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805+ Level|   Multiples and Factors|   Word Problems|                                       
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A school administrator will assign each student in a group 25 Jun 2014, 05:05
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Bunuel
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.
Show more

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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A school administrator will assign each student in a group 25 Jun 2014, 07:34
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LucyDang
Bunuel

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.

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!
Show more

Because for (2), m cannot have any factor of 13, thus for \(\frac{13n}{m}\) to be an integer n must be divisible by m.

For (1) m could be a multiple of 3, for example, 6, 9, or 12, and in this case n is not necessary to be divisible by m.
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A school administrator will assign each student in a group 15 Oct 2015, 10:44
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Bunuel
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.
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I have to start to think like bunuel, amazing! :) So easy if you read it...
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A school administrator will assign each student in a group 05 May 2016, 21:31
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Dear Bunuel,

I am still not clear with the line of reasoning used to prove if n/m is an integer with the given option statements.
For option 1 we are saying that 3n/m could give us an integer and also could not give us an integer basis values of 3<m<13,such as 3*15/5=Integer ,however,3*15/6 is not an integer. Then ,in the second option statement we are considering that since m is greater than 13 then n/m will definitely give us an integer.Why are we changing the thought process for option 2 statement.Even in this case,then 13* 20/5=Integer and 13*21/5 is not an integer.

Please clarify my doubt.

Regards,
Bhamini
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A school administrator will assign each student in a group 05 May 2016, 21:45
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bhamini1
Dear Bunuel,

I am still not clear with the line of reasoning used to prove if n/m is an integer with the given option statements.
For option 1 we are saying that 3n/m could give us an integer and also could not give us an integer basis values of 3<m<13,such as 3*15/5=Integer ,however,3*15/6 is not an integer. Then ,in the second option statement we are considering that since m is greater than 13 then n/m will definitely give us an integer.Why are we changing the thought process for option 2 statement.Even in this case,then 13* 20/5=Integer and 13*21/5 is not an integer.

Please clarify my doubt.

Regards,
Bhamini
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Hi,
the statement II tells us that 13N/M is an integer...
But M is less than 13 and 13 is a PRIME number, so there are no common factors of 13 and M..
so ONLY possiblity is that N/M is an integer..
so suff..

But statement I tells us that 3N/M is an integer...
and M is between 3 and 13, so there can be common factors between 3 and M like 3 and 6 and they can be co-primes 3 and 5..
so BOTH possiblity that N/M is an integer and not an integer exist..
so insuff..
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A school administrator will assign each student in a group 24 Oct 2016, 02:27
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Hey guys,

might be a little bit off topic.
But do I have to understand (be able to solve) these very hard questions if I am aiming for a low 600?
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A school administrator will assign each student in a group 24 Oct 2016, 11:09
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AK125
Hey guys,

might be a little bit off topic.
But do I have to understand (be able to solve) these very hard questions if I am aiming for a low 600?
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No, you absolutely don't. All you need to do with these problems is be able to recognize them. The only reason to recognize them is so you can guess on them and move on quickly, rather than mistaking them for simpler problems and spending too much time trying to solve before giving up.
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A school administrator will assign each student in a group 25 Oct 2016, 06:19
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AK125
Hey guys,

might be a little bit off topic.
But do I have to understand (be able to solve) these very hard questions if I am aiming for a low 600?
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Hi AK125

You need to understand whether your strength is Quant or Verbal

If your strength is Quant and your verbal section is weak then you must try all ways to get as many questions as possible however if your Strength area is Verbal then such hard questions can be done away with.

