Question

Let $M$ be the maximum value of the product of two positive integers when their sum is 66. Let the sample space $S =\left\{x \in Z : x(66-x) \geq \frac{5}{9} M\right\}$ and the event $A =\{x \in S : x$ is a multiple of 3$\}$. Then $P ( A )$ is equal to

• A. $\frac{15}{44}$
• B. $\frac{1}{3}$
• C. $\frac{7}{22}$
• D. $\frac{1}{5}$

Answer 1

The Correct Option is B

$M=33\times 33$
$x(66-x)\ge \frac{5}{9}\times 33\times 33$
$11\le x\le 55$
$A:\{12,15,18,\dots 54\}$
$P(A)=\frac{15}{45}=\frac{1}{3}$

Answer 2

1. The product of two numbers whose sum is \(66\) is given by: \[ P = x(66 - x). \] The maximum product occurs when \(x = 33\), i.e.: \[ M = 33 \cdot 33 = 1089. \]
2. The condition \((66 - x)x \geq \frac{5}{9}M\) becomes: \[ (66 - x)x \geq \frac{5}{9} \cdot 1089 = 605. \]
3. Rewrite as: \[ x^2 - 66x + 605 \leq 0. \] Solve the quadratic equation: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}, \quad a = 1, b = -66, c = 605. \] \[ x = \frac{66 \pm \sqrt{66^2 - 4 \cdot 605}}{2}. \] \[ x = \frac{66 \pm \sqrt{4356 - 2420}}{2} = \frac{66 \pm \sqrt{1936}}{2}. \] \[ x = \frac{66 \pm 44}{2}. \] \[ x = 55 \quad \text{or} \quad x = 11. \]
4. Therefore, \(x \in [11, 55]\). The total number of integers in \(S\) is: \[ 55 - 11 + 1 = 45. \]
5. The multiples of \(3\) in this range are: \[ 12, 15, 18, \ldots, 54. \] This is an arithmetic sequence with first term \(12\), last term \(54\), and common difference \(3\). The total terms are: \[ n = \frac{54 - 12}{3} + 1 = 15. \]
6. The probability \(P(A)\) is: \[ P(A) = \frac{\text{Number of favorable outcomes}}{\text{Total outcomes}} = \frac{15}{45} = \frac{1}{3}. \]
Thus, \(P(A) = \frac{1}{3}\). The maximum product of two numbers occurs when they are equal or close to equal. For the probability calculation, count the multiples of \(3\) within the specified range.