PuneetSood wrote:
Hi,
Can anyone explain, how can we have 1 as a solution using statement 2, since 2^0 X 3^0 + 1 = 2, so we can never reach 1. So only options left are 2 & 3 so ans should be statement B.
Dear
PuneetSood,
I'm happy to respond.
We CAN'T reach 1. That's the point. For any value of r, if r is prime and we can reach it using that formula, it's a Pierpont prime. But if r = 1, which is allowed by Statement #2, then we can't reach that form, and it's not prime anyway, so we get an answer of "no."
If r = 1, is r a Pierpont prime? No. By convention, 1 is not a prime number at all.
If r = 2, is r a Pierpont prime? Yes
If r = 3, is r a Pierpont prime? Yes
See:
https://magoosh.com/gmat/2012/gmat-math- ... me-number/Keep in mind the exact logical arrangement. We are
not guaranteed that the r we pick will be a Pierpont prime. Instead, we are going to pick
any possible r's in that range, and for each one, ask the question, "Is it a Pierpont prime?"
Does this make sense?
ronr34 wrote:
As I see it, the only possibilities are:
2,3,4,5
So the correct answer is A.
Can anyone confirm this?
For statement #1, 1 < r < 5, the possible values of r are {2, 3, 4}. The value r = 5 is not a possibility. For these three values, we are asking the question: if r is this value, is it a Pierpoint prime?
If r = 2, is r a Pierpont prime? Yes
If r = 3, is r a Pierpont prime? Yes
If r = 4, is r a Pierpont prime? NO! It's not a prime number at all.
In order to be a Pierpoint prime, a number must
(a) be a prime number, and
(b) satisfy that equation.
There are plenty of non-prime numbers that satisfy that equation --- for starters, every power of 3 plus 1 (4, 10, 28, 82, 244, etc). If a number isn't prime, it can't be a Pierpont prime.
Does this make sense?
Mike