gwiz87 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.
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 \(\frac{n}{m}\) 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 here (3n) is divisible by the number of classrooms (m), implying that \(\frac{3n}{m}\) is an integer.
Since we need to determine whether m will always divide into n, plug in EXTREME values for m.
m=4:
It's possible that m=4 and n=16, with the result that \(\frac{3n}{m} = \frac{3*16}{4} = 12\).
In this case, then \(\frac{n}{m} = \frac{16}{4} = 4\), which is an integer.
m=12:
It's possible that m=12 and n=16, with the result that \(\frac{3n}{m} = \frac{3*16}{12} = 4\).
In this case, \(\frac{n}{m}= \frac{16}{12} = \frac{4}{3}\), which is NOT an integer.
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 here (13n) is divisible by the number of classrooms (m), implying that \(\frac{13n}{m}\) is an integer.
It is not possible that m divides into 13, since the only factors of 13 are 1 and 13, and m must be BETWEEN 3 and 13.
Thus, for \(\frac{13n}{m}\) to be an integer, m must divide into n, implying that \(\frac{n}{m}\) is an integer.
SUFFICIENT.
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