clinton wrote:Does anyone have a better way of attacking this problem. I'm not really comprehending the book's explanation:


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 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 13n students to one of m classrooms so that each classroom has the same number of students assigned to it.

We first need to sort out what the question stem is really asking. In order to "evenly" divide our n students up among m classes, it must be the case that n is some multiple of m. We can try some numbers to see the relationship: If there are 4 classrooms, then 1 student per class means n=4, 2 students per class means n=8, 3 students per class means n=12. Notice that in all cases n is a multiple of m. So the real question is the following:

Is n/m an integer?

** Now, remember that in order for n/m to be an integer, it must be the case that all of the factors of m are absorbed into n. (i.e. 8/4= 2 = integer because all of 4 was able to cancel with 8).

Statement (1): (same rephrasing steps as above) = 3n/m is an integer.

The easier thing to do here is to remember that we are in a Yes/No question and try to find numbers that allow us to answer both Yes and No to the stem. Also, don't forget the constraints on m and n. Because m is an integer between 3 and 13, we should think of values that will allow 3n/m to be divisible by m but allow both a Yes and a No answer. If we allow m to be 3, then the denominator is completely absorbed into the numerator. Then we could choose n=6, so 3(6)/3 = 6, and 6/3 = 2 (n/m is an integer). We could also choose n=5, so 3(5)/3 = 5, but 5/3 is not an integer. INSUFFICIENT

Statement (2): (same rephrasing as above) = 13n/m. Now notice that for the m to be absorbed into the numerator regardless of the value of n, it would need to be 13 because 13 is prime, but m is restricted to a number between 3 and 13 (not inclusive) so m cannot be 13. Therefore, in order to absorb the denominator m completely into the numerator n it must be the case that n contains the terms that completely absorb the denominator (i.e. if 13(17)/n, then n must be 1 or 17 for the fraction to be an integer.) Therefore n/m must be an integer. SUFFICIENT


Whit