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# A school administrator will assign each student in a group

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Re: is n divisible by m [#permalink]  08 Mar 2011, 10:13
Expert's post
Merging similar topics.
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Re: Divisibility problem - n students in m classes (+/-700 lvl) [#permalink]  17 Mar 2011, 10:27
Could someone explain how we've reached the n/m divisibility statement. I couldn't understand that part ? Thanks.
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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.

Trickyyyyyyy..
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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.

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.
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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...
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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
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Re: a hard one [#permalink]  06 Jun 2011, 15:35
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Re: a hard one [#permalink]  07 Jun 2011, 02:36
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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.

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Re: a hard one [#permalink]  07 Jun 2011, 13:25
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.
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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.

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 ?
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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
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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

I came across this explanation at =>

http://bt.www.beatthegmat.com/gmat-12th ... tml#331474

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 ?
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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

I came across this explanation at =>

http://bt.www.beatthegmat.com/gmat-12th ... tml#331474

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
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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?
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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
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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.

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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.

Ans: "B"
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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
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Re: a hard one [#permalink]  09 Jun 2011, 20:56
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If 3 < m < 13, Is n/m an integer?

(1) 3n/m is an integer

(2) 13n/m is an integer

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.
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To find n/m= int?

a) 3n/m = int.
If a*b/c = int and gcd(a,c) = 1 then b/c = int

Hence 3n/4 = int but 3n/6 is not int

b) 13n/m = int as m<13
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