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OG DS Q128

Expert replies
by krnverma » Wed Jul 13, 2011 2:18 am
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.


I could easily understand how statement (1) is not sufficient, however, I have a doubt. The OG explained that if we can show an example which proves that 3n is divisible by m and n alone is not then we can say that the statement is not sufficient.

As for statement (2), after reading the explanation in the OG I felt that the solution there might not strike me intuitively under exam pressure. I preferred the solution explained by the Grockit OG TV, they simply explained that if 13n is divisible by m then m has to be a factor of 13n (as we know m cannot be a factor of 13, subsequently, it is pretty easy to see that n has to be divisible by m.)

Ideally, this should be it and I can move forward to reviewing the next question, but I am not convinced. What if I apply the same logic to statement (1)? That is, if 3n is divisible by m then as we know m cannot be 3 and, also as 3 is a prime number, n should be divisible by m but IT IS NOT.

Could some one throw some light and make things clearer?

Thanks in advance.

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Source: — Data Sufficiency |

by pemdas » Wed Jul 13, 2011 3:53 am
13/m will always give you fraction without n -> st(2) so n is divisible by m Sufficient, but 3/m will and/or will not give you fraction depending on n -> st (1) Not Sufficient.
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by Jim@Knewton » Wed Jul 13, 2011 8:28 am
Certain DS questions are (over)loaded with words!
In cases like this it is best to start with the Q stem:3 < m < 13 < n => m could be any of(4,5,6,7,8,9,10,11 or 12)
=> To conclude on the possibility of equal assignment of n to m rooms, n must be conclusively shown to be a multiple of m (of ALL possibilities of m, or the specific value of m if indicated directly / indirectly in the statements)
OR
n must be conclusively shown to NOT be a multiple of m (of ALL possibilities of m, or the specific value of m if indicated directly / indirectly in the statements)

From #1: 3n/m = integer => m could be 6, 9 or 12
=> we cannot conclude if n by itself is divisible by 6, 9 and 12
Hence insufficient

From #2: 13n/m = integer => m cannot be 13(nor a multiple of it) and n>13 (because "3 < m < 13 < n")
=> n must be divisible by m
=> #2 is sufficient by itself
Hence B
"That is, if 3n is divisible by m then as we know m cannot be 3 and, also as 3 is a prime number, n should be divisible by m but IT IS NOT."
The point is not that 13 is prime, but that 13 cannot be a factor of m which enables us to conclude that n/m = integer. 3 could be a factor of m for m=6 or 9 or 12.

Hope this helps... :-)
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