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A school administrator will assign each student [#permalink]
03 Jan 2011, 17:48
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. _________________
Re: A school administrator will assign each student [#permalink]
04 Jan 2011, 01:05
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.
OG DS 128 Arithmetic Properties [#permalink]
23 Feb 2011, 21:19
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. _________________
is n divisible by m [#permalink]
08 Mar 2011, 10:10
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. _________________
The proof of understanding is the ability to explain it.
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
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?
gmatclubot
Re: School Administrator
[#permalink]
08 Jun 2011, 23:30
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