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

Question

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

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Solution

Sure, let's break down the problem step by step.

  1. Understanding the Average:

    • The average of 100 numbers is given as zero.
    • The formula for the average of a set of numbers is the sum of the numbers divided by the count of the numbers.
    • Therefore, if the average is zero, the sum of these 100 numbers must also be zero.
  2. Sum of the Numbers:

    • Let’s denote the 100 numbers as a1,a2,a3,,a100a_1, a_2, a_3, \ldots, a_{100}.
    • The sum of these numbers is a1+a2+a3++a100=0a_1 + a_2 + a_3 + \ldots + a_{100} = 0.
  3. Positive and Negative Numbers:

    • To find out how many of these numbers can be greater than zero, we need to consider the properties of positive and negative numbers.
    • If some numbers are greater than zero, the sum of these positive numbers must be balanced by the sum of the negative numbers to ensure the total sum is zero.
  4. Maximizing Positive Numbers:

    • To maximize the number of positive numbers, we need to minimize the number of negative numbers.
    • The smallest possible negative number is a very small negative number, but for simplicity, let's consider the smallest possible negative number as 1-1.
  5. Balancing the Sum:

    • Suppose kk numbers are greater than zero and the remaining 100k100 - k numbers are less than or equal to zero.
    • To balance the sum to zero, the sum of the positive numbers must equal the absolute value of the sum of the negative numbers.
  6. Calculation:

    • If kk numbers are positive, their sum can be any positive value.
    • The remaining 100k100 - k numbers must sum to the negative of that value.
    • The smallest possible sum for the negative numbers, if each is 1-1, is (100k)-(100 - k).
  7. Setting Up the Equation:

    • Let’s denote the sum of the kk positive numbers as SS.
    • Then, the sum of the 100k100 - k negative numbers must be S-S.
    • If each of the 100k100 - k negative numbers is 1-1, then (100k)=S-(100 - k) = -S.
  8. Solving for kk:

    • S=100kS = 100 - k.
    • Since SS is the sum of kk positive numbers, the maximum value of SS occurs when each positive number is as small as possible but still positive.
    • The smallest positive number is a very small positive number, but for simplicity, let’s consider each positive number as 11.
    • Therefore, S=kS = k.
  9. Final Equation:

    • k=100kk = 100 - k.
    • Solving for kk, we get 2k=1002k = 100, so k=50k = 50.

Therefore, at most, 50 of the 100 numbers may be greater than zero.

This problem has been solved

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