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A school administrator will assign each student in a group

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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

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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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hey ya......

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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.

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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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"To dream anything that you want to dream, that is the beauty of the human mind. To do anything that you want to do, that is the strength of the human will. To trust yourself, to test your limits, that is the courage to succeed."

- Bernard Edmonds

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

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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

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28 Jun 2006, 03:25

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