The average of 39 numbers is zero. Out of then, how many may be greater than zero, at the most?

Options

A. 0
B. 38
C. 20
D. 39

Answer

To solve the problem, we need to determine how many of the 39 numbers can be greater than zero while still maintaining an average of zero.

### Step-by-Step Solution:

1. **Understanding Average**:
The average of a set of numbers is calculated by dividing the sum of the numbers by the total count of the numbers. In this case, the average of 39 numbers is given as zero.

2. **Setting Up the Equation**:
Let’s denote the 39 numbers as \( x_1, x_2, x_3, \ldots, x_{39} \). The average is given by:
\[ \text{Average} = \frac{x_1 + x_2 + x_3 + \ldots + x_{39}}{39} = 0 \]
This implies:
\[ x_1 + x_2 + x_3 + \ldots + x_{39} = 0 \]

3. **Considering Positive and Negative Numbers**:
Since the sum of all numbers is zero, if some numbers are positive, there must be enough negative numbers to balance them out.

4. **Maximizing Positive Numbers**:
To find the maximum number of positive numbers, we can assume that \( k \) numbers are greater than zero and the remaining \( 39 - k \) numbers are negative or zero.

5. **Finding the Maximum \( k \)**:
If we want to maximize \( k \), we can set \( k = 38 \). This means we can have 38 positive numbers. For the sum to remain zero, the last number (the 39th number) must be negative and equal to the sum of the first 38 numbers.

For example, if we take:
\[ x_1 = 1, x_2 = 1, \ldots, x_{38} = 1 \quad (\text{38 times}) \]
Then:
\[ x_{39} = -38 \]
The sum becomes:
\[ 1 + 1 + \ldots + 1 - 38 = 0 \]

6. **Conclusion**:
Therefore, the maximum number of numbers that can be greater than zero is 38.

### Final Answer:
The maximum number of numbers that can be greater than zero is **38**.

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