Being an instructor I would atleast suggest or motivate you to try to understand such question. After giving it a try even if you fail to solve such questions in actual test then atleast it will help you build concepts and confidence in cracking comparatively easier question

Getting 650 with aspiration of 600 is not a bad idea anyway... ;)
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A school administrator will assign each student in a group 25 Oct 2016, 06:27
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AK125
Hey guys,

might be a little bit off topic.
But do I have to understand (be able to solve) these very hard questions if I am aiming for a low 600?
Show more

Here is an effort to make the calculation of question a little easy for you

Quote:
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.
Show more

It will be possible to assign each student in classrooms equally only if No. of Students (N) is a factor of No. of classrooms (M)
i.g. 20 students can be assigned in 5 room because 5 is factor of 20 but 20 students can't be assigned in 6 room because 6 is not a factor of 20

Question REPHRASED: Is N/M=Integer???

Statement 1: 3N/M is an Integer
Now I wish to prove this statement insufficient therefore I take two cases
Case 1: N=20 and M=5, such that 3N/M is an Integer And also N/M is and Integer
Case 2: N=20 and M=6, such that 3N/M is an Integer BUT N/M is NOT and Integer
NOT SUFFICIENT

[P.S. I could prove this statement Insufficient because I could take denominator as multiple of 3 in order to use the coefficient of N given in numerator]


Statement 2: 13N/M is an Integer
Now I wish to prove this statement insufficient BUT since M<13 and 13 is a prime Number so I can't choose the value for denominator which can have any factor of 13 so for 13N/M to be an Integer N/M must also be an Integer
SUFFICIENT

Answer: option B
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A school administrator will assign each student in a group 21 May 2017, 23:05
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Bunuel
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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.

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.
Show more

Hi Bunuel,

\(3<M<13<N\)

Possible value of M = 4, 5, 6, 7, 8, 9, 10, 11, 12
Possible value of N = 14, 15, 16, 17, 18, 19

I could not understand the solution for statement (2). \(\frac{13N}{M} = integer\) What if M=4, N=14? \(\frac{13(14)}{4} ≠ integer\). Could you help to explain?

Shall we use the value as below?

M= 7___8___9__
N=14__16__18__

\(\frac{13N}{M}=13(\frac{14}{7})=26\)
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A school administrator will assign each student in a group 22 May 2017, 00:16
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hazelnut
Bunuel
gwiz87
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.

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.

Hi Bunuel,

\(3<M<13<N\)

Possible value of M = 4, 5, 6, 7, 8, 9, 10, 11, 12
Possible value of N = 14, 15, 16, 17, 18, 19

I could not understand the solution for statement (2). \(\frac{13N}{M} = integer\) What if M=4, N=14? \(\frac{13(14)}{4} ≠ integer\). Could you help to explain?

Shall we use the value as below?

M= 7___8___9__
N=14__16__18__

\(\frac{13N}{M}=13(\frac{14}{7})=26\)
Show more

The question asks whether \(\frac{n}{m}=integer\).

(i) From stem we are given that \(3<m<13<n\).
(ii) From (2) we are given that \(\frac{13n}{m}=integer\).

When testing values, we should choose those which satisfy (i) and (ii) and test whether they give definite answer to the question: is \(\frac{n}{m}=integer\). All values that satisfy (i) and (ii) will give an YES answer to the question "is \(\frac{n}{m}=integer\)?".

Does this make sense?
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A school administrator will assign each student in a group 16 Jun 2017, 09:39
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manimgoindowndown
Bunuel
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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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Hello. I had the same doubt as you to this question. According to me, the question simply asks "is it possible?", which means that are there any values of n and m which can solve this problem. Since for Statement (1), we do have possible values, shouldn't the answer be "D"?
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A school administrator will assign each student in a group 16 Jun 2017, 10:10
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bhavikagoyal2009

Hello. I had the same doubt as you to this question. According to me, the question simply asks "is it possible?", which means that are there any values of n and m which can solve this problem. Since for Statement (1), we do have possible values, shouldn't the answer be "D"?
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The point is that m and n are some specific numbers there. So, what the question means is that CAN we 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?
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A school administrator will assign each student in a group 19 Apr 2020, 04:26
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Hi Bunuel

The wording, "is it possible?", is a little confusing for me. In statement (1) it is also "possible" for n/m to be an integer such as n=15 and m=5. Can you explain how wording, "is it possible?", in the question should be interpreted?

