For each positive integer k, let \(\mathrm{a_k} = (1 + \frac{1}{\mathrm{k}+1})\). Is the product a_1a_2 ldots a_n an integer? n...

Understanding the Question

We need to determine if the product \(\mathrm{a_1 \cdot a_2 \cdot ... \cdot a_n}\) is an integer, where each \(\mathrm{a_k = 1 + \frac{1}{k+1}}\).

Let's first simplify \(\mathrm{a_k}\). We can rewrite it as:

\(\mathrm{a_k = \frac{k+1+1}{k+1} = \frac{k+2}{k+1}}\)

So our product becomes:

\(\mathrm{a_1 \cdot a_2 \cdot a_3 \cdot ... \cdot a_n = \frac{3}{2} \cdot \frac{4}{3} \cdot \frac{5}{4} \cdot ... \cdot \frac{n+2}{n+1}}\)

Key insight: This is a telescoping product! Watch what happens when we multiply:

• The 3 in the numerator of \(\frac{3}{2}\) cancels with the 3 in the denominator of \(\frac{4}{3}\)
• The 4 in the numerator of \(\frac{4}{3}\) cancels with the 4 in the denominator of \(\frac{5}{4}\)
• This pattern continues...

After all the cancellations, we're left with just: \(\frac{n+2}{2}\)

Therefore, our question simplifies to: Is \(\frac{n+2}{2}\) an integer?

This fraction is an integer when \(n+2\) is divisible by 2, which means \(n\) must be even.

Analyzing Statement 1

Statement 1 tells us that \(n + 1\) is a multiple of 3.

This means \(n + 1 = 3k\) for some positive integer \(k\), so \(n = 3k - 1\).

Let's check if this guarantees that \(n\) is even:

• If \(k = 1\): \(n = 3(1) - 1 = 2\) (even) → \(\frac{n+2}{2} = \frac{4}{2} = 2\) ✓ Integer
• If \(k = 2\): \(n = 3(2) - 1 = 5\) (odd) → \(\frac{n+2}{2} = \frac{7}{2} = 3.5\) ✗ Not an integer

Since we get different answers (sometimes yes, sometimes no), Statement 1 is NOT sufficient.

[STOP - Not Sufficient!] This eliminates choices A and D.

Analyzing Statement 2

We now analyze Statement 2 independently, forgetting Statement 1 completely.

Statement 2 tells us that \(n\) is a multiple of 2.

This means \(n = 2m\) for some positive integer \(m\), so \(n\) is even.

Since \(n\) is even, we can write \(n + 2 = 2m + 2 = 2(m + 1)\).

Therefore:

\(\frac{n+2}{2} = \frac{2(m+1)}{2} = m + 1\)

Since \(m\) is a positive integer, \(m + 1\) is also a positive integer.

This means whenever \(n\) is even, our product is guaranteed to be an integer.

Statement 2 is sufficient.

[STOP - Sufficient!] This eliminates choices C and E.

The Answer: B

We've determined that:

• Statement 1 alone is NOT sufficient (allows both even and odd values of \(n\))
• Statement 2 alone IS sufficient (guarantees \(n\) is even, making the product equal to the integer \(\frac{n+2}{2}\))

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