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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

"A school administrator..." [#permalink]
10 Jun 2011, 19:45

Hello. I hope you can explain the answer to this problem. It's from the Official Guide for GMAT Review (12th edition). I found the Guide's explanation to be confusing.

# 128, Data Sufficiency 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.

Re: "A school administrator..." [#permalink]
10 Jun 2011, 21:43

howlingwolf4 wrote:

Thanks for putting the link, fluke.

You're welcome!!!

It's easy to search a question. Just type in few unique looking words from the question in the rectangular box located at top-right corner and hit the magnifying glass in it. It will give the results showing posts from past threads.

e.g. For the question above, you could have typed "A school administrator will assign each"

Re: Students per classroom [#permalink]
01 Jul 2011, 09:53

ssr7 wrote:

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

(1) Given 3n is divisible by m Lets assume n=14 ; 3*[highlight]14[/highlight]=42 => m= [highlight]6[/highlight] or [highlight]7[/highlight] (Since 3n is divisible by m; 3<m<13<n) Now pick the above used numbers to verify whether n is divisible by m. So n=14 => m=7 (Is divisible) and m=6 (Is not divisible) [So we cant be sure that it is divisible ; [highlight]NOT SUFFICIENT[/highlight]]

(2) Given 13n is divisible by m Lets assume n=14 ; 13*[highlight]14[/highlight]=182 => m=[highlight]7[/highlight] (Since 13n is divisible by m ; 3<m<13<n) Now pick the above used numbers to verify whether n is divisible by m. So n=14 => m=7 (Is divisible)[So we can be sure that it is divisible ; [highlight]SUFFICIENT[/highlight]]

Similarly try plugging in other values of n and corresponding value for m to determine the sufficiency.

Re: Divisibility problem - n students in m classes (+/-700 lvl) [#permalink]
03 Jul 2011, 21:54

I can think of an easier solution than the LCM method. Let's see it.

Just repeating the statements given

(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.

From (1) 3n is divisible by m i.e. 3n/m is an integer. If 3n/m is an integer, then decimal part of n/m has to be 0 or 1/3 or 2/3. If it is is zero, then n is divisible by m. To see if this info is sufficient to say n/m is an integer. For that we need to prove that there is NO combination of values of n and m such that the remainder of n/m would be 1 or 2. For example, n = 14 and m = 6. The remainder of n/m is 1/3. As we found at least one combination of values to n and m such that n%m is not a zero, we don't have sufficient info from (1) to answer the question.

From (2) 13n is divisible by m i.e. 13n/m is an integer. If 13n/m is an integer, then the decimal part of n/m is one of 0, 1/13, 2/13, 3/13, 4/13 ... , 11/13 and 12/13. As m has to be less than 13 (from the info given), there is no way n%m can be 1 or 2 or 3 or 4 ... 12. So n%m must to be 0 (zero). As n%m is zero, (2) is sufficient.

OG 12ed Data Sufficiency question - CONFUSION [#permalink]
25 Sep 2011, 12:15

Hi GMATers,

I'm new to GMATClub so would appreciate any guidance on the following OG question:

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.

My idea is that, after rephrasing, the question is essentially asking whether "it is possible that n is divisible by m", NOT whether "n IS divisible by m". So given (1) & (2), both statements can lead to the conclusion that it is POSSIBLE that n is divisible by m, not n is ABSOLUTELY divisible by m. So I think the correct answer should be (D) - each statement alone is sufficient.

Does this make sense? Anyone could share some insights on this argument?

Thanks a lot! _________________

musikalisch und Jazz

gmatclubot

OG 12ed Data Sufficiency question - CONFUSION
[#permalink]
25 Sep 2011, 12:15