Let the sum of two positive integers be 24. If the probability, that their product is not less than 3/4

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 positive product, is \(\frac{m}{n}\), where gcd(m, n) = 1, then n – m equals :

(1) 9 

(2) 11 

(3) 8 

(4) 10

Answer 1

Correct option is (4) 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 x + y = 24 is 23

Probability \(\frac{13}{23}= \frac{m}{n} \)

n – m = 10