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A school administrator will assign each student in a group

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DS - School Admin [#permalink] New post 07 Sep 2008, 07:19
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.

Not able to solve it can some one do it for me????
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Re: DS - School Admin [#permalink] New post 07 Sep 2008, 08:29
vivektripathi 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.

Not able to solve it can some one do it for me????


The question is essentially asking if n is divisible by m.

(1) Insufficient

Here it is saying 3n/m is an integer. Does that tell us n/m is definitely an integer?
No.

Examples:
n = 20
m = 12
3*20/12 = 5

OR

n = 24
m = 12
3*24/12 = 6

(2) Sufficient
13n/m is an integer.

We know 3 < m < 13.
Thus, we know that 13 is not divisible m, since 13 is a prime number.
From this, we know n/m must be an integer, since 13n/m is an integer.
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Re: DS - School Admin [#permalink] New post 07 Sep 2008, 09:01
one more B.

stat 1 says that 3n is a multiple of m. so if n=16, then 3*16/12(m=12 say).but 16/12=???
insuff

stat2) 13n is a multiple of m, but m is less than 13 and gretaer than 3. so the 13n can be divisible by that value of m which should divide n. in other words n/m. suff
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Re: DS - School Admin [#permalink] New post 07 Sep 2008, 09:46
vivektripathi 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.

Not able to solve it can some one do it for me????


good one:

B.

1: if m is 9 and n is 15, 3n is divided equally among m classes.
if m is 10 and n is 20, 3n also divided equally among m classes.
so nsf.

2: if 13n is equally divided among m classes, then n is also divided among m classes because 13 is a prime and m is smaller then 13. suff......
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a hard one [#permalink] New post 10 Sep 2009, 05:18
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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.

[Reveal] Spoiler:
OA is B
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Re: a hard one [#permalink] New post 10 Sep 2009, 06:18
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1) insuff. it means,
3n can be divided by m. it means m can divide n or m can divide some of n's factors*3.
e.g.
n=14, m=6
n cannot be divided by m. But 3n can be ;)
or let say
n=15, m=5
not only n but also 3n is divisible by m
2)suff.
it means 13n can be dividen by m. it means m can divide n or m can divide some of n's factors*13.
in first case, n can be divided by m. in latter case, since m is less than 13, m must divide n.
B
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Divisibility problem - n students in m classes (+/-700 lvl) [#permalink] New post 31 Oct 2009, 06:33
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I was hoping somebody could take a crack at this problem and specifically explain how you beat any useful info out of statement (1) - that statment in specific has me LOST.


A school administrator will assigen 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 the 3n students to one of the 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: Divisibility problem - n students in m classes (+/-700 lvl) [#permalink] New post 31 Oct 2009, 08:57
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We are given that 3<m<13<n. this means m has to be between 4-12 or must necessarily be lesser than 13. Also, the number of students n can be assigned to each of the classrooms m equally which means n/m is an intger.

Statement 1 says - (1) It is possible to assign each of the 3n students to one of the m classrooms so that each classroom has the same number of students assigned to it. This means the number of students in each classroom after dividing 3n/m is an integer. This possibility states the number n, when multiplied by 3 is divisble by m. since m>3, this data is sufficient to answer the question - 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?

Stetement 2 says - (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. Going by the same logic mentioned above, the number of students 13n/m. However, n>13 and m<13, the data is not sufficient for the required question. For.e.g 52 students and 4 classroom is divisible into equal numbers, but 169 students can not be divided equally into any number of given m. Hence statement 2 is not sufficient.

I would go with ans A.
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Re: Divisibility problem - n students in m classes (+/-700 lvl) [#permalink] New post 31 Oct 2009, 10:11
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Hi,

you can solve it by Picking Numbers strategy, but I suggest you another, which for me is less confusing.

"...so that each classroom has the same number of students assigned to it" means n is divisible by m. Therefore, m is a factor of n.

The stem asks "If 3<m<13<n, is n divisible by m?"

To be divisible by any of 3<m<13, the number should be at least a LCM of all 3<m<13. To find LCM write all possible m by prime factors...

4=2*2*1
5=5*1
6=3*2*1
7=7*1
8=2*2*2*1
9=3*3*1
10=5*2*1
11=11*1
12=3*2*2*1

So, the LCM of all 3<m<13 is 2*2*2*3*3*5*7*11.

Let's test the statements.

1) 3n is divisible by m.
From the expression of LCM you can see that if 3n is at least a multiple of 2*2*2*3*3*5*7*11, n is at least a multiple of 2*2*2*3*5*7*11, which could or could not be divisible by 3<m<13. So, (1) is NOT SUFFICIENT.

