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# A school administrator will assign each student in a group

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A school administrator will assign each student in a group [#permalink]  13 Feb 2012, 21:10
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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.
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Re: A school admin will assign each student in a group of N [#permalink]  13 Feb 2012, 21:15
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gwiz87 wrote:
Hi,

I'm new to this forum and I'm hoping someone can help me understand the problem below:

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.

Can you please explain this to me?

Welcome to GMAT Club. Below is a solution for your problem. Please don't hesitate to ask in case of any question.

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.

Hope its' clear.
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Re: A school admin will assign each student in a group of N [#permalink]  13 Feb 2012, 21:24
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Thanks! Your explanation was much clearer than the guide.
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Re: A school administrator will assign each student in a group [#permalink]  05 Dec 2012, 01:52
pls, explain why from 13n/m= interger we can have n/m= interger. only by picking numbers we have this conclusion.

why from 13/m=no integer, we can have n/m=integer.

why b is correct. I can get B correct by picking numbers but the reasoning must be clear.
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Re: A school administrator will assign each student in a group [#permalink]  05 Dec 2012, 02:02
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thangvietnam wrote:
pls, explain why from 13n/m= interger we can have n/m= interger. only by picking numbers we have this conclusion.

why from 13/m=no integer, we can have n/m=integer.

why b is correct. I can get B correct by picking numbers but the reasoning must be clear.

For (2) we have that $$\frac{13n}{m}=integer$$ and $$3<m<13$$. Now, 13 is a prime number thus it has only 2 factors 1 and 13, which means that $$m$$ cannot be a factor of 13 (since $$3<m<13$$, then it's neither 1 nor 13). Therefore in order $$\frac{13n}{m}$$ to be an integer, $$m$$ must be a factor of $$n$$.

Hope it's clear.
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Re: A school administrator will assign each student in a group [#permalink]  19 Dec 2012, 15:21
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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, 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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Re: A school administrator will assign each student in a group [#permalink]  19 Dec 2012, 20:56
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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, 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?

For St 2 , we are given that 13N/M= Integer and that 3<M<13<N. Now since M, N are integers, we get Value of M can be from (4,5,6...12) and N can be any no greater than 13.Out of the possible values for M, not a single number can divide 13 because it is prime and its only divisor will be 1 and 13. Then for 13N/M to be an integer N has to be a multiple of M or else 13N/M cannot be an Integer which will contradict the given statement itself.

for St1, we are given that 3N/M =Integer and from Q. stem we get 3<M<13<N. Now In possible values of M that can make 3/M in fraction form will be 6,9 and 12 which will result in 3/M value as 1/2,1/3 and 1/4 respectively.

Now if N=15, and M=3 we get 3N/M =15 as integer and N/M also as Integer(Y).
If N=16 and M=3, we get 3N/M=16 as Integer but N/M is not an Integer(N) and hence not sufficient.

Therefore from St 1 we can get 2 ans and hence not sufficient.

Thanks
Mridul
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Re: A school administrator will assign each student in a group [#permalink]  20 Dec 2012, 03:08
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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, 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?

Not so.

(2) implies that $$\frac{13n}{m}=integer$$ but we also know that $$3<m<13$$. If we were not told that, then it would be possible m to be for example 26 and in this case n may or may not be a multiple of m, for example consider n=2 and n=26.

Hope it's clear.
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Re: A school administrator will assign each student in a group [#permalink]  27 Feb 2013, 00:59
I want to follow this posting.

not easy at all.
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Re: A school administrator will assign each student in a group [#permalink]  27 Feb 2013, 03:01
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Explained perfectly by Bunuel. Still, those who have a problem visualizing it, just try to plugin numbers for N and M. Given that 3<M<13<N.

Let N=15, M=5. Thus "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" NOW means is it possible to assign each of the 15 students to one of the classrooms(out of 5), the no of students being the same in each class. Thus, as we see that we can send 3 students to each of the 5 classes, we know that YES it is possible. Thus to prove whether the given generalized statement is possible, we have to prove that M is a factor of N.

From F.S 1, we know M is a factor of 3N. We have to find out whether M is a factor for N also. We can not say this with confidence. Assume M=6, and N =16. Here, M is not a factor of N. Yet,M(6) is a factor of 3N(48). Not sufficient.

From F.S 2, we know that M is a factor of 13N. Now, as 3<M<13, we know that M is co-prime to 13. Thus, it can only be a factor to N. Hence sufficient.

B.
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Re: A school administrator will assign each student in a group [#permalink]  26 Mar 2013, 10:17
The hardest part of this problem wasn't the math. It was the prompt.

Could someone break it down for me?

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?

"is it possible" to me implied whether we can choose an two values within the problem's given inequality to make n/m an integer, at all which we could for condition 1.
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Re: A school administrator will assign each student in a group [#permalink]  27 Mar 2013, 05:30
Expert's post
manimgoindowndown wrote:
The hardest part of this problem wasn't the math. It was the prompt.

Could someone break it down for me?

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?

"is it possible" to me implied whether we can choose an two values within the problem's given inequality to make n/m an integer, at all which we could for condition 1.

