1. ## Set of numbers

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

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

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

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

2. Originally Posted by Apprentice123
It is known that the number $\displaystyle 2^{13-1}$ is prime.

Is $\displaystyle n=2^{17-16}$
??$\displaystyle 2^{17-16}= 2^1= 2$. Did you mean $\displaystyle 2^{17-1}$?

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?

3. Is $\displaystyle 2^{13-1}$ prime?

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

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

Do you also mean something like this:

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

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

4. Originally Posted by Showcase_22
Is $\displaystyle 2^{13-1}$ prime?

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

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

Do you also mean something like this:

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

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

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