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A school administrator will assign each student in a group

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DS [ Number ] [#permalink] New post 27 Dec 2006, 05:39
A school administrator will assign each student in agroup of n students to one of m classroom. If 3<m<13<n. is it possible to assign each of the n students to one of the m classroom so that each classrom has the same number of students assigned to it?

1) It is possible to assign each of 3n students to one of m classroom 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 classroom so that each classroom has the same number of students assigned to it.

can somebody explain me the short method for this problem?

For me its too time consuming... :(
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 [#permalink] New post 27 Dec 2006, 08:28
problem:
3<m<13<n
=> 3<m<13
=>n>13
we need to see if n/m=int
1) 3n/m = int
2)13n/m =int

from (1)
n= 14 and m=7
3n/m= 3*14/7= 6 ; and n/m= 14/7= int
n=16,m=6
3n/m = 3*16/6 = 8; but n/m = 16/6 = not int
so (1) insuff

from (2)
13n/m =int
as 3<m<13 and n>13
n/m will be int as 13n/m =int.
so (2) is suff
ANS (B)
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DS - assigning students (HARD) [#permalink] New post 25 Apr 2007, 16:42
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.

One of the harder ones I suppose?
:?
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 [#permalink] New post 25 Apr 2007, 18:59
St1:
3n/m = whole integer. m cann be 3 since m>3, so m must be a factor of n. so n/m = whole integer. sufficient.

St2:
13n/m = whole integer. m cannot be 13, since m<13. So m must be a factor of n. So n/m = whole integer. Sufficient.

I think it's D
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 [#permalink] New post 25 Apr 2007, 19:11
Answer B??

Question asked Is m a factor of n?

Statement 1 says m is a factor of 3n and since 3<m<13<n, there is a possibility where m=6 and n=14 and this condition (3n/m =INTEGER) be true INSUFF

Statement 2 says m is a factor of 13n and since 13 is a prime number and denominator m <13, for this condition to be true m should be a factor of n.
SUFF
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 [#permalink] New post 25 Apr 2007, 19:20
vijay2001 wrote:
Answer B??

Question asked Is m a factor of n?

Statement 1 says m is a factor of 3n and since 3<m<13<n, there is a possibility where m=6 and n=14 and this condition (3n/m =INTEGER) be true INSUFF

Statement 2 says m is a factor of 13n and since 13 is a prime number and denominator m <13, for this condition to be true m should be a factor of n.
SUFF


ah yes, I neglected that fact. B should be it.
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School admin + DS [#permalink] New post 20 Jun 2007, 18:38
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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 [#permalink] New post 20 Jun 2007, 20:07
from the given info,

No of classrooms(M) can be one of {4, 5, 6, ..., 12}
No of students(N) is 14 or more

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, this asks if N is a multiple of M.

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.
this says 3N is a multiple of M.

taking values for M and N,
M=9, N=15(sothat 3N is multiple of M)? N is not multiple of M.
M=9, N=27? in this case, N is a multiple of M.

So 1) is not enough.

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.

this says 13N is a multiple of M.

looking at the range of values for M(4-12) and N(>=14), since 13 is not divisible by any possible value of M, this can be only possible if N is a multiple of M.

So 2) is enough

Ans B)
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DS n students, m classrooms [#permalink] New post 26 Sep 2007, 10:17
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.













OA: B- please explain. thanks!
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Re: DS n students, m classrooms [#permalink] New post 26 Sep 2007, 10:42
prospective mba 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.
OA: B- please explain. thanks!



B

This question is really asking: is N divisible by M.

3n is disivible by M. (3n/m = integer) and 3<m<13<n.

S1: n=22, and 6 =m (satifies main statement)

22/6 <--- not divisible

22*3/6? <---- is divisible = 11.

Don't need to find values that do work here, b/c we know there are many.

S2:
13n/m = integer.

Since 13 is a prime number and 3<m<13<n

id say B is suff. n has to be divisible by n since m cannot be 13 itself (or another number divisible by 13).
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Re: DS n students, m classrooms [#permalink] New post 26 Sep 2007, 11:41
GK_Gmat wrote:
prospective mba 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.

I'm getting B. Used the plug in numbers strategy.

Stat 1:

Let n=15, 3n = 45
Let m = 5; 3n can be divided equally and n can be divided equally; Yes.
Let m= 9; 3n can be divided equally but n cannot be divided equally; No.

Insuff.

Stat 2:
Let n = 20; 13n = 260
Let m = 4; 13n can be divided equally (65 each) and n can be divided equally (5 students in each class)

Let n = 15; 13n = 195
Let m = 5; 13n can be divided equally (39 students in each class) and n can be divided equally (3 students in each class)

Let n = 14; 13n = 182
Let m = 7; 13n can be divided equally (26 students in each class) and n can be divided equally (2 students in each class).

