If c and d are integers, is `c-3d` even?
(1) c and d are odd.
(2) `c-2d` is odd.

Answer

To determine whether \( c - 3d \) is even, we will analyze the two statements provided.

### Step 1: Analyze Statement 1
**Statement 1:** \( c \) and \( d \) are both odd integers.

- An odd integer can be represented as \( 2k + 1 \) for some integer \( k \).
- Let \( c = 2m + 1 \) and \( d = 2n + 1 \) for integers \( m \) and \( n \).

Now, we can substitute these into the expression \( c - 3d \):

\[ c - 3d = (2m + 1) - 3(2n + 1) \]
\[ = (2m + 1) - (6n + 3) \]
\[ = 2m + 1 - 6n - 3 \]
\[ = 2m - 6n - 2 \]
\[ = 2(m - 3n - 1) \]

Since \( 2(m - 3n - 1) \) is clearly a multiple of 2, \( c - 3d \) is even.

**Conclusion for Statement 1:** This statement is sufficient to conclude that \( c - 3d \) is even.

### Step 2: Analyze Statement 2
**Statement 2:** \( c - 2d \) is odd.

From this statement, we know that:

\[ c - 2d \text{ is odd.} \]

Since \( 2d \) is always even (as it is an integer multiplied by 2), for \( c - 2d \) to be odd, \( c \) must be odd. Thus, we can conclude that \( c \) is odd.

However, we do not have any information about \( d \). It could be either odd or even.

- If \( d \) is odd (let's say \( d = 2n + 1 \)):
\[ c - 3d = c - 3(2n + 1) = c - 6n - 3 \]
Since \( c \) is odd, \( c - 6n - 3 \) will be even (odd minus odd).

- If \( d \) is even (let's say \( d = 2n \)):
\[ c - 3d = c - 3(2n) = c - 6n \]
Since \( c \) is odd, \( c - 6n \) will also be odd (odd minus even).

Thus, depending on whether \( d \) is odd or even, \( c - 3d \) could be either even or odd.

**Conclusion for Statement 2:** This statement is not sufficient to conclude whether \( c - 3d \) is even.

### Final Conclusion
- **Statement 1 alone is sufficient.**
- **Statement 2 alone is not sufficient.**

The answer is that only Statement 1 is sufficient to determine whether \( c - 3d \) is even.