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PRIME

by Gurpinder » Tue Aug 24, 2010 10:54 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[spoiler] (B)[/spoiler]
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by clock60 » Tue Aug 24, 2010 11:31 am
here the problem simply asks
is (n/m)=integer
(1)(3n)/m=integer
try inserting numbers,
as n>13, let it be n=14, and m=7, (3*14)/7=integer
and n/m=14/7=integer
so the answer is yes
but if n=15, m=9, (3*15)/9=integer,
n/m=15/9 not integer the answer is no
1 st insufficient
(2) (13n)/m=integer
m<13, and 13 is prime so 13 is not divisible by m as m>1 and m<13 ( and 13=1*13). so for (13n)/m be integer -n must be divisible by m
B looks fine
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by Gurpinder » Tue Aug 24, 2010 2:33 pm
clock60 wrote:here the problem simply asks
is (n/m)=integer
(1)(3n)/m=integer
try inserting numbers,
as n>13, let it be n=14, and m=7, (3*14)/7=integer
and n/m=14/7=integer
so the answer is yes
but if n=15, m=9, (3*15)/9=integer,
n/m=15/9 not integer the answer is no
1 st insufficient
(2) (13n)/m=integer
m<13, and 13 is prime so 13 is not divisible by m as m>1 and m<13 ( and 13=1*13). so for (13n)/m be integer -n must be divisible by m
B looks fine
I understand your reasoning for the first statement.

You lost me on 2.

13n/m = int. that means that 13n = will be a multiple of 13. there are some values of M (4...12) that wont give int result when divided by a multiple of 13. so how is statement 2 sufficient?
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by debmalya_dutta » Tue Aug 24, 2010 3:52 pm
Gurpinder I think you were almost there.......

So 13n is divisible by m per the statement...
but m lies between 3 and 13(not including)... so that means that for m to be a factor of 13n ...m has to be a factor of n...and that in turn proves that I can divide the n students equally into m classes
Gurpinder wrote:
clock60 wrote:here the problem simply asks
is (n/m)=integer
(1)(3n)/m=integer
try inserting numbers,
as n>13, let it be n=14, and m=7, (3*14)/7=integer
and n/m=14/7=integer
so the answer is yes
but if n=15, m=9, (3*15)/9=integer,
n/m=15/9 not integer the answer is no
1 st insufficient
(2) (13n)/m=integer
m<13, and 13 is prime so 13 is not divisible by m as m>1 and m<13 ( and 13=1*13). so for (13n)/m be integer -n must be divisible by m
B looks fine
I understand your reasoning for the first statement.

You lost me on 2.

13n/m = int. that means that 13n = will be a multiple of 13. there are some values of M (4...12) that wont give int result when divided by a multiple of 13. so how is statement 2 sufficient?
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by Gurpinder » Tue Aug 24, 2010 7:44 pm
thanks for your reply! do you see my confusion though....

stmt (1)

3n/m=int so 3n would be a multiple of 3 which is divisible by some M's like 6,9,12 and not others.

Stmt(2)

in the same logic

13n has to be a multiple of 13. none of the values of M are factors of 13 because 13 is a prime number. So how is this tell us that you can divide the students equally.

i see that you guys mention that since 13n/m=int that must mean n/m=int. how? can you explain how you guys are doing these steps!

thanks!
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by thirst4edu » Tue Aug 24, 2010 8:02 pm
Gurpinder wrote:thanks for your reply! do you see my confusion though....

stmt (1)

3n/m=int so 3n would be a multiple of 3 which is divisible by some M's like 6,9,12 and not others.

Stmt(2)

in the same logic

13n has to be a multiple of 13. none of the values of M are factors of 13 because 13 is a prime number. So how is this tell us that you can divide the students equally.

i see that you guys mention that since 13n/m=int that must mean n/m=int. how? can you explain how you guys are doing these steps!

thanks!
I think the key here is to understand that m could be multiple of 3 in case of statement 1 hence n/m may not be necessarily true as shown in examples above.

Where as in case of statement 2, m cannot have multiple of 13 (since 3 < m < 13) , hence in order to make statement 2 true, m has to divide n equally.

I hope I haven't confused you more :)
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by debmalya_dutta » Tue Aug 24, 2010 8:21 pm
Gurpinder wrote:thanks for your reply! do you see my confusion though....

stmt (1)

3n/m=int so 3n would be a multiple of 3 which is divisible by some M's like 6,9,12 and not others.

Stmt(2)

in the same logic

13n has to be a multiple of 13. none of the values of M are factors of 13 because 13 is a prime number. So how is this tell us that you can divide the students equally.

i see that you guys mention that since 13n/m=int that must mean n/m=int. how? can you explain how you guys are doing these steps!

thanks!
Quoted from your post -
i see that you guys mention that since 13n/m=int that must mean n/m=int. how? can you explain how you guys are doing these steps!
look at this statement closely
assign each of the X students to one of the Y classrooms so that each classroom has the same number of students assigned to it

what does this mean ?
it means that I can divide X equally into Y parts... So what does that mean ... so this means that when we divide X into Y ..we get a quotient and surely no remainder (because the statement says that I can divide X equally into Y classrooms... )
for example ... if X=100 ..then Y is 10, 20 , 50 .... Y cannot be say 30.... because 100/30 has a remainder ...
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by Gurpinder » Wed Aug 25, 2010 8:14 am
thirst4edu wrote: I think the key here is to understand that m could be multiple of 3 in case of statement 1 hence n/m may not be necessarily true as shown in examples above.

