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EMAN
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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.
Explanations:
(1) Given that 3n is divisible by m, then n is divisible by m if m = n = 9 (note that 3n = 27 and m = 9, so 3n is divisible by m) and n is not divisible by m if m = 9 and n = 12 (note that 3n = 36 and m = 9m so 3n is divisible by m); NOT SUFFICIENT
Comments: Okay, so according to the problem, n has to be greater than 13, so how the hell can m = n = 9? Couldn't m also be 6 and n be 18. Next they show that if m is 9, n cannot be 12. Well if cannot be 12 if the first place because that is less than 13? Am I missing something here? I am thoroughly confused as I am not a math wiz by any means.
(2) Given that 13n is divisible by m, then 13n = qm, or n/m = q/13, for some integer q. Since 13 is a prime number that divides qm (because 13n = qm) and 13 does not divide m (m < 13), it follows that 13 divides q. Therefore q/13 is an integer and since n/m = q/13, then n / m is an integer. Thus n is divisible by m. SUFFICIENT.
Comments: This one makes a little more sense but if anyone can dumb this explanation down that would also be appreciated. My takeaway here is that since 13 satisfies the criteria of n > 13 and if dividing by m produces an integer, this would be sufficient.
GRRR...!
(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.
Explanations:
(1) Given that 3n is divisible by m, then n is divisible by m if m = n = 9 (note that 3n = 27 and m = 9, so 3n is divisible by m) and n is not divisible by m if m = 9 and n = 12 (note that 3n = 36 and m = 9m so 3n is divisible by m); NOT SUFFICIENT
Comments: Okay, so according to the problem, n has to be greater than 13, so how the hell can m = n = 9? Couldn't m also be 6 and n be 18. Next they show that if m is 9, n cannot be 12. Well if cannot be 12 if the first place because that is less than 13? Am I missing something here? I am thoroughly confused as I am not a math wiz by any means.
(2) Given that 13n is divisible by m, then 13n = qm, or n/m = q/13, for some integer q. Since 13 is a prime number that divides qm (because 13n = qm) and 13 does not divide m (m < 13), it follows that 13 divides q. Therefore q/13 is an integer and since n/m = q/13, then n / m is an integer. Thus n is divisible by m. SUFFICIENT.
Comments: This one makes a little more sense but if anyone can dumb this explanation down that would also be appreciated. My takeaway here is that since 13 satisfies the criteria of n > 13 and if dividing by m produces an integer, this would be sufficient.
GRRR...!
