.3010

therefore 1000000*.3010 ~ 300000 * 6.97 = 2.091*10^6

However I think that would be considered multiplying it out, I just couldn't figure out any other way.

Aidan I appreciate the explanation, just one question log(10) = 1 correct? assuming its base 10 and logb(b) = 1.
It doesn't matter what base you use, as long as you use the same base for log(2) and log(10). You could choose base 2 or base 10, or base e -- the ratio will remain constant.

In fact, unless a base is specifically given, you should always assume the base is e in a mathematical context.