A group of n students can be divided into equal groups of 4

Yes you can do the way you started by listing the possible values of n for both patterns and then picking first two matching numbers from these lists. Since we are dealing with easy and small numbers this approach probably would be the fastest one.

A group of n students can be divided into equal groups of 4 with 1 student left over --> n=4q+1 --> n can be: 1, 5, 9, 13, 17, 21, 25, 29, 33, 37, ... (basically an evenly spaced set with common difference of 4)

A group of n students can be divided into equal groups of 5 with 3 students left over --> n=5p+3 --> n can be: 3, 8, 13, 18, 23, 28, 33, 38, ... (basically an evenly spaced set with common difference of 5)

Therefor two smallest possible values of n are 13 and 33 --> 13+33=46.

Answer: B.

Else you can derive general formula based on n=4q+1 and n=5p+3.

Divisor will be the least common multiple of above two divisors 4 and 5, hence 20.

Remainder will be the first common integer in above two patterns, hence 13 --> so, to satisfy both conditions, n must be of a type n=20m+13: 13, 33, 53, ... (two two smallest possible values of n are for m=0 and for m=1, so 13, and 33 respectively) --> 13+33=46.

Answer: B.

For more about this concept see:
https://gmatclub.com/forum/manhattan-re ... ml#p721341
https://gmatclub.com/forum/when-positiv ... ml#p722552
https://gmatclub.com/forum/when-the-pos ... l#p1028654

Hope it helps.