Set of numbers

• Mar 20th 2009, 12:09 PM
Apprentice123
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
• Mar 20th 2009, 12:59 PM
HallsofIvy
Quote:

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}\$?

Quote:

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?
• Mar 20th 2009, 01:01 PM
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).
• Mar 20th 2009, 08:13 PM
matheagle
Quote:

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