Yesterday, Manhattan GMAT posted a GMAT question on our blog. Today, they have followed up with the answer:
Listing out the first several prime numbers (2, 3, 5, 7, 11, etc.), we should notice that after 2, every prime number is odd. The only even prime number is 2.
Let’s now list out the values of P(n) for several values of n, and let’s also note whether the result is even or odd.
If n = 1, then P(1) = 2 = even.
If n = 2, then P(2) = 2 + 3 = 5 = odd.
If n = 3, then P(3) = 2 + 3 + 5 = 10 = even.
If n = 4, then P(4) = 2 + 3 + 5 + 7 = 17 = odd.
We should notice by now that the result flip-flops back and forth between even and odd. After the first value, every new prime that we add is odd, making the new sum flip from odd to even or even to odd. In other words, the outcome depends on whether n is even or odd. Studying the exact pattern, we can conclude that P(even) = odd, and P(odd) = even.
So P(10) is odd—we never need to figure out its exact value. We can eliminate (D) as an answer. Now look at quantity II, and replace the function from the inside out,
P(P(10)) = P(odd) = even. This eliminates (B) and (E) as answers.
Finally, look at quantity III.
P(P(P(10))) = P(P(odd)) = P(even) = odd.
The correct answer is C.
