A school administrator will assign each student in a group of N students to one of M classrooms. If 3...

Understanding the Question

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:

• N students total
• M classrooms available
• \(3 < \mathrm{M} < 13 < \mathrm{N}\) (so M could be 4, 5, 6, 7, 8, 9, 10, 11, or 12)

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.

Analyzing Statement 1

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.

The Key Question: Does This Guarantee N Works Too?

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

• If N = 14 students, then \(3\mathrm{N} = 42\) students
• \(42 ÷ 6 = 7\) students per room ✓ (works perfectly)
• But \(14 ÷ 6 = 2.33...\) students per room ✗ (doesn't work)
• Answer: NO, we cannot distribute N students evenly

Example 2: M = 5 classrooms

• If N = 15 students, then \(3\mathrm{N} = 45\) students
• \(45 ÷ 5 = 9\) students per room ✓ (works perfectly)
• And \(15 ÷ 5 = 3\) students per room ✓ (also works!)
• Answer: YES, we can distribute N students evenly

Why the Difference?

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.

Conclusion for Statement 1

Since we get different answers (YES in some cases, NO in others), Statement 1 is NOT sufficient.

This eliminates choices A and D.

Analyzing Statement 2

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.

The Crucial Difference: Why 13 is Special

Here's what makes 13 different from 3:

• 13 is a prime number (only divisible by 1 and itself)
• 13 is larger than any possible value of M (remember \(\mathrm{M} < 13\))
• Therefore, 13 shares no common factors with any possible M value

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.

The Logic Made Simple

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.

Conclusion for Statement 2

This gives us a definitive YES answer—we can assign the students equally.

[STOP - Sufficient!] Statement 2 is sufficient.

The Answer: B

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."