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
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).