7/2 + a/2 + b/3 = (7 + a)/2 + b/3

If a is odd, 7 + a is even, so 7 + a is divisible by 2, and (7 + a)/2 is an integer. So we're adding some integer to b/3, and if we want this sum to be an integer, b/3 will need to be an integer, so b will need to be divisible by 3. Statement 2 guarantees that, but Statement 1 does not, so the answer is B.

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