EEA wrote:A school administrator will assign each student in a group of "n" students to one of "m" classrooms. If 3<m<13<n, 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?

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.

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.


Answer: B

Please explain.

We can rephrase the question in simpler terms as:
Does n=m*x? for n>13 and 3<m<13 ,x is a positive integer.

The first statement tells us that 3n=m*b (where b is a positive integer).
that means that either m or b (or both) has to have 3 as one of its factor.
if b has 3 as its factor than b/3 is an integer and therefore n=(b/3)*m, where b/3 is an integer multiple of m, satisfies the condition.
But if m has 3 as one of its factors and b does not, than b/3 will not be an integer, and the condition is not satisfied. (not sufficient).

The second statement tells us that 13n=m*c (where c is a positive integer). That mean that either m or c or both has to have 13 as one of its factors. since m is less than 13 and 13 is a prime number, m cannot have 13 as one of its factors. Therefore c must have 13 as one of its factors and n=(c/13)*m where c/13 is an interger. Satisfies the condition. Sufficient.

Answer is B.