Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

Re: Divisibility problem - n students in m classes (+/-700 lvl) [#permalink]
17 Mar 2011, 19:00

The n/m being an integer part is because we are told basically that we can divide n by m without a remainder.

I dont get how doing the method recommended is time saving at all :s

Yes like was written, for (2) 13 is a prime number, since m cant be 13 or a multiple of it, then it is sufficient intuitively that it is true.

For (1), we just see that 3 is also a prime number. The trap is that although m cant=3, it can=6,9,12 which are multiples of it, so we don't know if the statement is true.

Re: Students per classroom [#permalink]
16 Apr 2011, 03:15

Quote:

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.

Please help me understand what is wrong in this approach. I just used number. 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.

You picked one combination (n=14 and m=7) for which 13n is divisible by m and another combination (n=14 and m=5) for which 13n is NOT divisible by m. To show that statement (2) is insufficient to answer the original posed question, you should be able to pick two combinations of n and m for which 13n is divisible by m, but for one of them n should be divisible by m and for the other n should NOT be divisible by m. That would be the only way to say that the truth of statement two is not sufficient to answer the original question.

Re: Students per classroom [#permalink]
16 Apr 2011, 03:24

Quote:

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...

Shouldn't the "prime" argument used in (2) for 13 be applicable to (1) for 3. I understand that (1) is easily deemed insufficient when you consider two sets of values for m and n that satisfy the divisibility of the divisibility of 3n by m, but only one set of which satisfies the divisibility of m by n. However, I don't understand how the "prime" argument used for (2) can't be used for (1)....really confused...

Re: Students per classroom [#permalink]
20 Apr 2011, 19:23

Quote:

Quote: 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...

Shouldn't the "prime" argument used in (2) for 13 be applicable to (1) for 3. I understand that (1) is easily deemed insufficient when you consider two sets of values for m and n that satisfy the divisibility of the divisibility of 3n by m, but only one set of which satisfies the divisibility of m by n. However, I don't understand how the "prime" argument used for (2) can't be used for (1)....really confused...

Mods or anyone else have a way to clear up my confusion? :D

if you rephrase the question he is asking if n/m = 0

case 1: 3n/m = 0 case 2: 13n/m = 0

in case 1: consider n = 15 3n = 45 and 45 is only divisible by 9 coz 3<n<13 now consider n = 18 3n = 54 now 3n is divisible by 6,9 and both are between 3&12 So Insuff

case 2 : 13n/m = 0 since 13 is prime and m<13 n/m will always be zero.

n/m remainder should be 0. Stat 1 20/6 R is not 0 . However, 3*20/6 R is 0................(NO answer) 60/6 R is not 0 . However, 3*60/6 R is 0................(Yes answer) Stat 2 13n/m R is 0 given. Since 3<m<13 n must be divisible my m and R must be 0. B is the answer.
_________________

Consider me giving KUDOS, if you find my post helpful. If at first you don't succeed, you're running about average. ~Anonymous

School Administrator [#permalink]
08 Jun 2011, 21:35

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.

How do you know if the question asks you to see if M is a factor of N or N is a factor of M? or m/n or n/m ?

Re: School Administrator [#permalink]
08 Jun 2011, 22:35

siddhans 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.

How do you know if the question asks you to see if M is a factor of N or N is a factor of M? or m/n or n/m ?

the simple logic here is both n and m must be integers.. st.2 says 13*n/m is possible as 13 is a prime n/m must be an integer thus it is possible to arrange n students to m classrooms. Has to be B

Re: School Administrator [#permalink]
08 Jun 2011, 22:38

sudhir18n wrote:

siddhans 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.

