Let M be the maximum value of the product of two positive integers whose sum is 66.

Question

Let $M$ be the maximum value of the product of two positive integers whose sum is 66. Let $S=\{x\in\mathbb Z : x(66-x)\ge \dfrac59 M\}$ and $A=\{x\in S: x	ext{ is a multiple of }3\}$. Then $P(A)$ is:

Accepted Answer

$M = 33\times33$. Condition: $x(66-x)\ge \dfrac59(33)^2 \Rightarrow 11\le x\le55$, giving $S=\{11,\dots,55\}$, 45 elements.

$A=\{12,15,18,\dots,54\}$, which has 15 elements.

$$P(A) = \frac{15}{45}=\frac13$$

Solutions · 1