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Instead of counting two sums (of odd and even digits) and returning their difference, you could use a single sum value, subtracting a digit's value if it's even and adding if it's odd. An added benefit of this approach is that you are more protected from integer overflows: if odd and even digits are interleaved, the two-sum approach will be much less likely to overflow.
We know that it cannot be Option 3 because it doesn't include Option 1 but does include Option 2 which does include Option 1. For Option 3 to be valid it would have to include Option 1. We know it cannot be Option 2 because it is then not the only, and exactly one, correct answer. Option 3 fails on this account as well, but I like the argument for 3 above better. Option 1 references only itself, and is exactly one answer, and is thus correct.
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