The mean of six positive integers is 16. The range of the six numbers is 5. When two numbers are removed, the mean of the remaining four numbers is 14.5. Based on the given information, what is the greatest possible number from the original six numbers?

Let's solve the problem step by step.

### Step 1: Understanding the problem

We have six positive integers, and their mean is given as 16. This means:
\[
\text{Sum of the six numbers} = 6 \times 16 = 96
\]

The range of the six numbers is 5, so the difference between the maximum and minimum numbers is 5.

When two numbers are removed, the mean of the remaining four numbers is 14.5. Therefore:
\[
\text{Sum of the remaining four numbers} = 4 \times 14.5 = 58
\]
This means the sum of the two removed numbers is:
\[
96 - 58 = 38
\]

### Step 2: Exploring the range constraint

Let's denote the six numbers as \( a_1, a_2, a_3, a_4, a_5, a_6 \) where \( a_1 \leq a_2 \leq a_3 \leq a_4 \leq a_5 \leq a_6 \). The range condition implies:
\[
a_6 - a_1 = 5
\]

### Step 3: Maximizing the greatest possible number

We want to maximize \( a_6 \). Since the sum of all six numbers is 96, and we need to remove two numbers with a sum of 38, we should consider which numbers can be removed to keep \( a_6 \) as large as possible.

To achieve the highest \( a_6 \), ideally, \( a_6 \) should not be removed. Therefore, the pair \( (a_1, a_5) \), \( (a_2, a_4) \), or \( (a_3, a_2) \) might be candidates for removal.

### Step 4: Analyzing possible cases

#### Case 1: Removing \( a_1 \) and \( a_5 \)
- If \( a_6 = 16 \) and \( a_1 = 11 \) (since the range is 5), the sum of \( a_5 \) and \( a_1 \) needs to be 38. This implies \( a_5 = 27 \), which is impossible since \( a_6 = 16 \).

#### Case 2: Removing \( a_1 \) and \( a_4 \)
- Let \( a_1 = a_6 - 5 \), and \( a_6 \) be the largest number. If we remove \( a_1 = 11 \) and \( a_4 = 27 \), we have the remaining sum of 58 for \( a_2, a_3, a_5, \) and \( a_6 \). This case does not work since the mean would not hold.

#### Case 3: Removing \( a_2 \) and \( a_6 \)
- \( a_6 = 19 \) leads to \( a_2 = 38 - 19 = 19 \), which isn't allowed.

### Conclusion:
The optimal scenario to maximize \( a_6 \) would be removing the smallest and one of the larger middle values. If \( a_6 = 19 \), then removing \( a_1 = 14 \) and \( a_4 = 24 \) can satisfy the condition, making it work.

**Answer**: The greatest possible number \( a_6 \) can be **19**.

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Would you like further explanation on any step? Here are some related questions to consider:

1. What are the possible values for the removed numbers?
2. What is the sum of the smallest and largest numbers?
3. How would the scenario change if one of the removed numbers was the largest?
4. Can the remaining four numbers have a different mean? 
5. How does the condition of positive integers impact the solution?

**Tip**: When working with constraints, consider different cases based on range, sum, and conditions given to explore all possibilities effectively.

Learn how to maximize the greatest possible number from a set of six positive integers with specific constraints. This problem involves calculating means, ranges, and sums of integers, suitable for Grades 10-12 students.

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