Enough time spent. At this point I chose B and move on. If wrong, I couldn't come up with a better strategy!










OA: B- please explain. thanks!


pretty much did the same thing came up with around 4-5 examples, all were in favor of B so i just said B at this point. Also 13 is prime so that tells me that this probably B.
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 [#permalink] New post 26 Sep 2007, 19:27
St1:
If m = 4, and n = 16, then each of the 3n students can be assigned to one of the m classrooms so that each classroom has the same number of students assigned to it. Also, each of the n students can be assigned to one of the m classrooms so that each classroom has the same number of students assigned to it.

If m = 3, and n = 14, then each of the 3n students can be assigned to one of the m classrooms so that each classroom has the same number of students assigned to it. But not each of the n students can be assigned to one of the m classrooms so that each classroom has the same number of students assigned to it.

Insufficient.

St2:
13n/m = integer. Since 13 is a prime and m must be between 3 and 13, so n must be a multiple of m. So we can assign each of n students to one of m classrooms so that each of the classrooms has the same number of students assigned to it. SUfficient.

Ans B
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integers [#permalink] New post 01 Mar 2008, 23:20
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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Re: integers [#permalink] New post 02 Mar 2008, 07:46
ANSWER IS D

First statement :
suppose 3n= 33 and m=11 so each room can be assigned with 3 students
Therefore n=11 and m=11 so each room can be filled with 1 student each
so possible

Second statement : suppose 13n=156 and m=12 so each room with 13 students
therefore n=12 and m=12 so each room with 1 student
again possible
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Re: integers [#permalink] New post 02 Mar 2008, 09:06
neeraj.kaushal wrote:
ANSWER IS D

First statement :
suppose 3n= 33 and m=11 so each room can be assigned with 3 students
Therefore n=11 and m=11 so each room can be filled with 1 student each
so possible

Second statement : suppose 13n=156 and m=12 so each room with 13 students
therefore n=12 and m=12 so each room with 1 student
again possible


I can't follow you.

In other words, we must find if m is a factor of n, or if n is divisible by m.
1. says that 3n is divisible by m. in the case m=6 and n is divisible by 2 we would have remainder 0 but still won't know anything about divisibility b/w n and m.

2. says that 13n is divisible by m. since m<13 n must be divisible by m. suff

OA is B
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Re: integers [#permalink] New post 03 Mar 2008, 10:59
neeraj.kaushal wrote:
ANSWER IS D

First statement :
suppose 3n= 33 and m=11 so each room can be assigned with 3 students
Therefore n=11 and m=11 so each room can be filled with 1 student each
so possible

Second statement : suppose 13n=156 and m=12 so each room with 13 students
therefore n=12 and m=12 so each room with 1 student
again possible



It states N is greater than 13 in the stem, so N cannot be 11
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Re: integers [#permalink] New post 03 Mar 2008, 11:56
marcodonzelli 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.


This is actually a really simple problem. I got it incorrect initially. I just couldnt make sense of it at first.

Essnetially S1 reworded says that 3n/m= an integer. So Now you can see why we have a YEs and No scenario here. b/c if N= a prime number and m is = 3 then it works for S1, but n/m will not work.

S2: 13n/m since M cannot be 13. That means n/m is an integer. Suff.
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DS: Nice one [#permalink] New post 18 Jul 2008, 02:49
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 classrooms 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 the 13n students to one of m classrooms so that each classroom has the same number of students assigned to it.



Please explain your answer.
thanks
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Re: DS: Nice one [#permalink] New post 18 Jul 2008, 03:14
If n is divisible by m, so we can assign each of the n students to one of the m classrooms so that each classrooms has the same number of students assigned to it.
1. INSUF
2. 13n is divisible by m. 13 is prime factor. So n is divisible by m. ==> SUF

B is correct answer.
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Re: DS: Nice one [#permalink] New post 18 Jul 2008, 06:43
tarek99 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 classrooms 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 the 13n students to one of m classrooms so that each classroom has the same number of students assigned to it.

Please explain your answer.
thanks


Question: is n divisible by m

1) tells us that 3*n is divisible by m where 3<m<13 and n>13
for m=12 and n = 16... 3*n is divisible by m , but n is not divisible by m.
for m=5 and n = 15.. 3*n as well as n are divisible by m.

i.e if m has 3 as a factor then it's not guaranteed that n is divisible by m.
but if m is prime its possible that n is divisible by m

insuff

2) tells that 13*n is divisible by m where 3<m<13 and n>13

since for any m (such that 3<m<13) won't have 13 as a factor n must be divisible by m

sufficient

Thus B
Re: DS: Nice one   [#permalink] 18 Jul 2008, 06:43
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