Let the sum of two positive integers be 24. I | vedclass

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(D) Let the two positive integers be $x$ and $y$. Given $x+y=24$,where $x, y \in \mathbb{N}$.The product $P = xy$. By the $AM-GM$ inequality,$\frac{x+

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$10$

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(D) Let the two positive integers be $x$ and $y$. Given $x+y=24$,where $x, y \in \mathbb{N}$.The product $P = xy$. By the $AM-GM$ inequality,$\frac{x+y}{2} \geq \sqrt{xy}$,so $\sqrt{xy} \leq 12$,which implies $xy \leq 144$. The greatest positive product is $144$ (when $x=12, y=12$).We need the probability that $xy \geq \frac{3}{4} 	imes 144$,which means $xy \geq 108$.Since $y = 24-x$,we have $x(24-x) \geq 108$,or $24x - x^2 \geq 108$,which simplifies to $x^2 - 24x + 108 \leq 0$.Solving $x^2 - 24x + 108 = 0$ using the quadratic formula: $x = \frac{24 \pm \sqrt{576 - 432}}{2} = \frac{24 \pm \sqrt{144}}{2} = \frac{24 \pm 12}{2}$.Thus,$x = 6$ or $x = 18$. The inequality holds for $6 \leq x \leq 18$.The possible values for $x$ are ${6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18}$.There are $18 - 6 + 1 = 13$ such values.The total number of pairs $(x, y)$ such that $x+y=24$ for $x, y \in \mathbb{N}$ is $23$ (since $x$ can range from $1$ to $23$).The probability is $\frac{13}{23} = \frac{m}{n}$.Thus,$m=13$ and $n=23$. Then $n-m = 23-13 = 10$.

Let the sum of two positive integers be $24$. If the probability,that their product is not less than $\frac{3}{4}$ times their greatest positive product,is $\frac{m}{n}$,where $\operatorname{gcd}(m, n)=1$,then $n-m$ equals :

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