Foreword: In my solutions, I’ll try to explain what goes on in my mind as I read the question. As you read the solution, you’ll realize that I assimilate and, at times, manipulate the information as I read.

A school administrator will assign each student in a group of N students – We have a group of N students who will be assigned by a school administrator.

to one of M classrooms. – Those N students will be assigned to one of M classrooms. (So, N needs to be a multiple of M. Is it? No. If the same number of students are assigned to each classroom, then N needs to be a multiple of M. Otherwise, N need not be a multiple of M)

If 3<M<13<N – So, 4 ≦ M ≦ 12 AND 14 ≦ N (In my mind, I’ll transform the inequality this way so that I don’t end up taking 3 and 13 to be possible values of M or 13 to be a possible value of N. I’ve used the understanding here that M and N are integers since we can’t have the number of students or the number of classrooms non-integers)

is it possible to assign each of the N students to one of the M classrooms so that each classroom has the same number of students assigned to it? – The question is whether we can assign each of the N students to one of the M classrooms. The constraint is that each classroom has the same number of students assigned to it. (We have M classrooms and N students. For each classroom to have an equal number of students, N needs to be a multiple of M) So, essentially, the question is whether N is a multiple of M.

(1) It is possible to assign each of 3N students to one of M classrooms so that each classroom has the same number of students assigned to it.

This statement tells us that if there were 3N students, we could assign the same number of students to each of the M classrooms. From this info, we get that 3N is a multiple of M. However, our question was whether N is a multiple of M.

(Since we are solving a DS question, our objective while evaluating each information statement is to prove the statement insufficient by coming up with two different answers to the given question, unless of course, we can sense or feel that the statement is sufficient, in which case we try to ‘prove’ that the statement is sufficient. Since in this case, I do not feel that the statement is sufficient, my objective is to prove this statement insufficient by coming up with two different answers to the given question.)

Can we get a Yes answer? i.e. Can N be a multiple of M?

N can easily be a multiple of M since whenever N is a multiple of M, 3N will also be a multiple of M (Thus, our constraint in this statement will be satisfied whenever N is a multiple of M). Thus, Statement 1 can generate a YES.

Can we get a No? I.e. Can N not be a multiple of M?

Here, we have to think a bit. We are given that 3N is a multiple of M. Is it possible that 3N is a multiple of M and, at the same time, N is not a multiple of M?

Uhhh.. Yes. If M has 3 as a factor whereas N does not. For example: If M=6 and N=14, N is not a multiple of M since M contains 3 as a factor whereas N does not. However, 3N i.e. 42 is a multiple of M. (Keep in mind that except a single 3, all other factors of M should be factors of N as well for us to satisfy the constraint in Statement 1. For example, 2 was the other factor of M in this case. 2 was also the factor of N. If 2 had not been a factor of N, then 3N would also not have been a multiple of M)

Thus, statement 1 can also generate a NO answer.

Since we are able to generate two different answers to the question from this statement, Statement 1 is insufficient.

(2) It is possible to assign each of 13N students to one of M classrooms so that each classroom has the same number of students assigned to it.

This statement tells us that if there were 13N students, we could assign the same number of students to each of the M classrooms. From this info, we get that 13N is a multiple of M. However, our question was whether N is a multiple of M.

Can we get a Yes answer? i.e. Can N be a multiple of M?

N can easily be a multiple of M since if N is a multiple of M, 13N will also be a multiple of M (Thus, our constraint in this statement will be satisfied whenever N is a multiple of M). Thus, statement 1 can generate a YES.

Can we get a No? I.e. Can N not be a multiple of M?

Here, we have to think a bit. We are given that 13N is a multiple of M. Is it possible that 13N is a multiple of M and, at the same time, N is not a multiple of M?

Uhhh.. Yes. If M has 13 as a factor whereas N does not. Oh! We have a problem. The maximum value of M is 12. Thus, M cannot have 13 as a factor. So, Statement 2 cannot generate a NO answer.

Thus, we realize that for 13N to be a multiple of M, N also needs to be a multiple of M. Therefore, Statement 2 will always give a YES answer and is thus sufficient.

Thus, the answer is B.

Additional Notes: From this question, we can understand that if ab (where a and b are two integers) is a multiple of c (another integer), then a should also be a multiple of c if b and c do not have any factor in common (other than 1). Think about it!