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.
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