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 possible product, is $\frac{m}{n}$, where $\operatorname{gcd}(m, n)=1$, then $n-m$ equals

[JEE Main 2024, 8 Apr (Shift 1)]

Question

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 possible product, is $\frac{m}{n}$, where $\operatorname{gcd}(m, n)=1$, then $n-m$ equals

[JEE Main 2024, 8 Apr (Shift 1)]

YESPYQ · Accepted Answer

Correct answer: b) 10. $ \mathrm{x}+\mathrm{y}=24, \mathrm{x}, \mathrm{y} \in \mathrm{~N} $ $ \mathrm{AM}>\mathrm{GM} \Rightarrow \mathrm{xy} \leq 144 $ $ x y \geq 108$ Favorable pairs of $(x, y)$ are ${(13,11),(12,12),(14,10),(15,9),(16,8),(17,7),\ (18,6),(6,18),(7,17),(8,16),(9,15),(10,14),(11,13)}$ i.e. $13$ cases Total choices for $\mathrm{x}+\mathrm{y}=24$ is $23$ Probability $\frac{13}{23}=\frac{m}{n}$ $n-m=10$

Let the sum of two positive integers be \(24.\) If the probability, that their product is not less than \(\frac{3}{4}\) …

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