Problem Setup

We are given 6 positive integers. Let these integers be $x_1, x_2, x_3, x_4, x_5, x_6$, ordered such that $x_1 \le x_2 \le x_3 \le x_4 \le x_5 \le x_6$.

Key information provided:

• The average of the 6 integers is 45.
• The difference between the largest ($x_6$) and the smallest ($x_1$) is 18. So, $x_6 - x_1 = 18$.
• All integers are positive, meaning $x_i \ge 1$ for all $i$.

The goal is to find the maximum possible value for the largest integer, $x_6$.

Calculate Sum of Integers

The average is calculated as the sum of numbers divided by the count of numbers. $ \text{Average} = \frac{\text{Sum}}{N} $ Here, Average = 45 and N = 6. $ \text{Sum} = \text{Average} \times N $ $ \text{Sum} = 45 \times 6 = 270 $ So, $x_1 + x_2 + x_3 + x_4 + x_5 + x_6 = 270$.

Formulate Equations and Constraints

We have the following equations and inequalities:

1. $x_1 + x_2 + x_3 + x_4 + x_5 + x_6 = 270$
2. $x_6 - x_1 = 18 \implies x_1 = x_6 - 18$
3. $x_1 \ge 1$ (positive integers)
4. $x_1 \le x_2 \le x_3 \le x_4 \le x_5 \le x_6$

Strategy for Maximizing the Largest Integer

To maximize $x_6$, we need to minimize the sum of the other integers ($x_1 + x_2 + x_3 + x_4 + x_5$), while satisfying all the constraints.

From constraint (2), $x_1 = x_6 - 18$. Substitute this into the sum equation (1): $ (x_6 - 18) + x_2 + x_3 + x_4 + x_5 + x_6 = 270 $ $ 2x_6 + x_2 + x_3 + x_4 + x_5 = 270 + 18 $ $ 2x_6 + x_2 + x_3 + x_4 + x_5 = 288 $ To maximize $x_6$, we need to minimize the term $x_2 + x_3 + x_4 + x_5$. According to constraint (4), the smallest possible values for $x_2, x_3, x_4, x_5$ occur when they are equal to the smallest number, $x_1$. So, we set $x_2 = x_3 = x_4 = x_5 = x_1$. This means $x_1 = x_2 = x_3 = x_4 = x_5$. The minimum value for this set is $x_1 = x_6 - 18$. Therefore, we set $x_2 = x_3 = x_4 = x_5 = x_6 - 18$.

Solve for the Largest Integer

Substitute $x_1 = x_2 = x_3 = x_4 = x_5 = x_6 - 18$ into the sum equation:

Verification

Let's check if this value satisfies all conditions:

• If $x_6 = 60$, then $x_1 = x_6 - 18 = 60 - 18 = 42$.
• The set of integers becomes $42, 42, 42, 42, 42, 60$.
• Are they positive integers? Yes, 42 and 60 are positive integers.
• Is the average 45? Sum = $5 \times 42 + 60 = 210 + 60 = 270$. Average = $270 / 6 = 45$. Yes.
• Is the difference between the largest and smallest 18? $60 - 42 = 18$. Yes.
• Are the integers ordered correctly? $42 \le 42 \le 42 \le 42 \le 42 \le 60$. Yes.

All conditions are met. The maximum possible value for the largest integer ($x_6$) is 60.