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.

To assign the same number of students to each classroom, the number of students (n) must be divisible by the number of classrooms (m).

Question rephrased: Is n/m an integer?

Statement 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.
In other words, the number of students (3n) is divisible by the number of classrooms (m).
Implication:
(3n)/m = 3(n/m) = integer.

Case 1: n/m = integer
It's possible that n=16 and m=4, with the result that n/m = 16/4 = 4.
Here, 3(n/m) = 3(16/4) = 12.

Case 2: n/m = k/3, where k is not a multiple of 3
In this case, since n/m = k/3, m must be a multiple of 3.
It's possible that n=14 and m=6, with the result that n/m = 14/6 = 7/3.
Here, 3(n/m) = 3(7/3) = 7.

Since n/m is an integer in Case 1 but not in Case 2, INSUFFICIENT.

Statement 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.
In other words, the number of students (13n) is divisible by the number of classrooms (m).
Implication:
(13n)/m = 13(n/m) = integer.

Case 3: n/m = integer
It's possible that n=16 and m=4, since 16/4 = 4.
Here, 13(n/m) = 13(16/4) = 52.

Case 4: n/m = k/13, where k is not a multiple of 13
In this case, since n/m = k/13, m must be a multiple of 13.
Not possible, since 3 < m < 13.

Since only Case 3 is possible, n/m = integer.
SUFFICIENT.

The correct answer is B.

aditiniyer wrote:A school administrator will assign each student in a group on n students to one of m classes.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 the 3n students to one of m classes so that each classroom has the same number of students assigned to it.

2)It is possible to assign each of the 13n students to one of m classes so that each classroom has the same number of students assigned to it.

We are given that each student in a group of n students is going to be assigned to one of m classrooms. We are being asked whether it is possible to assign each of the n students to one of the m classrooms so that each classroom has the same number of students.

Thus, we need to determine whether n/m = integer.

Statement One Alone:

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.

Statement one is telling us that 3n is evenly divisible by m. Thus, 3n/m = integer.

However, we still do not have enough information to answer the question. When n = 16 and m = 4, n/m DOES equal an integer; however, when n = 20 and m = 6, n/m DOES NOT equal an integer.

Statement Two Alone:

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 is telling us that 13n is divisible by m. Thus, 13n/m = integer.

What is interesting about this statement is that we know that n is greater than 13 and that m is less than 13 and greater than 3. Thus, we know that m could equal any of the following: 4, 5, 6, 7, 8, 9, 10, 11, or 12. We see that none of those values (4 through 12) will divide evenly into 13.

Knowing this, we can say conclusively that m will never divide evenly into 13. Thus, in order for m to divide into 13n, m must divide evenly into n.

Answer: B