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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.

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"

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]

Show Tags

03 Jul 2011, 22:54

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.

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.

Question: n/m = integer? (i.e. is m a divisor of n?)

From Statement 1 3n/m = integer factors of 3n: 3n, 3, n, 1 Since m>3 and m<13, m can be divisor of either 3n or n. Insufficient.

From Statement 2 13n/m = integer factors of 13n: 13n, 13, n, 1 Since m>3 and m<13, m can only be divisor of n. therefore, n/m is an integer. Sufficient!

Answer: B _________________

"The best day of your life is the one on which you decide your life is your own. No apologies or excuses. No one to lean on, rely on, or blame. The gift is yours - it is an amazing journey - and you alone are responsible for the quality of it. This is the day your life really begins." - Bob Moawab

OG 12ed Data Sufficiency question - CONFUSION [#permalink]

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25 Sep 2011, 13:15

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?

Re: OG 12ed Data Sufficiency question - CONFUSION [#permalink]

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25 Sep 2011, 23:31

1

This post received KUDOS

Expert's post

tititalin wrote:

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!

Think of it this way:

My question to you: "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?" The statement will be sufficient if you can answer with a 'YES' or a 'NO' It will be insufficient if you answer with a 'MAY BE'.

When will you say 'YES'? You will say yes when it will be possible to assign students the way I want. When will it be possible to assign students the way I want? When n will be divisible by m. If you do not know whether n is divisible by m and I ask you, "Can you assign students the way I have asked?" what will you say? You will say, "May be. If n is divisible, then yes, if it is not divisible then no." You will not say yes. You answered my original question with a 'may be'. Statement 1 gives you 'may be it is possible to completely divide n by m'. So your answer is 'may be we can assign the students as you requested' which makes the statement insufficient. _________________

Re: OG 12 DS 128- Assigning students per classroom [#permalink]

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29 Sep 2011, 07:08

1

This post received KUDOS

nakib77 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.

We want to find out whether n is divisible m 1) 3n/m is integer For, n = 15, m = 9, 3n/m is integer but n/m is not For n =14, n=7, 3n/m and n/m are both integers Insufficient

2) 13n/m is integer since m can be max 12 and 13 is prime, n/m will always be integer _________________

Thanks and Regards, GM.

gmatclubot

Re: OG 12 DS 128- Assigning students per classroom
[#permalink]
29 Sep 2011, 07:08

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