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students / classrooms

by hpgmat » Sun Aug 23, 2009 7:53 pm
a school adminstrator 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 n students to one of 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 adssigned to it.



my guess is each statement alone is sufficient.
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by PussInBoots » Sun Aug 23, 2009 9:07 pm
Rephrase the question:

Is 3 < m < 13 < n. Is n divisible by m?

(1) 3n is divisible by m
(2) 13n is divisible by m.

(1) does not give us anything.
(2) 13n is divisible by m. 13 is prime number and it cannot be a factor of m, hence n/m has to be an integer, hence the answer is B
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by hpgmat » Sun Aug 23, 2009 9:32 pm
PussInBoots wrote:Rephrase the question:

Is 3 < m < 13 < n. Is n divisible by m?

(1) 3n is divisible by m
(2) 13n is divisible by m.

(1) does not give us anything.
(2) 13n is divisible by m. 13 is prime number and it cannot be a factor of m, hence n/m has to be an integer, hence the answer is B
3 IS A PRIME NUMBER AS WELL, NEITHER DIVISIBLE BY M . SO N/M HAS TO BE AN INTEGER. SO EACH STATEMENT IS SUFFICIENT BY ITSELF. PLS COMMENT
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by shanrizvi » Mon Aug 24, 2009 9:13 am
Hmm. 3<m<13 and n>13, is n divisible by m? is n/m an integer?

Statement 1: 3n is divisible by m

3n/m = 3*18/9 is an integer. 18/9 is also an integer.

3*14/6 is an integer but 14/6 is not.

Doesn't that mean statement 1 is not sufficient?

Statement 2: 13n is divisible by m

13*(15/5) is an integer and so is 15/5.
13*(16/4) is an integer and so is 16/4.

I think the key point here is that statement 2 is sufficient because m is between 3 and 13 and there are multiples of 3 in that range but not multiples of 13.
Last edited by shanrizvi on Mon Aug 24, 2009 11:22 am, edited 1 time in total.
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by navalpike » Mon Aug 24, 2009 11:12 am
shanrizvi wrote: I think the key point here is that statement 1 is sufficient because m is between 3 and 12 and there are multiples of 3 in that range but not multiples of 13.
I think you mean to say this is why statement 2 is sufficient
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by shanrizvi » Mon Aug 24, 2009 11:20 am
Oops. Yes. Sorry. I've corrected it now.
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by hpgmat » Tue Aug 25, 2009 12:25 am
navalpike wrote:
shanrizvi wrote: I think the key point here is that statement 1 is sufficient because m is between 3 and 12 and there are multiples of 3 in that range but not multiples of 13.
I think you mean to say this is why statement 2 is sufficient
THANK YOU FOR CLARIFICATION. OTHER THAN TRIAL AND ERROR METHOD, IS THERE ANY OTHER WAY ( FASTER WAY ) TO COME TO THE CONCLUSION ON THIS TYPE OF QUESTIONS?
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by Blues » Mon Aug 31, 2009 1:15 am
hpgmat wrote:
navalpike wrote:
shanrizvi wrote: I think the key point here is that statement 1 is sufficient because m is between 3 and 12 and there are multiples of 3 in that range but not multiples of 13.
I think you mean to say this is why statement 2 is sufficient
THANK YOU FOR CLARIFICATION. OTHER THAN TRIAL AND ERROR METHOD, IS THERE ANY OTHER WAY ( FASTER WAY ) TO COME TO THE CONCLUSION ON THIS TYPE OF QUESTIONS?
Sorry that I'm only seeing this now, a few days after you posted it. Hopefully, this helps a bit.

Trial and error is not necessary in this problem (although testing numbers is helpful in checking your logic). All we need to determine is whether or not the prime factors of m are present in n.

In the first statement, 3 is not divisible by m, but there is no reason 3 cannot be a factor of m. In other words, m may or may not be divisible by 3. From the original statement, we know that m could be 4, 5, 6, 7, 8, 9, 10, 11, or 12. Since we cannot tell whether 3 and m share any prime factors, we cannot determine if all of the prime factors of m are in n. If we want to test numbers, n could be 14, in which case m could be 6 or 7. Six is not a factor of 14, while 7 is. As we can see, this statement is insufficient.

In statement two, we know for a fact that m and 13 do not share any prime factors (since m has to be smaller than 13). Since none of the prime factors of m are in 13, all of the prime factors of m MUST be in n in order for 13n to be divisible by m.

Now...if only I could puzzle this out in the two minute time frame allotted for each question. Out of curiosity, do any of the experts on the board have an idea of the approximate difficulty level of this question?
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