If x and y are positive integers, is x an even integer? \(\mathrm{x(y + 5)}\) is an even integer. 6y^2...

Understanding the Question

We need to determine whether x is even, given that x and y are positive integers.

What We Need to Determine

The question asks: "Is x an even integer?" This is a yes/no question, which means we need sufficient information to answer either:

• "Yes, x is definitely even," OR
• "No, x is definitely odd"

Given Information

• x and y are positive integers (\(\mathrm{x, y > 0 \text{ and } x, y ∈ ℤ⁺}\))
• We need to determine if x is even (can be written as \(\mathrm{x = 2k}\) for some integer k)

Key Insight

For a product of integers to be even, at least one factor must be even. This principle will guide our analysis of how the statements constrain x's parity.

Analyzing Statement 1

Statement 1: \(\mathrm{x(y + 5)}\) is an even integer.

Since we have a product that equals an even number, at least one factor must be even. This gives us two possibilities:

• Case 1: x is even (then the product is even regardless of whether y + 5 is even or odd)
• Case 2: x is odd AND y + 5 is even

Let's test concrete examples to see if we can determine x's parity:

Example 1: Let \(\mathrm{x = 2}\) (even) and \(\mathrm{y = 1}\) (odd)

• Then \(\mathrm{y + 5 = 6}\) (even)
• So \(\mathrm{x(y + 5) = 2 × 6 = 12}\) (even) ✓

Example 2: Let \(\mathrm{x = 1}\) (odd) and \(\mathrm{y = 1}\) (odd)

• Then \(\mathrm{y + 5 = 6}\) (even)
• So \(\mathrm{x(y + 5) = 1 × 6 = 6}\) (even) ✓

Both examples satisfy Statement 1, but they give us different answers about whether x is even. In Example 1, x is even; in Example 2, x is odd.

Therefore, Statement 1 alone 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: \(\mathrm{6y² + 41y + 25}\) is an even integer.

Let's analyze this expression by examining the parity of each term:

• \(\mathrm{6y²}\) is always even (since 6 is even, and even × anything = even)
• \(\mathrm{41y}\) has the same parity as y (since 41 is odd: odd × odd = odd, odd × even = even)
• \(\mathrm{25}\) is odd

So our expression becomes: even + (parity of y) + odd

For the sum to be even, we need an even total number of odd terms:

• If y is odd: even + odd + odd = even ✓ (two odd terms)
• If y is even: even + even + odd = odd ✗ (one odd term)

This means Statement 2 tells us that y must be odd. However, knowing that y is odd doesn't tell us anything about whether x is even or odd.

Therefore, Statement 2 alone is NOT sufficient.

This eliminates choice B.

Combining Both Statements

From Statement 1: \(\mathrm{x(y + 5)}\) is even
From Statement 2: y is odd

Since y is odd, we know that y + 5 is even (odd + odd = even).

So Statement 1 becomes: x × (even number) = even

This equation is satisfied whether x is even or odd:

• If x is even: even × even = even ✓
• If x is odd: odd × even = even ✓

Let's verify with concrete examples:

• \(\mathrm{x = 2}\) (even), \(\mathrm{y = 3}\) (odd): \(\mathrm{x(y + 5) = 2 × 8 = 16}\) (even) ✓
• \(\mathrm{x = 3}\) (odd), \(\mathrm{y = 3}\) (odd): \(\mathrm{x(y + 5) = 3 × 8 = 24}\) (even) ✓

Even with both statements combined, we cannot determine whether x is even or odd. Different values of x (some even, some odd) can satisfy both conditions.

Therefore, both statements together are NOT sufficient.

This eliminates choice C.

The Answer: E

Neither statement alone nor both statements together provide sufficient information to determine whether x is even.

Answer Choice E: "The statements together are not sufficient."