rajatvmittal wrote:A Wagstaff prime is a prime number p such that p=((2^q)+1)/3, when q is another prime. If p and q are positive integers, is p a Wagstaff prime? (q is a power to 2)

1) p=q
2) q=3


i chose D

Note: Use "^" to indicate powers.

Target question: Is p a Wagstaff prime?

Given: p=((2^q)+1)/3
p and q must be a prime

Statement 1: p=q
Replace q with p to get: p=((2^p)+1)/3
Multiply both sides by 3 to get: 3p=(2^p)+1
Rearrange to get: 3p - 1 = 2^p
There are 2 possible solutions here: p=1 and p=3. Let's examine each:
p=1: No good, since we're told that p and q must be prime.
p=3: Good, since this means p and q are both prime.
Since we can answer the target question with certainty, statement 1 is SUFFICIENT

Statement 2: q=3
If q=3, then p must equal 3 (which is prime)
Since we can answer the target question with certainty, statement 2 is SUFFICIENT.

Answer = D

Cheers,
Brent

rajatvmittal wrote:i got d too.

but rxplanation says: -
Solution: B

Start with the easier statement first. If p = q, p and q could be any integer, but to answer the question we must know if both p and q are prime; statement (1) is INSUFFICIENT. Statement(2:q = 3. Now plug it into the formula given. ((2^3)+1)/3 = 9/3=3, so p = 3. This is prime, so statement (2) is SUFFICIENT on its own; (B).

I don't quite follow this solution.
If we say that statement 1 is insufficient, then there must be more than 1 possible value of p, such that one value is prime and the other value is not prime. What are these two possible values?

Cheers,
Brent

rajatvmittal wrote:Hi Brent

I got the same question in my mind when i looked at the explanation. I too dint quite understand it and that is the reason I posted it on the forum.

So can i say that answer is D not B

Yes, I think the answer is D.
Unless I'm overlooking something (wouldn't be the first time ).

Cheers,
Brent

This question makes no logical sense, or at least it doesn't if the OA is not C. It may be easiest to illustrate why it is nonsensical with a simpler example. If a question reads:

An odd integer is an integer k such that k = 2m + 1, where m is an integer. If k and m are integers, is k odd?
1. m = 5
2. k = 11

Notice here that the first part of the question, which tells us k is odd if k = 2m + 1 for some integer m, is just a definition of an odd number. In a definition, the letters are just placeholders; it doesn't matter if the definition uses 'k' and 'm' or if it uses 'x' and 'y'. The first part of the question could just as well read "An odd integer is an integer x such that x = 2y + 1 where y is an integer". It is most certainly not a fact about k and m; if you assume it is, then you're assuming in advance precisely what you are supposed to prove, and you'd be committing the 'begging the question' fallacy. So in the question above, the answer is B, since the value of m has nothing to do with anything.

So back to the question above, it starts by defining a Wagstaff prime as one which is equal to (2^q + 1)/3, where q is a prime. But this is just the definition of a Wagstaff prime; it most certainly is not a relationship we should assume holds between p and q, and if we do, we're assuming the answer to the question before we get started. Since letters in definitions are just placeholders, logically, the question is the identical to this one:

A Wagstaff prime is a prime number z such that z=((2^y)+1)/3, when y is another prime. If p and q are positive integers, is p a Wagstaff prime?
1. p=q
2. q=3

from which the answer is very clearly C.