Set of numbers

**Apprentice123** · March 20th 2009, 12:09 PM

It is known that the number $2^{13-1}$ is prime.

Is $n=2^{17-16}$

With all the natural numbers, the number of divisors of n is:

a) 6
b) 10
c) 5
d) 8

**HallsofIvy** · March 20th 2009, 12:59 PM

Originally Posted by Apprentice123

It is known that the number $2^{13-1}$ is prime.

Is $n=2^{17-16}$

?? $2^{17-16}= 2^1= 2$ . Did you mean $2^{17-1}$ ?

With all the natural numbers, the number of divisors of n is:

a) 6
b) 10
c) 5
d) 8

??Is n still [tex]2^{17- 16}? And what does "with all the natural numbers" mean?

**Showcase_22** · March 20th 2009, 01:01 PM

Is $2^{13-1}$ prime?

$2^{13-1}=2^{12}$ which is divisible by 2.

Do you mean $2^{13}-1$ ?

Do you also mean something like this:

$2^{17}-16=2^{17}-2^4=2^4(2^{13}-1)$

Hence it's divisible by 10 numbers: $2,2^2,2^3,2^4$ , $2^{13}-1$ and $2^{13}-1$ multiplied by $2,2^2$ and $2^3$ AND itself and 1 (forgot those two first time round :s).

**matheagle** · March 20th 2009, 08:13 PM

Originally Posted by Showcase_22

Is $2^{13-1}$ prime?

$2^{13-1}=2^{12}$ which is divisible by 2.

Do you mean $2^{13}-1$ ?

Do you also mean something like this:

$2^{17}-16=2^{17}-2^4=2^4(2^{13}-1)$

Hence it's divisible by 10 numbers: $2,2^2,2^3,2^4$ , $2^{13}-1$ and $2^{13}-1$ multiplied by $2,2^2$ and $2^3$ AND itself and 1 (forgot those two first time round :s).

I was trying to figure out how $2^{12}$ could possibly be prime.
I had a nasty injection today and I thought maybe my mind was affected by it.
Whew

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