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# A school admin will assign each student in a group of N

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Manager
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A school admin will assign each student in a group of N [#permalink]  06 May 2006, 20:16
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A school admin 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-admin-will-assign-each-student-in-a-group-of-n-127509.html
[Reveal] Spoiler: OA

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Last edited by Bunuel on 22 Jun 2012, 16:44, edited 1 time in total.
Edited the question and added the OA. Topic is locked.
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Re: DS - assigning students [#permalink]  06 May 2006, 21:31
The question stmt asks whether each of the n students can be assigned to the m classroom so that equal number of students are accomodated in the classrooms.
statement 1: 3n students can be accomodated in the m classrooms, which means that 3n is divisible m. This does not necessarily tell us that n should also be divisible by m. Hence the question string can be either yes or no.
statement 2: 13n should be divisible by m, which implies n should be definately divisible by m. hence above stmt can be answered in yes.

Hence 2nd stmt alone is sufficient to answer the question. Hence B
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[#permalink]  07 May 2006, 00:41
I think the answer is E.

i) Students can be divided evenly only if 3n/m is a positive integer. If n is odd then 3 n = odd x odd = odd. For m=even nos. statement 1 is not possible but for some m odd nos it is. Hence (i) is insufficient.

ii) Same as (i). 13n will be odd for n odd nos. and maybe possible for some m odd nos. Hence (ii) is also insufficient.

(i) & (ii) will also be insufficient.

Thus E.
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[#permalink]  07 May 2006, 04:16
I would say B

1) 3xn / m = k with k integer
and we cannot conclude anything at all.

2) 13xn / m = j with j integer
As we know that m < 13, it means that 13 cannot be a prime factor of m

which means all prime factors from m are prime factors of n, in other words n/m is an integer

>>> Sufficient
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[#permalink]  07 May 2006, 12:15
Agree with B
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[#permalink]  08 May 2006, 08:36
it is B

Statement 1 is insufficient: Though 3n is divisible by m, that doesn’t mean that n is also divisible by m. Consider n=15, m=9 or n=15, m=5.

Statement 2 is sufficient: Because 13 is prime, and 13n is divisible by m, than n must also be divisible by m.
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[#permalink]  08 May 2006, 10:21
I agree with B Also!

Nice question... Makes me feel good.
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Re: A school administrator will assign each student in a group [#permalink]  22 Jun 2012, 16:36
Can any one provide a much elaborate solution, did not understand the above's....
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Re: A school administrator will assign each student in a group [#permalink]  22 Jun 2012, 16:43
Expert's post
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riteshgupta wrote:
Can any one provide a much elaborate solution, did not understand the above's....

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?

Basically the question asks whether $$n$$ (# of students) is a multiple of $$m$$ (# of classrooms), or whether $$\frac{n}{m}=integer$$, because if it is then we would be able to assign students to classrooms so that each classroom has the same number of students assigned to it.

Given: $$3<m<13<n$$.

(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 --> $$\frac{3n}{m}=integer$$, from this we can not say whether $$\frac{n}{m}=integer$$. For example $$n$$ indeed might be a multiple of $$m$$ ($$n=14$$ and $$m=7$$) but also it as well might not be ($$n=14$$ and $$m=6$$). Not sufficient.

(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 --> $$\frac{13n}{m}=integer$$, now as given that $$3<m<13$$ then 13 (prime number) is not a multiple of $$m$$, so $$\frac{13n}{m}$$ to be an integer the $$n$$ must be multiple of $$m$$. Sufficient.

Answer: B.

Hope its' clear.

OPEN DISCUSSION OF THIS QUESTION IS HERE: a-school-admin-will-assign-each-student-in-a-group-of-n-127509.html In case of any questions please post there.
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Re: A school administrator will assign each student in a group   [#permalink] 22 Jun 2012, 16:43
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# A school admin will assign each student in a group of N

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