If p is a positive integer, is 2^p + 1 a prime number?

(1) p is a prime number.
2^2 + 1 = 5 = Prime
2^3 + 1 = 9 = Not Prime
Insufficient

(2) p is an even number.
2^2 + 1 = 5 = Prime
2^4 + 1 = 17 = Prime
2^6 + 1 = 65 = Not Prime
Insufficient

Combining both :
We only know of one number that is both Even & Prime i.e, 2
2^2 + 1 = 5 = Prime
Sufficient

Hence, C.
_________________

Given p is a positive integer.
Question stem:- Is \(2^p + 1\) a prime number?

St1:-p is a prime number
We don't have sufficient info to determine whether \(2^p + 1\) is a prime number or not.
If p=2, then \(2^p+1=5\),which is prime.
If p=3, then \(2^p + 1=9\), which is not prime
Insufficient.

St2:- p is an even number
If p=2 then \(2^p + 1=5\), which is prime.
If p=0, then \(2^p + 1 =2^0+1=1+1=2\), which is prime
If p=6, then \(2^p + 1 =2^6+1=64+1=65\), which is not prime
Insufficient.

Combining, p has to be both even and prime number; we know p=2 is the only even prime number.
Now, \(2^p+1=2^2+1=4+1=5\), which is prime.
Sufficient.

Ans. (C)
_________________

Statement 1: Test the data. plug value for p. NOT sufficient.

Statement 2: This one is bit tricky. For even integer 0, 2 , 4 we get prime. But if we stop after that we surely miss something . even integer 6 changes the scenario. NOT sufficient.


Combining both options:

Both option together limited our choice. Only 2 is valid here. 2 is even and prime. Sufficient .

The best answer is C.

my logic:
a. 2^p + 1 = prime for only for p= 1,2,4 etc.. non prime for p=3.. 2 answers.. Yes/no -->NS
b. p = even.. this can be checked using p =2 (yes) p=3 (no)

combined .. p = even prime (only ) 2.

2^p + 1 = not prime
C

Here is my method for DS, which I call pre-workout. In most of the times, when I see formulae in question stem of DS I try first to modify them.

2^p + 1 a prime number? We know that any prime number can be expressed as 6n+1 or 6n-1, so

a. 2^p + 1=6n+1, 2^p =6n, 2^p =2*3*n - never a case, because 2^p can not have 3.
b. 2^p + 1=6n-1, 2^p =6n-2, 2^p =2*(3n-1), (3n-1) - has to be 2, 8, 32 ..., or p=2,4,6... - we have to find an option that gives us p=even

(1)+(2) states that p=2 - the only prime number that is even. Answer (C)

P.S. From the beginning this method seems to be slower, but it certanly has several benefits:

1. It allows you to narrow down to information that you are looking for, and not to be confused with fluff in options
2. In some cases it's easier to conduct calculations with only one formula in the beginning
3. Sometimes plug-in method can be very messy, because you have a lot of scenarios.

Instead of solving -- (2^p)+ 1, I started solving for 2^ (p+1).
How to distinguish between the two if the question is written the way its mentioned above.

2^p + 1
If it was to mean 2^(p + 1), there would always be a bracket for p+1.
That is why they say GMAT quant is not much tough, but in hurry and pressure you might misinterpret the nuances.
So, be careful and always be on the watch out for such confusions.
_________________

Thanks Sudarshan.. it means we should not complicate the statement. Solve what is appears to be at first sight. But literally, writing everything in same line may confuse people...

You can’t use p=0 because the stem says that p is a positive integer. 0 is neither negative nor positive.

Bunuel

can we solve this question without testing values??

Thanks!

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________