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# A school administrator will assign each student in a group

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SVP
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A school administrator will assign each student in a group [#permalink]

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07 Oct 2005, 14:46
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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.

OPEN DISCUSSION OF THIS QUESTION IS HERE: a-school-administrator-will-assign-each-student-in-a-group-127509.html
[Reveal] Spoiler: OA
Director
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07 Oct 2005, 21:19
i think the answer is D

Question basically asks whether n/m is an integer.
from A, we can say 3(n/m) is an integer. So, n/m must be an integer.
This will fail if n=1 and m=3 and other lower numbers, but the range of values for n and m exclude this possibility.
Similarly with B.
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08 Oct 2005, 10:22
I don't think it is D.

A) n=17 & m=7 --> 3n/m is not an integer
B) n=15 & m=4 ---> 13n/m is not an integer

Both cases wont work

Is it E?

Last edited by gsr on 08 Oct 2005, 15:47, edited 3 times in total.
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08 Oct 2005, 12:53
I think ans should be B.

13n/m is an integer. And the question stem says 3<m<13<n

Suppose n = 14 [any no. greater than13] then for 13n/m to be an integer m has to be a factor of n. And so i think the statement alone is sufficient.

Lemme know if i am wrong.
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09 Oct 2005, 11:42
OA is A.

can anybody has any explnation.
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04 Feb 2006, 16:37
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.
Director
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04 Feb 2006, 17:49
B?

If 13n is assigned to m classes... then the number n is surely divisible by m since m cannot be 13 or 1.. So we are sure that m can be divided into n..

Is that right?

OA?
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05 Feb 2006, 01:32
I could not solve! but convinced with willget800 explaination!

btw where did you get this question!
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05 Feb 2006, 02:19
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willget800 wrote:
B?

If 13n is assigned to m classes... then the number n is surely divisible by m since m cannot be 13 or 1.. So we are sure that m can be divided into n..

Is that right?

OA?

Good explanation.
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05 Feb 2006, 11:53
yeah! agree with willget800 , seems simple yet a good question!
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05 Feb 2006, 14:00
i am stuck here

using the same reasoning can't I be sufficent also?

3n/m , 3 is prime so n has to be divisible by m?
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05 Feb 2006, 14:36
joemama142000 wrote:
i am stuck here

using the same reasoning can't I be sufficent also?

3n/m , 3 is prime so n has to be divisible by m?

For n = 14 and m = 6
3n/m is divisible but n/m is not!

For n=15, and m = 5 both 3n/m and n/m are divisible.

Hence 1 is INSUFF.

HTH
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Another good DS [#permalink]

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02 May 2006, 03:00
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 eachclassroom 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 eachclassroom has the same number of students assigned to it.
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02 May 2006, 03:43
Hallo,
Think that A is insufficient
From A) 3n=K*m now n-15 then m can be 5 or 9 which makes A insuff
From B) 13n=K*m then m can not be a prime bigger than 13 , n=15 m can be 3 or 5, n=20 m can be 2,4,10
So think that B Is sufficient
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26 Jun 2006, 16:48
CAREFUL !
OA is not A but B !!!!
See http://www.gmatclub.com/phpbb/viewtopic ... inistrator for explanation !!!
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26 Jun 2006, 17:12
#1. 3n/m is an integer.
Since m > 3, m could be a multiple of 3. However, n may or may not be a multiple of m.
e.g. n = 16, m = 12. or n = 15, m = 5

#2. 13n/m is an integer.
Since 13 > m, m cannot be a multiple of 13. Hence m has to be a factor of n. Sufficient.

B.
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Uh uh. I know what you're thinking. "Is the answer A, B, C, D or E?" Well to tell you the truth in all this excitement I kinda lost track myself. But you've gotta ask yourself one question: "Do I feel lucky?" Well, do ya, punk?

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26 Jun 2006, 18:15
Good question.
shd be (B)

exactly as paddyboy explained..
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27 Jun 2006, 09:49
Ive just decided to start studying for the gmat, so im a rookie here, but....
The question asks: is it possible to do so. and i think that (D) is correct b/c they both are sufficient to recognizing that it is possible.
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27 Jun 2006, 14:35
mthizzle wrote:
Ive just decided to start studying for the gmat, so im a rookie here, but....
The question asks: is it possible to do so. and i think that (D) is correct b/c they both are sufficient to recognizing that it is possible.

This should be B.

In A if 3n = 42 and m = 6 then stem fails but if 3n = 48 and m = 4 then it works. So its INSUFF.

In B it works always.
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28 Jun 2006, 03:25
B.

1) Say there are 14 students hence 42 students can be divided into m classrooms

M can be 6, 7 etc but with 6 classrooms, each classroom won't have equal number of students

2) Say there are 14 students hence 182 students can be divided into m class rooms

here we only get 7 classrooms....
Pick any other value for students and u will see you only get classrooms that are factors of students
28 Jun 2006, 03:25

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