Sum of Prime numbers...Answer doesn't match

[gmatfeel]

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

I need help with the solution to the above question.
My approach was:
e=even prime integer
o=odd prime integer

P(10) = e+o+o+o+o+o+o+o+o+o
P(10) = (e+o)+(o+o)+(o+o)+(o+o)+(o+o)
RULE: even+even = even
odd+even = odd
odd+odd = even
Using the above rule-
P(10) = (o)+(e)+(e)+(e)+(e)
P(10) = (o)+[(e)+(e)]+[(e)+(e)]
P(10) = (o)+[e]+[e]
P(10) = odd prime integer

so p(10) is odd

Now for solving,
P(P(10)) = (P(10)+P(10)+P(10)+P(10)+P(10)+P(10)+P(10)+P(10)+P(10)+P(10))
P(P(10)) = (o+o+o+o+o+o+o+o+o+o)
P(P(10)) = even prime integer

Now for solving,
P(P(P(10))) = (P(P(10))+P(P(10))+P(P(10))+P(P(10))+P(P(10))+P(P(10))+P(P(10))+P(P(10))+P(P(10))+P(P(10)))
P(P(P(10))) = (e+e+e+e+e+e+e+e+e+e)
P(P(P(10))) = even prime integer

So per my approach I get "(A)" as the answer but the correct answer is "(C)"
I believe there is a flaw in my approach.
Would someone please help me understand the question better.

Thanks...

[goalevan]

P(n) is the sum of the smallest n prime numbers. The question stem asks which of the three choices are odd integers.

The only even prime number is 2, so the pattern will be E + O + O + O + ...

If n is an odd number, we will have an even number (2) and (n - 1)/2 groups of two odd numbers. Since the sum of two odd numbers is even, we will have E + E = E.

P(odd) = E + [(O + O) + (O + O) + ...]
P(odd) = E + [E] = E

If n is an even number, we will have an even number (2), (n - 2)/2 groups of two odd numbers, and another odd number, so E + E + O = O.

P(even) = E + [(O + O) + (O + O) + ...] + O
P(even) = E + [E] + O = O

I. P(10) = P(even) = odd
II. P(P(10)) = P(P(even)) = P(odd) = even
III. P(P(P(10))) = P(P(P(even))) = P(P(odd)) = P(even) = odd

I and III are odd, so C

[gmatfeel]

Got it....thanks very much.
That makes a lot of sense

[amit2k9]

2 is only even prime number.
sum of two odd primes = even
sum of 3 odd primes = odd

thus sum of 9 odd primes = odd
odd sum+ 2 = odd. P(10) = odd. 1 is correct.

sum of 10 primes = 129 where 128 odd primes = sum even + 2 = even. 2 is not correct.

thus P(P(10))) will be odd.

hence 1 and 3 are correct.

[MBA.Aspirant]

P(10) = even + (odd no. of odds = odd) = odd

P(odd) = even + (even no. of odds = even) = even

P(even) = even + (odd no. of odds) = odd