How do you know if the question asks you to see if M is a factor of N or N is a factor of M? or m/n or n/m ?

the simple logic here is both n and m must be integers.. st.2 says 13*n/m is possible as 13 is a prime n/m must be an integer thus it is possible to arrange n students to m classrooms. Has to be B

We first need to sort out what the question stem is really asking. In order to "evenly" divide our n students up among m classes, it must be the case that n is some multiple of m. We can try some numbers to see the relationship: If there are 4 classrooms, then 1 student per class means n=4, 2 students per class means n=8, 3 students per class means n=12. Notice that in all cases n is a multiple of m. So the real question is the following:

Is n/m an integer?

** Now, remember that in order for n/m to be an integer, it must be the case that all of the factors of m are absorbed into n. (i.e. 8/4= 2 = integer because all of 4 was able to cancel with 8).

Statement (1): (same rephrasing steps as above) = 3n/m is an integer.

The easier thing to do here is to remember that we are in a Yes/No question and try to find numbers that allow us to answer both Yes and No to the stem. Also, don't forget the constraints on m and n. Because m is an integer between 3 and 13, we should think of values that will allow 3n/m to be divisible by m but allow both a Yes and a No answer. If we allow m to be 3, then the denominator is completely absorbed into the numerator. Then we could choose n=6, so 3(6)/3 = 6, and 6/3 = 2 (n/m is an integer). We could also choose n=5, so 3(5)/3 = 5, but 5/3 is not an integer. INSUFFICIENT

Statement (2): (same rephrasing as above) = 13n/m. Now notice that for the m to be absorbed into the numerator regardless of the value of n, it would need to be 13 because 13 is prime, but m is restricted to a number between 3 and 13 (not inclusive) so m cannot be 13. Therefore, in order to absorb the denominator m completely into the numerator n it must be the case that n contains the terms that completely absorb the denominator (i.e. if 13(17)/n, then n must be 1 or 17 for the fraction to be an integer.) Therefore n/m must be an integer. SUFFICIENT

The question I have is why havent we chosen values of n > 13 since 3<m<13<n Doesnt it mean we need to choose values >13 to test statement 1 and 2 ?

Re: School Administrator [#permalink]
08 Jun 2011, 22:44

siddhans wrote:

sudhir18n wrote:

siddhans 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.

How do you know if the question asks you to see if M is a factor of N or N is a factor of M? or m/n or n/m ?

the simple logic here is both n and m must be integers.. st.2 says 13*n/m is possible as 13 is a prime n/m must be an integer thus it is possible to arrange n students to m classrooms. Has to be B

We first need to sort out what the question stem is really asking. In order to "evenly" divide our n students up among m classes, it must be the case that n is some multiple of m. We can try some numbers to see the relationship: If there are 4 classrooms, then 1 student per class means n=4, 2 students per class means n=8, 3 students per class means n=12. Notice that in all cases n is a multiple of m. So the real question is the following:

Is n/m an integer?

** Now, remember that in order for n/m to be an integer, it must be the case that all of the factors of m are absorbed into n. (i.e. 8/4= 2 = integer because all of 4 was able to cancel with 8).

Statement (1): (same rephrasing steps as above) = 3n/m is an integer.

The easier thing to do here is to remember that we are in a Yes/No question and try to find numbers that allow us to answer both Yes and No to the stem. Also, don't forget the constraints on m and n. Because m is an integer between 3 and 13, we should think of values that will allow 3n/m to be divisible by m but allow both a Yes and a No answer. If we allow m to be 3, then the denominator is completely absorbed into the numerator. Then we could choose n=6, so 3(6)/3 = 6, and 6/3 = 2 (n/m is an integer). We could also choose n=5, so 3(5)/3 = 5, but 5/3 is not an integer. INSUFFICIENT

Statement (2): (same rephrasing as above) = 13n/m. Now notice that for the m to be absorbed into the numerator regardless of the value of n, it would need to be 13 because 13 is prime, but m is restricted to a number between 3 and 13 (not inclusive) so m cannot be 13. Therefore, in order to absorb the denominator m completely into the numerator n it must be the case that n contains the terms that completely absorb the denominator (i.e. if 13(17)/n, then n must be 1 or 17 for the fraction to be an integer.) Therefore n/m must be an integer. SUFFICIENT

The question I have is why havent we chosen values of n > 13 since 3<m<13<n Doesnt it mean we need to choose values >13 to test statement 1 and 2 ?