2) 13n is divisible by m.
Since expression of LCM of all 3<m<13 doesn't contain 13 it is necessary for n to be at least a multiple of 2*2*2*3*3*5*7*11. Therefore, n is divisible by m. So, (2) is SUFFICIENT.

The answer is B.

KUDOS if you find it helpful ;)
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Re: Divisibility problem - n students in m classes (+/-700 lvl) [#permalink] New post 31 Oct 2009, 20:09
Vyacheslav wrote:
Hi,

you can solve it by Picking Numbers strategy, but I suggest you another, which for me is less confusing.



The answer is B.

KUDOS if you find it helpful ;)


Great explanation, Vya. I must confess, the LCM idea never crossed my mind. After examining the question a bit and looking at your advice (and finally seeing than we are trying to see if n / m), I took a bit of a conceptual approach, and plugged numbers in to verify my thoughts.

(i) This statement took most of my time - The fact that we are give a statement that 3n is divisible my m made me suspicious about n being divisible by m in all cases. It was relatively easy to come up with a value for n and m that satisfied both (3n/m) and (n/m) - such as n=14 and m=7.

To disprove it, I looked for a value of n greater than 13 that had NO 3 in the prime factorization - I started with 16, which is 2x2x2x2. For m, I settled on 6, which has a 3 and a 2 in prime factorization. Sure enough, 16x3 = 48, which is divisible by 6. 16 is NOT divisible by 6. We have both a possible right or wrong answer here, so clearly INSUFFICIENT.

(ii) This one is simply intuitive - 13 is a prime, and m cannot equal 13. n must be divisible by m PRIOR to being multiplied by 13. SUFFICIENT.

Yes, the answer is B. My method took about 3 minutes! Thanks for the LCM angle - when I see this problem again, I will adopt that and save time. Enojy the Kudos!
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Re: Divisibility problem - n students in m classes (+/-700 lvl) [#permalink] New post 07 Nov 2009, 13:29
Vyacheslav wrote:
Hi,

you can solve it by Picking Numbers strategy, but I suggest you another, which for me is less confusing.

"...so that each classroom has the same number of students assigned to it" means n is divisible by m. Therefore, m is a factor of n.

The stem asks "If 3<m<13<n, is n divisible by m?"

To be divisible by any of 3<m<13, the number should be at least a LCM of all 3<m<13. To find LCM write all possible m by prime factors...

4=2*2*1
5=5*1
6=3*2*1
7=7*1
8=2*2*2*1
9=3*3*1
10=5*2*1
11=11*1
12=3*2*2*1

So, the LCM of all 3<m<13 is 2*2*2*3*3*5*7*11.

Let's test the statements.

1) 3n is divisible by m.
From the expression of LCM you can see that if 3n is at least a multiple of 2*2*2*3*3*5*7*11, n is at least a multiple of 2*2*2*3*5*7*11, which could or could not be divisible by 3<m<13. So, (1) is NOT SUFFICIENT.

2) 13n is divisible by m.
Since expression of LCM of all 3<m<13 doesn't contain 13 it is necessary for n to be at least a multiple of 2*2*2*3*3*5*7*11. Therefore, n is divisible by m. So, (2) is SUFFICIENT.

The answer is B.

KUDOS if you find it helpful ;)


Nice explanation .. kudos
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OG question [#permalink] New post 29 Mar 2010, 19:09
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 didn't quite understand the explanation given in the OG. So I am hoping that someone can help me out.
Thanks!
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Re: OG question [#permalink] New post 29 Mar 2010, 22:47
nsp 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 didn't quite understand the explanation given in the OG. So I am hoping that someone can help me out.
Thanks!


To make n students equally divided into m classrooms, m has to be multiple of n. So indirectly we have to prove if m is the multiple of n

stmt1: 3n students can be assigned to m classrooms equally. which means m is multiple of 3n=> m is either 3 or multiple of n
since 3<m which means m <> 3 so m is multiple of n. Hence sufficient.
stmt2: 13n students can be assigned to m classrooms equally. which means m is multiple of 13n=> m is either 13 or multiple of n
since m<13 which means m <> 13 so m is multiple of n. Hence sufficient.
So d both stmts alone are sufficient
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Re: OG question [#permalink] New post 30 Mar 2010, 00:04
sidhu4u wrote:
nsp 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 didn't quite understand the explanation given in the OG. So I am hoping that someone can help me out.
Thanks!


