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OG problem

by iridebikes » Fri Nov 19, 2010 8:49 am
OG 12th edition #128 DS.

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.

I have the official guide explanation but I am still not understanding something. It says that 1 is INSUFFICIENT. If n=20 and m=6, n/m= 20/6= 10/3. 3n can cancel all the factors of m, but 20 can't. I don't understand why its necessary to look at if n/m are divisible by each other, if 3n is divisible by m because that is what the question is asking. Hopefully someone can clarify this for me. Thanks guys.
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by clawhammer » Fri Nov 19, 2010 9:31 am
Question asks you to divide all students (n) equally in all (m) classrooms. In other words it simply asks if n/m = integer.

3n can be divisible by m, but that does NOT guarantee that n is divisible by m.

For example: If you had 16 students, and 6 classrooms, will you be able to divide them evenly in all classrooms? 16/6 = not integer so no. But what about 3n/m or 3*16/6 ? Yes, this is divisible by m. So you see, even when 3n can be divisible by m, it doesn't necessarily mean n is. The reason for this is, you need to ensure ALL the factors of m can be taken out by n. In the above example, m had two primes: 3 and 2. Now when you have 3n, you can't be sure if the 3 (from m) is being canceled out by a 3 from n or by the 3 that has been externally multiplied. So (A) alone isn't sufficient.

Come to (B): 13n/m = integer. Now notice you got yourself a bonus because 13 is a prime. So in order for 13n to be divisible by m, you "could" have 13 in m right? But m<13, so there's no way you could cancel out or divide that 13. In order to make 13n divisible by m, you therefore MUST be able to divide the n, or, in other words, have primes in m that can all be taken out by n alone. That means n alone can be divided by m (all factors of m is also a factor of n). So this helps us being sure n/m = integer.
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by iridebikes » Sat Nov 20, 2010 6:19 pm
Thank you I appreciate your response I definitely understand it now. Its weird how it takes a bit for something to click sometimes but you definitely cleared everything up for me .
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by clawhammer » Sun Nov 21, 2010 12:38 am
No worries bro. I get stuck at much easier problems, happens all the time.
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