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 $\mathrm{P}(\mathrm{A})$ is equal to

JEE Main · 2023 · 25 Jan 2023 · Shift 1 · Probability

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 Z : x (66 - x) \geq \frac{5}{9} M}

and the event

A = {x \in S : x

is a multiple of 3

}

. Then

P (A)

is equal to

Answer: (B)

The product of two positive integers of fixed sum

66

is greatest when both equal

33

, so

M = 1089

. The condition

x (66 - x) \geq \frac{5}{9} \cdot 1089 = 605

gives

x^{2} - 66 x + 605 \leq 0

, i.e.

11 \leq x \leq 55

, so

S

has

45

integers. The multiples of

3

among them are

12, 15, \dots, 54

, which number

15

. Hence

P (A) = \frac{15}{45} = \frac{1}{3}

JEE Main · 2023 · 25 Jan 2023 · Shift 1 · Single correct option