Let the sum of two positive integers be 24 .

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

Accepted Answer

Take two numbers as $a$ and $b$

$a+b=24$

For product to be maximum

$$\begin{aligned}
& \frac{a+b}{2} \geq \sqrt{a b} \
& 144>a b
\end{aligned}$$

Maximum product is 144

Now, $$a b \geq \frac{3}{4} \cdot 144=108$$

Sample space $$=\{(23,1),(22,2), \ldots\}$$

Integer points on line in shaded region

$$\begin{aligned}
& \{(6,18),(7,17),(8,16), \ldots(18,6)\} \
& P(E)=\frac{n(E)}{n(S)}=\frac{13}{23}=\frac{m}{n} \Rightarrow n-m=10
\end{aligned}$$

Solutions · 1