Thank you!
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A school administrator will assign each student in a group 19 Apr 2020, 04:49
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Hi Bunuel

The wording, "is it possible?", is a little confusing for me. In statement (1) it is also "possible" for n/m to be an integer such as n=15 and m=5. Can you explain how wording, "is it possible?", in the question should be interpreted?

Thank you!
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I think this is explained many times on previous pages. "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?" means whether we CAN divide equally n students in m classrooms, so whether n/m is an integer.
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A school administrator will assign each student in a group 24 Apr 2020, 12:30
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kelleygrad05
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?


Hi kellygrad05,

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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"In 13n/m if you put n=15 and m=5, the you get an integer I think , and it both values satisfies the condition given in the question
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A school administrator will assign each student in a group 19 May 2020, 03:42
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Bunuel
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 cannot 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.
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Bunuel : Statement 2 is not sufficient since n/m may or may not be an integer. If n= 20, m=4 then 13*20/4 gives an integer. The answer to the question is 'Yes'. If, m=4 and n = 21, 13 * 21/4 is not an integer, the answer is a 'No' . Therefore we are getting a 'Yes' and a 'No', so the statement is NOT sufficient. So, we have to combine both statements and check. Please correct me if I am wrong.
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theleopride
Bunuel
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 cannot 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.

Bunuel : Statement 2 is not sufficient since n/m may or may not be an integer. If n= 20, m=4 then 13*20/4 gives an integer. The answer to the question is 'Yes'. If, m=4 and n = 21, 13 * 21/4 is not an integer, the answer is a 'No' . Therefore we are getting a 'Yes' and a 'No', so the statement is NOT sufficient. So, we have to combine both statements and check. Please correct me if I am wrong.
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m = 4 and n = 21 does not satisfy the second statement, so these values are not valid. B is correct.
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A school administrator will assign each student in a group 19 May 2020, 05:39
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Bunuel
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 cannot 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.
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Bunuel : Statement 2 is not sufficient since n/m may or may not be an integer. If n= 20, m=4 then 13*20/4 gives an integer. The answer to the question is 'Yes'. If, m=4 and n = 21, 13 * 21/4 is not an integer, the answer is a 'No' . Therefore we are getting a 'Yes' and a 'No', so the statement is NOT sufficient. So, we have to combine both statements and check. Please correct me if I am wrong.[/quote]

m = 4 and n = 21 does not satisfy the second statement, so these values are not valid. B is correct.[/quote]

In that case, I am thinking since the second statement says 'it is possible', it means n/m must be an integer.

In that case statement 1 should be sufficient, because it says 'it is possible', which means, n/m has to be an integer. Where am I going wrong?
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Quote:
theleopride
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 cannot 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.

Bunuel : Statement 2 is not sufficient since n/m may or may not be an integer. If n= 20, m=4 then 13*20/4 gives an integer. The answer to the question is 'Yes'. If, m=4 and n = 21, 13 * 21/4 is not an integer, the answer is a 'No' . Therefore we are getting a 'Yes' and a 'No', so the statement is NOT sufficient. So, we have to combine both statements and check. Please correct me if I am wrong.
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m = 4 and n = 21 does not satisfy the second statement, so these values are not valid. B is correct.[/quote]

In that case, I am thinking since the second statement says 'it is possible', it means n/m must be an integer.

In that case statement 1 should be sufficient, because it says 'it is possible', which means, n/m has to be an integer. Where am I going wrong?[/quote]

I'll try again.

The question asks: given that \(3<m<13<n\), is \(\frac{n}{m}=integer\)?

(1) says: \(\frac{3n}{m}=integer\). This is not sufficient to answer whether \(\frac{n}{m}=integer\).

(2) says: \(\frac{13n}{m}=integer\). This is sufficient to answer whether \(\frac{n}{m}=integer\).
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