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Number Properties | OG 12

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Number Properties | OG 12

by [email protected] » Sat Sep 15, 2012 10:06 pm
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.

Please do not give the OG Explanation. I did not understand that!!
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Source: — Data Sufficiency |
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by neelgandham » Sun Sep 16, 2012 12:03 am
Please refer to this thread - https://www.beatthegmat.com/a-school-adm ... 42237.html
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by GMATGuruNY » Sun Sep 16, 2012 3:41 am
[email protected] 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.

Please do not give the OG Explanation. I did not understand that!!
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 here (3n) is divisible by the number of classrooms (m), implying that 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 3n/m = (3*16)/4 = 12.
In this case, then n/m = 16/4 = 4, which is an integer.

m=12:
It's possible that m=12 and n=16, with the result that 3n/m = (3*16)/12 = 4.
In this case, n/m = 16/12 = 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 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 13n/m to be an integer, m must divide into n, implying that n/m is an integer.
SUFFICIENT.

The correct answer is B.
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by kelleygrad05 » Wed Dec 19, 2012 3:23 pm
I actually worked this problem out

I picked numbers for statement 1 and found it to be insufficient where:

N=15, M=5

3n/m = 45/5=15 ---> 15/5=3 -----> n is divisible by m or n/m is an integer


and N=15, M=9

3n/m = 45/9=5 ---> 15/9=1.3333 -----> n is NOT divisible by m or n/m is NOT an integer


For statement 2 I picked numbers:


N=14

13N = 182

M=4 no M=5 no M=6 no

I got lazy at that point and said 182 is an even number and isn't divisible by 4 or 6, probably nothing will work and chose answer E although unsure.

Upon re-examining it, I see 7 would have worked. However I agree if you can express this algebraically as you did and the OG's answers did, it does become much simpler than picking these vast arrays of numbers.

My problem with your logic (and the OG's) is that finding statement 2 sufficient is based solely on the fact that 13 is a prime number or that 13 is not divisible by m, is 3 not also a prime? What condition allows you to draw this conclusion for the statement that 13n/m = integer that you can't for 3n/m = integer?
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