Ask your own question, for FREE!
Mathematics 21 Online
jhonyy9 (jhonyy9):

1+(2 on exponent (n-1)) is always prime True or False ?

OpenStudy (anonymous):

true

OpenStudy (anonymous):

False

OpenStudy (anonymous):

It's always odd, but not prime

OpenStudy (ash2326):

yeah n=4 gives 9

OpenStudy (anonymous):

i swear i meant false! lol. i was thinking in my mind no way it can be true and accidently wrot down true

OpenStudy (ash2326):

no problem yasmeen

OpenStudy (anonymous):

\[1 + 2^{n-1}\] = \[(1 +2) (2^{n-2} - 2^{n-3} ... + 1)\]

OpenStudy (anonymous):

so it is always divisible by 3

jhonyy9 (jhonyy9):

so but 1+(2+(2) on exponent (n-1)) will be true ???

OpenStudy (anonymous):

can you type that in the equation?

jhonyy9 (jhonyy9):

i think that is false too but

OpenStudy (anonymous):

Do you know why the first one is false?

jhonyy9 (jhonyy9):

\[1+2*2^{(n-1)}\]

jhonyy9 (jhonyy9):

is always true

jhonyy9 (jhonyy9):

prime

OpenStudy (anonymous):

\[1+ 2^n\] no it's false

jhonyy9 (jhonyy9):

for n=3 is false

OpenStudy (anonymous):

\[2^n + 1 = (2 +1) (2^{n-1} - 2^{n-2} ... + 1)\]

jhonyy9 (jhonyy9):

can be writing the primes in the form of exponent ?

OpenStudy (anonymous):

but this only works if n is odd number So if n is an odd number, it must be divisible by 3

jhonyy9 (jhonyy9):

and than every primes can be writeing like ... ?

OpenStudy (anonymous):

nope for \[n \ge 3\] and n is a odd number, it must have a factor of 3. So it must not be a prime number

jhonyy9 (jhonyy9):

but i like getting one formula with exponent for primes do you lnow it ? please

jhonyy9 (jhonyy9):

or dont exist ?

jhonyy9 (jhonyy9):

do you kno

jhonyy9 (jhonyy9):

know

jhonyy9 (jhonyy9):

sorry for incorrect words

OpenStudy (anonymous):

Check out this wikipedia page http://en.wikipedia.org/wiki/Formula_for_primes

jhonyy9 (jhonyy9):

ok

jhonyy9 (jhonyy9):

thank you

OpenStudy (anonymous):

no problem

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!