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# A school administrator will assign each student in a group

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Re: This one is tough one... atleast for me [#permalink]

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19 Sep 2010, 01:46
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amitjash 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.

If each of the 13n students are assign to one of the m classes => 13n and m have HCF greater than 1.
=> since m <13 and it has hcf>1 with 13n, n and m also have hcf >1. Thus statement 2 is sufficient.

If you apply the same with statement 1, 3n and m have hfc >1. But it is possible that n and m does not have hcf>1 as m could be 3,6,9. Thus not sufficient.

Eg of statement 1: Suppose you have 30 students, you can distribute them in m classes only if 30 and m have hcf>1.
30 = 2*3*5. If m does not have any common factor how can you distribute? Say m = 7. 30 can not be distributed in 7 classes.

Eg of statement 2: say 13n = 13*10 =130 are distributed among m classes and since m<13 => for 13n students to be distributed in m classes, m must be multiple of 5.

Since n and m both have a common factor 5, n students can be distributed among m classes.
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Re: This one is tough one... atleast for me [#permalink]

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19 Sep 2010, 01:49
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Mathematically speaking, the question is: is n divisible by m !!

(1) 3n is divisible by m. It doesn't say if n is divisible by m because 3 < m < 13. Indeed, n can be equal to 14 and m = 6. In that case m does neither divide 3 nor n. But with m = 6 and n = 18, m divides n.
INSUFFICIENT

(2) 13n is divisible by n. But 13 is a prime number that m can't divide! Besides 3 < m < 13 so it can't be a combination of 13 and n (like 13* 2) that can be divided by 13.

ANS : B.

Hope it's clear enough...
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Re: This one is tough one... atleast for me [#permalink]

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19 Sep 2010, 08:17
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amitjash 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 can rephrase the question as :

3<m<13<n. Does m divide n ?
(1) m divides 3n
(2) m divides 13n

(1) : Not sufficient. Eg. m=6, n=14 m does not divide n. But m=6, n=18 m divides n. But each case m divides 3n. So clearly not sufficient

(2) : Since m<13 and 13 is prime. m cannot divide 13, hence m must divide n. So sufficient

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Re: This one is tough one... atleast for me [#permalink]

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19 Sep 2010, 16:18
amitjash 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.

chice 2: Let n=14..Hence 13X14=182. Now choosing m between 3 < m < 13 < n.
Say m=7. Then 182 is divisible by 7. But when m=5 it is not divisible. So (2) is not sufficient.
Some one please explain what is wrong in choosing numbers to solve this and where I am missing something.
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Re: This one is tough one... atleast for me [#permalink]

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19 Sep 2010, 16:20
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vigneshpandi wrote:
amitjash 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.

chice 2: Let n=14..Hence 13X14=182. Now choosing m between 3 < m < 13 < n.
Say m=7. Then 182 is divisible by 7. But when m=5 it is not divisible. So (2) is not sufficient.
Some one please explain what is wrong in choosing numbers to solve this and where I am missing something.

In (2) it is given that m divides 13n. In your example, 7 does divide 13n, ie, 182. But 5 does not, so it is not a valid choice for m
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08 Oct 2010, 18:57
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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08 Oct 2010, 20:32
So n has to be multiple of m.. lets say n = km where k > 1

1) states that 3n = lm (where each room has l number of people). as 3 being a prime number, lm must be divisible by 3. so is l/3 = k (an integer). we don't know as in all scenario its possibel

- l,m are divisible by 3 >> here l/3 is an integer
- only m is divisible by 3
- only l is divisible by 3 >> here l/3 is an integer
so not sufficient

2)13n = lm..so lm must be divisible by 13, 13 being a prime number and as m < 13 , l/13 must be an integer. That means n = km where k = l/13 Hence sufficient. Answer should be B
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08 Oct 2010, 22:27
B is my answer. if 13n is divisible by m then 3n has to be divisible by m.
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08 Oct 2010, 23:23
mergin similar topics
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09 Oct 2010, 10:46
shrouded1 wrote:
mergin similar topics

Thanks shrouded1!, have you become moderator of the forum?
I said this because you have merged the topics.
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09 Oct 2010, 11:32
metallicafan wrote:
shrouded1 wrote:
mergin similar topics

Thanks shrouded1!, have you become moderator of the forum?
I said this because you have merged the topics.

Yep !

Posted from my mobile device
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22 Nov 2010, 07:44
Hello everyone,

I need some help with an official question. #128 in the DS section of the 12th ed. official guide:

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 understand why (1) is not sufficient however I can not grasp why (2) is sufficient. Can someone please explain?
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22 Nov 2010, 08:19
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Expert's post
jferes2345 wrote:
Hello everyone,

I need some help with an official question. #128 in the DS section of the 12th ed. official guide:

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 understand why (1) is not sufficient however I can not grasp why (2) is sufficient. Can someone please explain?

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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22 Nov 2010, 08:43
Thanks a million. I was making this so much harder then it really was. Thanks again.
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22 Nov 2010, 13:44
Questions is asking indirectly whether n is divisible by m or not.

as per A, 3n is divisible by m. This does not necessarily mean n is divisible by m. For ex- n=12, m=4 and N=6 and m=9

as per B, 13n is divisible by m. notice that 13 is a prime number and so, the other factor, n has to be divisible by m.

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03 Jan 2011, 18:48
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.

Took a long time in this ques. any efficient way to solve this.
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04 Jan 2011, 02:05
The Question is If N/M is an integer ?

(1) 3N/M is an integer
Case 1 - N=15 M=3 it is clearly an integer when we substitute in above equ.
Case 2 - N=14 M=6 Satisfy the statement 2 but doesn't work for above equ.
Therefore Insuff.

(2) 13N/M= integer
13 is a prime number and since m is less than 13 therefore for this statement to hold true, N/M has to be integer. Hence Sufficient.

I hope that helps.
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17 Feb 2011, 08:33
Thanks Bunuel

I was looking for a clearer explanation
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OG DS 128 Arithmetic Properties [#permalink]

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23 Feb 2011, 22:19
128. 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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is n divisible by m [#permalink]

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08 Mar 2011, 11:10
128) A school administer 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) Its 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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is n divisible by m   [#permalink] 08 Mar 2011, 11:10

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