Let M be the maximum value of the product of two positive integers when their sum is 66. Let the sample space $S = \{x \in \mathbb{Z} : x(66-x) \geq \frac{5}{9}M\}$ and the event $A = \{x \in S : x \text{ is a multiple of } 3\}$. Then P(A) is equal to
Step-by-Step Solution
Key Concept: Find $M = 33 \times 33$, determine the integer range satisfying the inequality, count multiples of 3
$M = 33 \times 33 = 1089$. $x(66-x) \geq \frac{5}{9}\times 1089 = 605 \Rightarrow 11 \leq x \leq 55$. So $S = \{11,12,...,55\}$, $|S| = 45$. $A = \{12,15,18,...,54\}$: multiples of 3 from 12 to 54, count $= (54-12)/3+1 = 15$. $P(A) = 15/45 = 1/3$. Answer: (2)
Correct Answer: $\frac{1}{3}$