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Rooms-Students assignment

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Rooms-Students assignment

by kvcpk » Tue Jun 08, 2010 4:54 am
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.

OA is B. Can Somene please explain?
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Source: — Data Sufficiency |
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by liferocks » Tue Jun 08, 2010 4:59 am
From 1
3n=mk where k is any integer
or n=mk/3 now k/3 may or may not be integer depending on which n may or may not be a multiple of m..insufficient

From 2

13n=mk or n=mk/13
since 3<m<13,k/13 has to an integer so that mk/13 is integer..sufficient
Ans option B
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by Patrick_GMATFix » Tue Jun 08, 2010 5:24 am
This is a pretty tough divisibility question. It basically asks whether we can distribute n kids into m equal groups. In other words, REPHRASE: Is n divisible by m?

This is #128 in OG12. Detailed solution and take-away are attached below.

-Patrick
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by kvcpk » Tue Jun 08, 2010 6:09 am
liferocks wrote: From 2

13n=mk or n=mk/13
since 3<m<13,k/13 has to an integer so that mk/13 is integer..sufficient
Ans option B
Hi,

I understood why the first statement is insufficient. But the second statement is confusing. Can you please tell me why k/13 has to be an integer?
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by kvcpk » Tue Jun 08, 2010 6:11 am
Patrick_GMATFix wrote:This is a pretty tough divisibility question. It basically asks whether we can distribute n kids into m equal groups. In other words, REPHRASE: Is n divisible by m?

This is #128 in OG12. Detailed solution and take-away are attached below.

-Patrick
Hi patrick,

The link says that the access is denied. Can you please look into it?
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by Patrick_GMATFix » Tue Jun 08, 2010 6:14 am
kvcpk wrote:
Hi patrick,

The link says that the access is denied. Can you please look into it?
Hi kvcpk. Not sure why you can't see the solution I've written; 10 ppl have already downloaded it. Send me a PM and I'd be glad to help.
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by liferocks » Tue Jun 08, 2010 6:16 am
kvcpk wrote:
liferocks wrote: From 2

13n=mk or n=mk/13
since 3<m<13,k/13 has to an integer so that mk/13 is integer..sufficient
Ans option B
Hi,

I understood why the first statement is insufficient. But the second statement is confusing. Can you please tell me why k/13 has to be an integer?
13n=mk or n=mk/13
here LHS(n) is integer so RHS has to be integer
i.e mk/13 is integer
now m<13 and 13 is a prime so only way mk/13 is an integer when k/13 is integer.
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by kvcpk » Tue Jun 08, 2010 6:36 am
liferocks wrote:
kvcpk wrote:
liferocks wrote: From 2

13n=mk or n=mk/13
since 3<m<13,k/13 has to an integer so that mk/13 is integer..sufficient
Ans option B
13n=mk or n=mk/13
here LHS(n) is integer so RHS has to be integer
i.e mk/13 is integer
now m<13 and 13 is a prime so only way mk/13 is an integer when k/13 is integer.
Why cant we apply the same analysis to premise1:

3n=mk or n= mk/3
here also n is integer. so mk/3 has to be integer. m is already integer. So k/3 has to be an integer.

??
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by liferocks » Tue Jun 08, 2010 7:16 am
kvcpk wrote:
liferocks wrote:
kvcpk wrote:
liferocks wrote: From 2

13n=mk or n=mk/13
since 3<m<13,k/13 has to an integer so that mk/13 is integer..sufficient
Ans option B
13n=mk or n=mk/13
here LHS(n) is integer so RHS has to be integer
i.e mk/13 is integer
now m<13 and 13 is a prime so only way mk/13 is an integer when k/13 is integer.
Why cant we apply the same analysis to premise1:

3n=mk or n= mk/3
here also n is integer. so mk/3 has to be integer. m is already integer. So k/3 has to be an integer.

??
for 1 n=mk/3 now either m/3 or k/3 can be integer since 3<m
in which case n will not be divisible by 3
ex m=6,k=7 in that case mk=42=3*n ..so n=14 which is not divisible by m=6
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by kvcpk » Tue Jun 08, 2010 7:57 am
liferocks wrote:
kvcpk wrote:
liferocks wrote:
kvcpk wrote:
liferocks wrote:
for 1 n=mk/3 now either m/3 or k/3 can be integer since 3<m
in which case n will not be divisible by 3
ex m=6,k=7 in that case mk=42=3*n ..so n=14 which is not divisible by m=6
Thanks for your response and patience. I get it now. There cannot be any multiple of m below 13. It looks simple..but easily distracting..
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