This kind of question can give students fits. The key is to figure out what is being asked. The function P is summing up the first N prime numbers, so for example P(3) = 2 + 3 + 5. The total is 10. If I picked P(4), I'd get the same sum + 7, or 17.

Since the question hinges on whether the sum is even or odd, the only number that's unique in this circumstance is 2, as it is the only even prime number in an otherwise homogenous sea of odd numbers. Thus P(1) is even, P(2) is odd, P(3) is even again and P(4) is odd again. This is the pattern, so clearly P(10) will be 2 + nine odd numbers, so it will be odd. We can eliminate D.

P(P(10)) is where this starts getting interesting. You're doing the same test on a number we don't exactly know, but we know it must be odd. Since we know the pattern, the odd number will give us even. P(P(10)) will not be odd, eliminate B and E.

P(P(10))) will be the same function over a number we just calculated would be even. Hence it must be odd again. Eliminate A, the answer must be C.

Function questions are among the least understood questions on the GMAT, and this type of question can get people spending 3-4 minutes extrapolating numbers. If you understand the pattern using a small sample and reasoning, you can get this question right in under two minutes.

Hope this helps!
-Ron
_________________

2 is the only even prime number. Now, the sum of first 10 prime numbers = 9*odd + 2 = odd. Thus P(10) = odd.

Again, the sum of the first n -odd primes = (n-1)*odd+2 = even.Thus, p(P(10)) = even.

Similarly, the sum of the first m - even primes = (m-1)*odd+2 = odd.

I and III.

C.
_________________

Ah that makes sense... took me a while.

I. Even + 9 odds = odd.
II. Even + (odd * (odd - 1)) = even
III. Even + (odd * (even - 1)) = odd

Nice question +1. For this I would identify two cases: First we know that the only even number is 2. Therefore, if the number of prime integers that is n is even then the sum is odd, while if n is odd the sum is even.

In I we have that n is even therefore sum is odd. TRUE.

In II we have that the sum is odd so if we take that n is odd then the sum again will be even. FALSE.

In III, we get that the sum of n=10 is odd. Now the sum of odd is even and again the sum of even is odd. So III is TRUE as well.

Thus answer is C

Just to clarify

n=even, sum is odd
n=odd, sum is even

Hope this is clear
Cheers
J

I. P(10) = 2 + Σ(9 odd integers) = 2 + Σ(8 odd integers) + odd = 2 + even + odd = odd

II. P(P(10)) = P(odd) = 2+ Σ(even # of odd integers) = even + even = even

III. P(P(P(10)) = P(even) = 2 + Σ(odd # of odd integers) = 2 + odd = odd

C.

Thanks for the solution....
Putting numbers to solve this kind of question can take 5 minutes easily....
We must follow the right approach to solve the question quickly...
_________________

P(n) = Sum of 1st n prime numbers. Among these prime numbers only 2 i.e. first prime number is even.
P(10 )= Sum of 2 and 9 odd numbers = even +9*odd = odd
P(P(10)) = P(odd) = even + even *odd = even
P(P(P(10))) = P(even) = even + odd * odd = odd.

So I and III only
Hence Answer C
_________________

The question basically says that
\(P(n) = 2 + 3 + 5 + ... + n^{th} prime number\)
\(P(10) = 2 + 3 + 5 + ... + 29\)
= \(even + odd + odd + ... 10^{th}\) odd prime number
= even + nine odd numbers
= even + odd
= odd

So, I has to be there in the answer. Based on answer choices we are nowhere as I is in all of them.

\(P(odd) = 2 + 3 + 5 + ... + odd^{th} prime number\)
= even + even number of odd prime numbers
= even + even
= even

So, II must not be there in the answer.
B, D and E are eliminated.

\(P(even) = 2 + 3 + 5 + ... + even^{th} prime number\)
= even + odd number of prime numbers
= even + odd
= odd

So, III must be there. Hence II and III are the odd ones.

Answer D.
_________________