If the average of n consecutive integers equal to 1?
(1) n is even.
(2) If S is the sum of the n consecutive integers, then `0ltSltn`

Answer

To solve the problem of whether the average of \( n \) consecutive integers can equal 1, we will analyze the two statements provided.

### Step-by-Step Solution:

1. **Understanding the Average of Consecutive Integers**:
The average of \( n \) consecutive integers can be expressed as:
\[ \text{Average} = \frac{S}{n} \]
where \( S \) is the sum of the \( n \) consecutive integers.

2. **Statement 1: \( n \) is even**:
- If \( n \) is even, then the integers can be represented as:
\[ x, x+1, x+2, \ldots, x+n-1 \]
- The sum \( S \) of these integers is:
\[ S = nx + \frac{(n-1)n}{2} \]
- The average becomes:
\[ \text{Average} = \frac{S}{n} = x + \frac{(n-1)}{2} \]
- Since \( n \) is even, \( \frac{(n-1)}{2} \) is a half-integer, making the average \( x + \text{half-integer} \), which cannot equal 1 (an integer) for any integer \( x \).
- **Conclusion**: If \( n \) is even, the average cannot equal 1. Thus, Statement 1 is sufficient.

3. **Statement 2: \( 0 < S < n \)**:
- The average can also be expressed as:
\[ \text{Average} = \frac{S}{n} \]
- Given that \( S < n \), we can substitute:
\[ \text{Average} < \frac{n}{n} = 1 \]
- Since \( S > 0 \), we can also say:
\[ \text{Average} > 0 \]
- Therefore, the average must be greater than 0 but less than 1, which means it cannot equal 1.
- **Conclusion**: Statement 2 is also sufficient to conclude that the average cannot equal 1.

### Final Conclusion:
Both statements are sufficient to determine that the average of \( n \) consecutive integers cannot equal 1.