Loading...
A school administrator will assign each student in a group of \(\mathrm{N}\) students to one of \(\mathrm{M}\) classrooms. If \(3 < \mathrm{M} < 13 < \mathrm{N}\), is it possible to assign each of the \(\mathrm{N}\) students to one of the \(\mathrm{M}\) classrooms so that each classroom has the same number of students assigned to it?
Let's break down what we're asking: Can we distribute N students equally among M classrooms with no one left over?
In simpler terms, we need to know if N divided by M gives us a whole number. If it does, then yes—each classroom gets the same number of students. If not, then some classrooms would have to have more students than others.
Given information:
What "sufficient" means here: We need to definitively answer YES or NO to whether N divides evenly by M. Either answer works—we just need to be certain.
Statement 1 tells us: We can evenly distribute \(3\mathrm{N}\) students among M classrooms.
This means when we triple our student body to \(3\mathrm{N}\), they fit perfectly into M classrooms with the same number in each room.
Here's where it gets tricky. Just because 3 times something divides evenly doesn't guarantee the original number does! Let's see why:
Example 1: M = 6 classrooms
Example 2: M = 5 classrooms
The key insight: When M and 3 share a common factor (like 6 and 3 both being divisible by 3), multiplying by 3 can "help" make the division work even when N alone doesn't divide evenly. But when M and 3 share no common factors (like 5 and 3), then if \(3\mathrm{N}\) works, N must work too.
Since we get different answers (YES in some cases, NO in others), Statement 1 is NOT sufficient.
This eliminates choices A and D.
Now let's forget Statement 1 completely and analyze Statement 2 independently.
Statement 2 tells us: We can evenly distribute \(13\mathrm{N}\) students among M classrooms.
This means when we multiply our students by 13, they fit perfectly into M classrooms.
Here's what makes 13 different from 3:
Think of it this way: If 13 groups of N students can be arranged perfectly into M classrooms, and 13 doesn't "help" the division in any way (no shared factors), then 1 group of N students must also arrange perfectly.
Imagine you have 13 identical boxes, each containing N items. If these \(13\mathrm{N}\) items can be distributed evenly among M rooms, and 13 has nothing in common with M, then the items in just one box (N items) must also distribute evenly among the M rooms.
This gives us a definitive YES answer—we can assign the students equally.
[STOP - Sufficient!] Statement 2 is sufficient.
Statement 2 alone is sufficient because knowing that \(13\mathrm{N}\) divides evenly by M guarantees that N divides evenly by M (since 13 and M share no common factors).
Statement 1 alone is not sufficient because 3 can share common factors with M, leading to different possible answers.
Answer Choice B: "Statement 2 alone is sufficient, but Statement 1 alone is not sufficient."