Professor Newton divided his students into groups of $4$ and $2$ were left over. He divided his students into groups of 5 and one student was left out. If 15 students are women and there are more women than men, the number of male students is:
(a) 7
(b) 8
(c) 9
(d) 10
(e) 11
Solution 1: Since the number of male students is less than 15 and the number of female students is 15, we have

$$
15<\text { male students }+ \text { female students }<15+15=30
$$

that is: the number of students is between 15 and 30.

On the other hand, when we divide by 4, there are 2 students left, so the number of students is even. When we divide by 5, there is one left, so the last digit of the number is 1 or 6, but being even it can only be 6. Thus, we have only two possible values: 16 and 26. We discard 16 because it is divisible by 4. Therefore, the answer is 26.

Solution 2: As above, the number of students is between 15 and 30. Observe that the number 6 divided by 4 leaves a remainder of 2 and divided by 5 leaves a remainder of 1. Therefore, if we add to 6 a common multiple of 4 and 5, the resulting number will also have this property. The least common multiple of 4 and 5 is 20, so the possible values for the number of students are $6, 26, 46, 66, \ldots$ Given that the number of students is between 15 and 30, the solution is 26.