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
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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

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.

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
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