Manhattan GMAT Challenge Problem of the Week - 19 July 2011

by Manhattan Prep, Jul 19, 2011

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Question

If, for all positive integer values of n, P(n) is defined as the sum of the smallest n prime numbers, then which of the following quantities are odd integers?

I. P(10)

II. P(P(10))

III. P(P(P(10)))

(A) I only

(B) I and II only

(C) I and III only

(D) II and III only

(E) I, II, and III

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.

Lets now list out the values of P(n) for several values of n, and lets 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 oddwe 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.

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