St2. says 13 *n/m is an integer. u can take any value of n>13 such that it is divisible by m ( its a constraint of the problem) hence u cant take prime numbers.. take n = 24 but m should be either 4,6,8,12.

Is that what u meant? let me know if I havent answered what u mean to ask . Thanks

Re: School Administrator [#permalink]
08 Jun 2011, 23:30

What i mean is stmt 2 says : 13n/m should be an integer.... so we need to plug in values on n that are greater than 13 correct which satisfies this condition?

Re: School Administrator [#permalink]
08 Jun 2011, 23:34

siddhans wrote:

What i mean is stmt 2 says : 13n/m should be an integer.... so we need to plug in values on n that are greater than 13 correct which satisfies this condition?

More so because the question clearly mentions a constraint 3 < m < 13 < n so n can never be less than 13

Re: School Administrator [#permalink]
09 Jun 2011, 02:19

Hi both, interesting question, upon reading the questions I had seen the constraints N>13 m is between 4-12 I Failed to realize or comprehend that effectively both the statements are essentially asking for the answers to be integers or n evenly dividing into m , or the fact that they require factors of 3 and 13 with the constraints. Having said that, if we look at statement 1 it is insufficient because 3n assuming greater than 13 would give two separate answers for statement 1 and for statement none of the values between 4-12 are factors of 13. So I chose e as an answer. Can you please take me through this question by making it idiot proof for me.

Re: School Administrator [#permalink]
09 Jun 2011, 04:53

siddhans 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.

How do you know if the question asks you to see if M is a factor of N or N is a factor of M? or m/n or n/m ?

3 < m < 13 < n Q: Is m a factor of n?

(1) m is a factor of 3n. m=6; n=12; 3n=36; m is a factor of 36 AND m is ALSO a factor of 12. m=6; n=14; 3n=42; m is a factor of 36 BUT m is NOT a factor of 14. Not Sufficient.

(2) m is a factor of 13n. If m is a factor of 13n. m must also be a factor of n. Why so? We know; m is a value between 3 and 13, exclusive. m can be any of {4,5,6,7,8,9,10,11,12} None of these values has factors, except 1, in common that 13 has and it is given that m divides 13n evenly. Thus, it must divide n to make the statement correct because we know 13's factors is not the reason behind 13n's divisibility. Thus, "n" must have at least as many factors as "m" has. Sufficient.

Re: School Administrator [#permalink]
09 Jun 2011, 17:58

1) m=3n so 3n/m is an integer. Given that n can be any number bigger that 13, n can be 14, 15, 16, 17, 18, 19... . m can be 4, 5, 6, 7,8, 9, 10, 11, 12. If m = 10 and n=15, 3*15/10=4,5 which is not an integer, BUT if I have m=10 and n= 30, 3n/m is an integer. NOT SUFFICIENT

2) Gives 13n is divisible by m. So m is a factor of 13 ( 1 or 13) or n. But 3<m<13 , so m is only factor of n. hence satisfies distribution

IMO B..!

Nice question
_________________

Fight till you succeed like a gladiator..or doom your life which can be invincible

The only way way to ensure that n students can be grouped, without remainder, into m classrooms is that all (m) factors of 3n or 13n were also factors of n itself.

Stat (1)

Where n > 13 and 3n is an integer. 3n and n may or may not have the same factors within the range of m.

This is because 3 has multiples within the range, and these multiples could create factors that n does not have.

Insufficient.

Stat (2)

Where n > 13 and 3 < m < 13, 13n and n will have the same factors within the range of m. 13 has no factors within: 3 < m < 13

n/m is an integer

Last edited by Study1 on 14 Jan 2012, 08:44, edited 2 times in total.