2012 AMC 10B Problem 9

1. The problem gives running totals. Subtract to find what each new pair adds:
- first pair: $26$;
- second pair: $41 - 26 = 15$;
- third pair: $57 - 41 = 16$.
2. Recall the rules for adding two integers: even + even = even, odd + odd = even, even + odd = odd.
3. The first pair adds to $26$, which is even. Both numbers can be even, for example $12 + 14$. So this pair needs no odd numbers.
4. The third pair adds to $16$, also even. Take $8 + 8$. No odd numbers needed.
5. The second pair adds to $15$, which is odd. Two even numbers always add to an even number, so this pair must contain an odd number. One is enough, for example $7 + 8$.
6. So at least one odd integer is needed, and the example $12, 14, 7, 8, 8, 8$ has exactly one.

The answer is $\boxed{\textbf{(A)}\ 1}$.

Whether a sum is odd or even depends only on how many odd numbers are in it: an odd sum needs an odd count of odd numbers, and an even sum can have none. For "fewest odd numbers" questions, use zero odd numbers wherever the sum is even and one wherever it is odd. Look at what each new pair adds, not at the running totals.

Separate the three pairs by subtracting consecutive totals: the first pair sums to $26$, the second pair to $41 - 26 = 15$, and the third pair to $57 - 41 = 16$.

Two integers with an even sum can both be even, so the first and third pairs need no odd integers at all (for instance $13 + 13$ and $8 + 8$). Two integers with an odd sum must have opposite parity, so the second pair contains exactly one odd integer (for instance $7 + 8$).

Hence at least one odd integer is required, and one is enough.

The answer is $\boxed{\textbf{(A)}\ 1}$.