There are:
m classrooms where 3 < m < 13 < n.
n students where 3 < m < 13 < n.

We are asked to find whether we can divide equally n students in m classrooms, so whether n/m is an integer.

Does this make sense?
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Re: A school administrator will assign each student in a group [#permalink]  27 Mar 2013, 08:50
Bunuel wrote:
manimgoindowndown wrote:
The hardest part of this problem wasn't the math. It was the prompt.

Could someone break it down for me?

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?

"is it possible" to me implied whether we can choose an two values within the problem's given inequality to make n/m an integer, at all which we could for condition 1.

There are:
m classrooms where 3 < m < 13 < n.
n students where 3 < m < 13 < n.

We are asked to find whether we can divide equally n students in m classrooms, so whether n/m is an integer.

Does this make sense?

It is possible as you demonstrated in your solutions post to have a situation where premise I will make n/m an integer. However it is also possible that there is a situation where it wont be an integer. I guess I focused on the possibility part. Since its possible I would see I. As correct.
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Re: A school administrator will assign each student in a group [#permalink]  27 Mar 2013, 23:47
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gwiz87 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.

Given: 3<M<13<N

Question: Can N/M be an integer?

Statement 1: 3N/M is an integer. Not sufficient as this does not conclusively answer whether N/M is an integer. Either N/M is an integer or M is a multiple of 3. Both are possible. For example when 3N/M = (3*15)/5 then N/M is an integer but when 3N/M = (3*14)/6, M is a multiple of 3 and N/M is not an integer.

Statement 2: 13N/M is an integer. Either N/M has to be an integer or M is a multiple of 13. Because M<13, the latter is ruled out. Therefore N/M is an integer.

So the question can be answered from this statement alone.
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Re: A school admin will assign each student in a group of N [#permalink]  03 Jul 2013, 13:28
Bunuel wrote:
gwiz87 wrote:
Hi,

I'm new to this forum and I'm hoping someone can help me understand the problem below:

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.

Can you please explain this to me?

Welcome to GMAT Club. Below is a solution for your problem. Please don't hesitate to ask in case of any question.

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.
.

Hi Bunuel, I could not understand how you inferred whether \frac{n}{m}=integer[/m] from this question stem?

I tried to plug in few values to the question stem to help me understand your line of thought, but I could not. As always, thank you for your great help and support.
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Re: A school admin will assign each student in a group of N [#permalink]  03 Jul 2013, 13:47
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mitmat wrote:
Bunuel wrote:
gwiz87 wrote:
Hi,

I'm new to this forum and I'm hoping someone can help me understand the problem below:

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.

Can you please explain this to me?

Welcome to GMAT Club. Below is a solution for your problem. Please don't hesitate to ask in case of any question.

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.
.

Hi Bunuel, I could not understand how you inferred whether \frac{n}{m}=integer[/m] from this question stem?

I tried to plug in few values to the question stem to help me understand your line of thought, but I could not. As always, thank you for your great help and support.

The question asks whether we can divide n students into m classes so that each classroom has the same number of students. For example, if there are 20 students and 10 classrooms, then we can assign 2 students to each of the 10 classrooms but if there are 20 students and say 9 classrooms, then we cannot assign (divide) 20 students to 9 classrooms so that each classroom to have the same number of students.

So, we need to determine whether n will be divisible by m, or which is the same whether n/m=integer.

Hope it's clear.
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Re: A school admin will assign each student in a group of N [#permalink]  25 Nov 2013, 14:22
Hi Bunuel!

What about n=24 and m=4?

13*24/4 is an integer, so the divisibility of the classes would still be possible. Awaiting for your reply.
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Re: A school admin will assign each student in a group of N [#permalink]  25 Nov 2013, 14:24
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sebby05 wrote:
Hi Bunuel!

What about n=24 and m=4?

13*24/4 is an integer, so the divisibility of the classes would still be possible. Awaiting for your reply.

In this case too n (24) is a multiple of m (4).
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Re: A school administrator will assign each student in a group [#permalink]  19 Jan 2014, 09:31
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, 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?

Because 3<m<13<n. Therefore 13 cannot divide m.

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Re: A school administrator will assign each student in a group [#permalink]  25 Jun 2014, 04:05
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Bunuel wrote:
gwiz87 wrote:
Hi,

I'm new to this forum and I'm hoping someone can help me understand the problem below:

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.

Can you please explain this to me?

Welcome to GMAT Club. Below is a solution for your problem. Please don't hesitate to ask in case of any question.

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.

Hope its' clear.

Dear Bunuel,

I have concern here.

To explain for (2) 13n/m to be integer, you said that since 13 is prime and 3<m<13 --> 13/m = non-integer --> n/m must be integer, then 13n/m = integer.
So why the same reasoning cannot be applied for (1) as following:
--> (1): 3n/m = integer. Since 3<m<13 and 3 is prime --> 3/m = non-integer --> n/m must be integer, then 3n/m = integer --> sufficient

Please help me to clarify this point. Thank you so much!
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Re: A school administrator will assign each student in a group   [#permalink] 25 Jun 2014, 04:05

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