1. Key Concepts and Formulas

• Arithmetic Mean-Geometric Mean (AM-GM) Inequality: For any two non-negative real numbers $a$ and $b$, $\frac{a+b}{2} \ge \sqrt{ab}$. Equality holds if and only if $a=b$. This is used to find the maximum product for a fixed sum.
• Solving Quadratic Inequalities: A quadratic inequality of the form $(x-a)(x-b) \le 0$ (where $a<b$) is satisfied for values of $x$ such that $a \le x \le b$. This is crucial for defining the sample space.
• Counting Elements in a Set:
  • The number of integers in an inclusive range $[a, b]$ is $b - a + 1$.
  • The number of terms in an arithmetic progression with first term $a_1$, last term $a_n$, and common difference $d$ is given by $n = \frac{a_n - a_1}{d} + 1$.
• Classical Probability: The probability of an event A, $P(A)$, is given by the ratio of the number of favorable outcomes $n(A)$ to the total number of possible outcomes $n(S)$, i.e., $P(A) = \frac{n(A)}{n(S)}$.

2. Step-by-Step Solution

Step 1: Determine the maximum value M of the product.

• What we are doing: We need to find the maximum value of the product of two positive integers whose sum is 66. Let these integers be $x$ and $y$.
• Why we are doing this: The value of $M$ is a direct input to define the sample space $S$ in the next step.
• Mathematical application: We are given $x, y \in \mathbb{Z}^+$ and $x+y=66$. We want to maximize $P = xy$. Applying the AM-GM inequality to $x$ and $y$: $\frac{x+y}{2} \ge \sqrt{xy}$ Substitute $x+y=66$: $\frac{66}{2} \ge \sqrt{xy}$ $33 \ge \sqrt{xy}$ Squaring both sides (which is valid since both sides are non-negative): $(33)^2 \ge xy$ $1089 \ge xy$ The maximum possible value of the product $xy$ is 1089.
• Reasoning: The equality in the AM-GM inequality holds when $x=y$. For $x+y=66$ and $x=y$, we get $2x=66 \implies x=33$. Since $x=33$ and $y=33$ are positive integers, this condition is met. Thus, the maximum product $M = 1089$.

Step 2: Define the Sample Space S and calculate $n(S)$.

• What we are doing: We need to find the set of integers $x$ that satisfy the given inequality $x(66-x) \ge \frac{5}{9}M$, which forms our sample space $S$. Then we will count the number of elements in $S$.
• Why we are doing this: The sample space $S$ represents all possible outcomes, and its size $n(S)$ is the denominator for calculating probability.
• Mathematical application: Substitute $M=1089$ into the inequality: $x(66 - x) \ge \frac{5}{9} \cdot 1089$ Simplify the right side: $x(66 - x) \ge 5 \cdot \left(\frac{1089}{9}\right) = 5 \cdot 121 = 605$ Expand and rearrange the inequality to a standard quadratic form: $66x - x^2 \ge 605$ Move all terms to one side to make the $x^2$ coefficient positive: $0 \ge x^2 - 66x + 605$ Or, equivalently: $x^2 - 66x + 605 \le 0$ To solve this quadratic inequality, we find the roots of $x^2 - 66x + 605 = 0$. We look for two numbers that multiply to 605 and sum to -66. These numbers are -11 and -55. So, the quadratic factors as: $(x - 11)(x - 55) \le 0$
• Reasoning: For the product of two terms to be less than or equal to zero, $x$ must lie between or be equal to the roots. The roots are $x=11$ and $x=55$. Thus, the solution to the inequality is $11 \le x \le 55$. Since $x \in \mathbb{Z}$ (integers) and $x$ must be positive (as stated in the problem for the original $x$), the sample space $S$ includes all integers from 11 to 55, inclusive: $S = \{11, 12, 13, \ldots, 55\}$ To find the number of elements in $S$, $n(S)$, we use the formula for counting integers in an inclusive range: $n(S) = 55 - 11 + 1 = 44 + 1 = 45$

Step 3: Define Event A and calculate $n(A)$.

• What we are doing: We need to identify the elements within the sample space $S$ that are multiples of 3, which constitutes event $A$. Then we will count the number of elements in $A$.
• Why we are doing this: Event $A$ represents the favorable outcomes, and its size $n(A)$ is the numerator for calculating probability.
• Mathematical application: Event $A = \{ x \in S : x \text{ is a multiple of } 3 \}$. The smallest integer in $S$ is 11. The smallest multiple of 3 greater than or equal to 11 is $12$ ($3 \times 4$). The largest integer in $S$ is 55. The largest multiple of 3 less than or equal to 55 is $54$ ($3 \times 18$). So, the elements of event $A$ are: $A = \{12, 15, 18, \ldots, 54\}$ This is an arithmetic progression with first term $a_1=12$, last term $a_n=54$, and common difference $d=3$. To find the number of elements in $A$, $n(A)$, we use the formula for terms in an arithmetic progression: $n(A) = \frac{a_n - a_1}{d} + 1$ $n(A) = \frac{54 - 12}{3} + 1 = \frac{42}{3} + 1 = 14 + 1 = 15$
• Reasoning: By systematically identifying the first and last multiples of 3 within the range of $S$, we can precisely define event $A$ and accurately count its elements using the arithmetic progression formula.

Step 4: Calculate the Probability P(A).

• What we are doing: We will calculate the probability of event $A$ using the classical definition of probability.
• Why we are doing this: This is the final objective of the problem.
• Mathematical application: Using the values we found for $n(A)$ and $n(S)$: $P(A) = \frac{n(A)}{n(S)} = \frac{15}{45}$ Simplify the fraction: $P(A) = \frac{15 \div 15}{45 \div 15} = \frac{1}{3}$
• Reasoning: The probability is simply the ratio of favorable outcomes to total possible outcomes, assuming each outcome in $S$ is equally likely.

3. Common Mistakes & Tips

• AM-GM for Integers: When using AM-GM for integers, always check if the equality condition ($a=b$) yields integer values. If not (e.g., sum is odd), the maximum product for integers will occur when the integers are as close as possible (e.g., for sum 65, $x=32, y=33$). In this problem, $x=y=33$ are valid integers.
• Quadratic Inequality Direction: Be careful when manipulating quadratic inequalities. Ensure the $x^2$ term has a positive coefficient before determining the intervals of solution, or use a sign chart carefully.
• Counting in Ranges: When counting integers in an inclusive range $[a, b]$, always remember to add 1: $b-a+1$. Similarly, for arithmetic progressions, the formula is $\frac{\text{last} - \text{first}}{\text{common difference}} + 1$.

4. Summary

This problem integrates several core mathematical concepts to arrive at a probability. We began by using the AM-GM inequality to find the maximum possible product $M$ of two positive integers with a sum of 66. This value of $M$ was then used to establish a quadratic inequality, the solution of which defined the integer sample space $S$. Next, we identified the subset of $S$ consisting of multiples of 3, which formed event $A$. Finally, by calculating the number of elements in $A$ and $S$, we determined the probability $P(A)$ using the fundamental definition of probability.

The final answer is $\boxed{\frac{1}{3}}$, which corresponds to option (A).