Where as in case of statement 2, m cannot have multiple of 13 (since 3 < m < 13) , hence in order to make statement 2 true, m has to divide n equally.

I hope I haven't confused you more :)
We cannot say that M could be a multiple of 3 because we have to take statement INDEPENDENTLY. We can't use any information from the previous statement.

But I get the logic now. 13n/m=int. 13n cannot be a multiple of 13 because M < 13, and therefore will not be a factor of 13n. That must mean that n/m=int. we dont need to go any further because this was our original question: whether n/m

thanks to all!!
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by tomada » Wed Aug 25, 2010 9:23 am
I'm sure I'm overlooking something very basic, but I don't see how the sufficiency of Statement (1) is negated when using n=15 as an example.

First, we've checked n=14, and have confirmed that 3n=42 students can be equally allocated to 'm' classrooms, with 'm' satisfying the inequality. Similarly, n=14 students can be allocated evenly to 'm' classrooms, with 'm' satisfying the inequality.

If n=15, 3n= 45 students, which can be allocated evenly into 3, 5, 9, or 15 classrooms. However, I'll eliminate 3 and 15 because they don't satisfy the given inequality.

Suppose m=5. Can 45 students be allocated evenly into 5 classrooms? Yes.
Now, back to n=15. Can 15 students be allocated evenly into 5 classrooms? Yes, 3 students in each classroom.

While it's true that this doesn't hold for m=9, we only need to determine if 'n' students can be equally allocated into 'm' classrooms, such that 'm' satisfies the given inequality. We don't need to ensure that all possible values of 'm' are satisfied.

That said, when n=15, 3n=45, and 45 can be allocated equally to 'm' classrooms, such that 'm' satisfies the given inequality.

Now, is it possible for n=15 students to be allocated equally to 'm' classrooms, such that 'm' satisfies the given inequality?
The answer is yes, when 'm' = 5.

Therefore, Statement (1) cannot be deemed insufficient after checking n=14 and n=15.

At this point, Statement (1) is still sufficient by itself.
Can someone find a value of 'n' for which '3n' can be equally allocated into 'm' (satisfying the given inequality), but for which no value of 'm' exists that would allow 'n' students to be equally allocated into 'm' ?
clock60 wrote:here the problem simply asks
is (n/m)=integer
(1)(3n)/m=integer
try inserting numbers,
as n>13, let it be n=14, and m=7, (3*14)/7=integer
and n/m=14/7=integer
so the answer is yes
but if n=15, m=9, (3*15)/9=integer,
n/m=15/9 not integer the answer is no
1 st insufficient
(2) (13n)/m=integer
m<13, and 13 is prime so 13 is not divisible by m as m>1 and m<13 ( and 13=1*13). so for (13n)/m be integer -n must be divisible by m
B looks fine
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by thirst4edu » Wed Aug 25, 2010 9:33 am
tomada wrote: Can someone find a value of 'n' for which '3n' can be equally allocated into 'm' (satisfying the given inequality), but for which no value of 'm' exists that would allow 'n' students to be equally allocated into 'm' ?
condition is - 3 < m < 13 < n,

(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.

when m = 6, n = 14 <= here m, n both satisfies given condition

according to stmt1) 3n/m is int
3*14/6 = int
but 14/6 is not integer!

Point here is that , m "could be multiple of 3" and 3 in expression "3n" could be hiding one 3 hence it is difficult to see..
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by tomada » Wed Aug 25, 2010 9:48 am
However, the fact that 14/6 isn't an integer is extraneous, because we can still equally allocate 14 students into 'm' classrooms for some value of 'm'', such that 3 < m < 13 < n. If m=7, 14/7=2, which is all we need to know that it can be done. The fact that it can't be done for other values of 'm' doesn't negate its sufficiency.

thirst4edu wrote:
tomada wrote: Can someone find a value of 'n' for which '3n' can be equally allocated into 'm' (satisfying the given inequality), but for which no value of 'm' exists that would allow 'n' students to be equally allocated into 'm' ?
condition is - 3 < m < 13 < n,

(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.

when m = 6, n = 14 <= here m, n both satisfies given condition

according to stmt1) 3n/m is int
3*14/6 = int
but 14/6 is not integer!

Point here is that , m "could be multiple of 3" and 3 in expression "3n" could be hiding one 3 hence it is difficult to see..
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by thirst4edu » Wed Aug 25, 2010 12:29 pm
tomada wrote:However, the fact that 14/6 isn't an integer is extraneous, because we can still equally allocate 14 students into 'm' classrooms for some value of 'm'', such that 3 < m < 13 < n. If m=7, 14/7=2, which is all we need to know that it can be done. The fact that it can't be done for other values of 'm' doesn't negate its sufficiency.
You have brought a very good point, tomada ..

Question says -
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?
Yeah is it "possible" with statement 1 :) are we missing something? should the question/answer be flawed?
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by tomada » Wed Aug 25, 2010 12:33 pm
Does anyone know the source of this question?
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