If the average of 100 numbers is zero. Of them, at the most, how many may be greater than zero? Options 100 0 99 1

To solve this problem, let's break it down step by step:

1. Understanding the average: The average of 100 numbers is given as zero. This means the sum of all 100 numbers divided by 100 is zero.

2. Sum of the numbers: If the average is zero, the sum of all 100 numbers must be zero. Mathematically, if $x_1, x_2, \ldots, x_{100}$ are the 100 numbers, then: $\frac{x_1 + x_2 + \cdots + x_{100}}{100} = 0$ This implies: $x_1 + x_2 + \cdots + x_{100} = 0$

3. Positive and negative numbers: To have a sum of zero, the positive numbers must be balanced by the negative numbers.

4. Maximizing positive numbers: To find the maximum number of positive numbers, we need to consider that the sum of the positive numbers must be balanced by the sum of the negative numbers.

5. All but one positive: If we have 99 positive numbers, their sum can be any positive value. To balance this, we need one negative number whose value is the negative of the sum of the 99 positive numbers. This way, the total sum will still be zero.

6. Conclusion: Therefore, at most, 99 numbers can be greater than zero.

So, the correct answer is: 99