If k, m, and t are positive integers and k/6 + m/4 = t/12, do t and 12 have a...

Understanding the Question

We need to determine whether t and 12 have a common factor greater than 1.

Let's first simplify the given equation. We have \(\frac{\mathrm{k}}{6} + \frac{\mathrm{m}}{4} = \frac{\mathrm{t}}{12}\). Converting to a common denominator of 12:

• \(\frac{\mathrm{k}}{6} = \frac{2\mathrm{k}}{12}\)
• \(\frac{\mathrm{m}}{4} = \frac{3\mathrm{m}}{12}\)

So we get: \(\frac{2\mathrm{k}}{12} + \frac{3\mathrm{m}}{12} = \frac{\mathrm{t}}{12}\)

This gives us: t = 2k + 3m

What We Need to Determine

The question asks if t and 12 share a common factor greater than 1. Since \(12 = 2^2 \times 3\), this happens if and only if t is divisible by 2 or 3 (or both).

For this yes/no question, "sufficient" means we can definitively answer either YES (they do share a common factor > 1) or NO (they don't).

Key Insight

Since \(\mathrm{t} = 2\mathrm{k} + 3\mathrm{m}\), we need to understand when this sum inherits divisibility properties:

• For t to be divisible by 3: Since 3m is always divisible by 3, we need 2k to also be divisible by 3
• For t to be divisible by 2: Since 2k is always even, we need 3m to also be even

Analyzing Statement 1

Statement 1 tells us: k is a multiple of 3

This means \(\mathrm{k} = 3\mathrm{n}\) for some positive integer n.

Let's see what this tells us about t:

• \(\mathrm{t} = 2\mathrm{k} + 3\mathrm{m} = 2(3\mathrm{n}) + 3\mathrm{m} = 6\mathrm{n} + 3\mathrm{m} = 3(2\mathrm{n} + \mathrm{m})\)

Since \(\mathrm{t} = 3(2\mathrm{n} + \mathrm{m})\), we can see that t is definitely divisible by 3.

Therefore, t and 12 share the common factor 3, which is greater than 1.

The answer to our question is definitively YES.

[STOP - Statement 1 is SUFFICIENT!]

This eliminates choices B, C, and E.

Analyzing Statement 2

Now let's forget Statement 1 completely and analyze Statement 2 independently.

Statement 2 tells us: m is a multiple of 3

This means \(\mathrm{m} = 3\mathrm{p}\) for some positive integer p.

Now \(\mathrm{t} = 2\mathrm{k} + 3\mathrm{m} = 2\mathrm{k} + 3(3\mathrm{p}) = 2\mathrm{k} + 9\mathrm{p}\)

For t to be divisible by 3, we need the entire expression \(2\mathrm{k} + 9\mathrm{p}\) to be divisible by 3. Since \(9\mathrm{p}\) is always divisible by 3, we need \(2\mathrm{k}\) to also be divisible by 3. This happens only when k is divisible by 3.

But Statement 2 tells us nothing about k!

Testing Different Scenarios

Let's verify with specific values:

• If k = 1, m = 3: Then \(\mathrm{t} = 2(1) + 3(3) = 2 + 9 = 11\)
  • Is 11 divisible by 2 or 3? No
  • So \(\mathrm{gcd}(11, 12) = 1\)
  • Answer: NO
• If k = 3, m = 3: Then \(\mathrm{t} = 2(3) + 3(3) = 6 + 9 = 15\)
  • Is 15 divisible by 3? Yes \((15 = 3 \times 5)\)
  • So \(\mathrm{gcd}(15, 12) = 3 > 1\)
  • Answer: YES

Since we get different answers (NO when k = 1, YES when k = 3), we cannot determine a definitive answer.

Statement 2 is NOT sufficient.

This eliminates choices B and D.

The Answer: A

Statement 1 alone tells us that t is divisible by 3, which means t and 12 definitely share a common factor greater than 1. Statement 2 alone leaves us uncertain because the answer depends on whether k is divisible by 3.

Answer Choice A: "Statement 1 alone is sufficient, but Statement 2 alone is not sufficient."