To make n students equally divided into m classrooms, m has to be multiple of n. So indirectly we have to prove if m is the multiple of n

stmt1: 3n students can be assigned to m classrooms equally. which means m is multiple of 3n=> m is either 3 or multiple of n
since 3<m which means m <> 3 so m is multiple of n. Hence sufficient.
stmt2: 13n students can be assigned to m classrooms equally. which means m is multiple of 13n=> m is either 13 or multiple of n
since m<13 which means m <> 13 so m is multiple of n. Hence sufficient.
So d both stmts alone are sufficient


I think you swapped m and n. We need to find out if n is a multiple of m ( from the problem statement n>m). Here is how i approached it.

As per stmt1 3n is a multiple of m i.e. 3n is divisible by m. If there is no common factor between m and 3 then we can say for sure that n is also divisible by m. So when m is 4,5,7,8,10 or 11 n is divisible by m but if m is 6 or 12 m may or may not divide n (m=6 and n=14 3n is divisible by m but n is not , m=6 and n=42 both 3n and n are divisible by m)

This doesn't give a conclusive yes/no answer. So stmt1 alone is not sufficient

As per stmt2 13n is a multiple of m i.e. 13n is divisible by m
Since m value lies between 4 and 12 there won't be a common factor between m and 13 so n should be divisible by m which is what we are trying to prove. So stmt2 alone is sufficient.

Answer should be B
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12th edition official guide DS #128 Properties of numbers [#permalink] New post 13 Apr 2010, 03:22
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.

Correct answer is B.

I do not understand why the (1) statement is not also sufficient. Can someone explain it to me?
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Re: 12th edition official guide DS #128 Properties of numbers [#permalink] New post 13 Apr 2010, 07:11
hoffeman 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.

Correct answer is B.





I do not understand why the (1) statement is not also sufficient. Can someone explain it to me?


Assume m = 6 and n = 16

question 1 is saying that 16*3 = 48 is a multiple of 6 (6*8 = 48), which is correct. However 16/6 is not an integer so the answer is no

if m = 5 and n = 20 the answer is yes
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wrong explanations? [#permalink] New post 01 Aug 2010, 07:56
hi all,
here is the question from OG. i think part of the explanation provided is wrong. please comment!

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.
Explanation:
Determine if n is divisible by m.
(1) Given that 3n is divisible by m, then n is
divisible by m if m = n = 9 (note that 3n = 27
and m = 9, so 3n is divisible by m) and n is
not divisible by m if m = 9 and n = 12 (note
that 3n = 36 and m = 9, so 3n is divisible by
m); NOT sufficient.

(2) Given that 13n is divisible by m, then 13n =
qm, or n/m=q/13, for some integer q. Since 13
is a prime number that divides qm (because
13n = qm) and 13 does not divide m
(because m < 13), it follows that 13 divides q.
Therefore,q/13 is an integer, and since n/m=q/13, then
n/m is an integer. Thus, n is
divisible by m; SUFFICIENT.

even though the answer is B, i think the the example/explanation given for the S(1) is wrong because n can't be 9 or 12 becasue n>13 given in the question stem. am i correct?
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Re: wrong explanations? [#permalink] New post 02 Aug 2010, 10:36
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Hi there,

It looks to me that they do indeed use invalid numbers for n in the explanation for statement 1. However, the principle that they demonstrate is still applicable. The idea is that, because you'll end up with 3n students, the number of classrooms must itself be a multiple of 3. So the value of m can only be 6 or 9. The key is that n itself doesn't have to be divisible by 3 just because m is 6 or 9. n could be 18-->3n = 54-->54 is divisible by 6, and so is 18. But n could also be 22-->3n = 66-->66 is divisible by 6, but the original n, 22, is not. In fact, as long as n is even to begin with, it will always be divisible by 6 when you multiply n by 3. The same thing applies if m = 9. n could be divisible by 9 to begin with (like 18), or it could have just one factor of 3 (like 21), so that when you multiply it by 3 to get 3n, it becomes divisible by 9 thanks to the additional factor of 3.

This problem is testing your understanding of the properties of factors, which is why it doesn't really matter which numbers they use in the explanation for statement 1. That said, you do seem to have found a typo.

Good eyes!
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Re: wrong explanations? [#permalink] New post 03 Aug 2010, 17:24
i agree with you that the logic behind the explanation is good, but the explanation is against the condition given on the question stem. thanks and i am glad that we both agree on this issue!
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This one is tough one... atleast for me [#permalink] New post 19 Sep 2010, 00:33
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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.
This one is tough one... atleast for me   [#permalink] 19 Sep 2010, 00:33
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