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OG12 - DS - 128

Expert replies
by JeffB » Tue Jun 30, 2009 5:10 pm
A school admin 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

Anyone have a good/easy explanation?
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Source: — Data Sufficiency |

Re: OG12 - DS - 128

by ssmiles08 » Tue Jun 30, 2009 6:13 pm
JeffB wrote:A school admin 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

Anyone have a good/easy explanation?
IMO B.

If you translate the question in layman's terms, its really asking if n/m = integer.

1) translation: 3n/m = integer. This would be insufficient b/c n can be divisible by m or 3 is a factor of m. if 3 is a factor of m, them n would only be divisible by a factor of m and not m itself.

ex: n = 18 m = 6 or n = 14 and m = 6.

either case scenario, 3n/m = integer, but n/m would provide contradicting results.

2) 13n/m = integer.

This is interesting b/c 3<m<12. since 13 is a prime number and divisible by 1 and 13 only, we know that m is not a factor of of 13. so m has to be a factor of n for 13n/m to be an integer. we know from this that n/m = integer and thus is sufficient.

OA?
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by aj5105 » Tue Jun 30, 2009 8:26 pm
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hmm

by JeffB » Wed Jul 01, 2009 3:10 am
Thx ssmiles - that's the correct answer.

I searched the forums everywhere for this problem and couldn't find it, thx for the link aj.

It's really quite simple once you breakdown the problem. If it just said n/m = integer it would have been a lot easier :)
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