We can prime factorize:
1600 = 16*100 = (2^4)(2^2)(5^2) = 2^6 * 5^2
Any divisor of 1600 thus has to look like (2^a)(5^b), where a is between 0 and 6 inclusive, and b is between 0 and 2 inclusive. If our divisor must be even, we must have at least one '2', so a must be at least 1. If our divisor is not to be a multiple of 16, then a must be less than 4. So we have three possible values of a (1, 2 and 3), and three possible values of b (0, 1 and 2), and thus 3*3 = 9 options in total.
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