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inconsistent OG12 logic DS # 128

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inconsistent OG12 logic DS # 128

by c210 » Thu Apr 26, 2012 9:23 pm
I understand the reasoning for (2) being sufficient but can someone explain why (1) is insufficient?

prob #129

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.

(1) - FROM OG
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 = 9, so 3n is divisible by m); NOT sufficient


I don't get this logic since n cannot equal 12?? and n cannot equal 9 either?( both due to 3<m<13<n )

Also, shouldn't N be divisible by M because the when factoring 3N, 3 is not divisible by any number between 4 and 12 ( again due to 3<m<13<n) and therefore N should be divisible by M to make 13n/m an integer?

best,

c
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by shantanu86 » Thu Apr 26, 2012 11:02 pm
c210 wrote:I understand the reasoning for (2) being sufficient but can someone explain why (1) is insufficient?

prob #129

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.

(1) - FROM OG
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 = 9, so 3n is divisible by m); NOT sufficient


I don't get this logic since n cannot equal 12?? and n cannot equal 9 either?( both due to 3<m<13<n )

Also, shouldn't N be divisible by M because the when factoring 3N, 3 is not divisible by any number between 4 and 12 ( again due to 3<m<13<n) and therefore N should be divisible by M to make 13n/m an integer?

best,

c
Hi c210,

This question can be broken down to a simpler question-

Is n divisible by m given the constrain 3<m<13<n

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

from 1) you are completely right that no m can exist that divides 3, but the point you are missing is what if 3 and m have a common factor.. and in that case even if 3n is divisible by m, n may not be.

Example-

Let n = 3*i ,where i doesn't have 3 as factor.
Let m = 3*3

Now, 3n = 3*3*i (3n) will be divisible by 3*3 (m) but 3*i (n) will not be divisible by (n)
eg if you take i=7
n =21 , m=9

from 2) we can be sure that m and 13 can not have common factors as m<13 and 13 is a prime number.
Thus, 2 alone is sufficient.

Hope it helps!!

PS: Please put answer/solution under spoiler when you post questions.
If you feel like it, hit thanks :)
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