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 $$\gcd(m, n) = 1$$, then $$n - m$$ equals

Two positive integers sum to 24. Find the probability their product is at least $$\frac{3}{4}$$ of the greatest possible product.

Since $$x + y = 24$$ with $$x, y \in \mathbb{Z}^+$$, the product can be written as $$P = xy = x(24 - x)$$, which is a quadratic in $$x$$ that attains its maximum when $$x = 12$$. This gives $$P_{\max} = 144$$.

From the condition $$xy \geq \tfrac{3}{4}\cdot144 = 108$$, we require $$x(24 - x) \geq 108$$, so that $$24x - x^2 \geq 108$$ and hence $$x^2 - 24x + 108 \leq 0$$. The roots of the equation $$x^2 - 24x + 108 = 0$$ are found by $$x = \frac{24 \pm \sqrt{576 - 432}}{2} = \frac{24 \pm 12}{2}$$, yielding $$x = 6$$ or $$x = 18$$, and therefore $$6 \leq x \leq 18$$.

Since $$x$$ can be any integer from 1 to 23 (so that $$y = 24 - x > 0$$), there are 23 total equally likely outcomes. Substituting the range for $$x$$, the favorable values are $$x = 6, 7, 8, \dots, 18$$, giving $$18 - 6 + 1 = 13$$ outcomes.

This gives the probability $$\frac{13}{23}$$. Because $$\gcd(13,23) = 1$$, we have $$m = 13$$ and $$n = 23$$, so that $$n - m = 23 - 13 = 10$$.

The correct answer is Option (1): 10