Mathematics
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jhonyy9 (jhonyy9):
1+(2 on exponent (n-1)) is always prime
True or False ?
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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
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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?
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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
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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
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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
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jhonyy9 (jhonyy9):
know
jhonyy9 (jhonyy9):
sorry for incorrect words
jhonyy9 (jhonyy9):
ok
jhonyy9 (jhonyy9):
thank you
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OpenStudy (anonymous):
no problem