Let the sum of two positive integers be 24 . If the probability, that their product is not less than…
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
10
9
11
8
Solution
$\begin{aligned} & x+y=24, x, y \in N \\ & A M>G M \Rightarrow x y \leq 144 \\ & x y \geq 108\end{aligned}$
Favorable pairs of $(x, y)$ are
$\begin{aligned} & (13,11),(12,12),(14,10),(15,9),(16,8), \\ & (17,7),(18,6),(6,18),(7,17),(8,16),(9,15), \\ & (10,14),(11,13)\end{aligned}$
i.e. 13 cases Total choices for $\mathrm{x}+\mathrm{y}=24$ is 23
$\begin{aligned} & \text { Probability }=\frac{13}{23}=\frac{m}{n} \\ & n-m=10\end{aligned}$