Find three consecutive largest positive integers such that the sum of one-third of first, one-fourth of second and one-fifth of third is at most 20.

Let three consecutive positive integers be x, x + 1 and x + 2.

Given, sum of one-third of first, one-fourth of second and one-fifth of third is at most 20

$\therefore \dfrac{1}{3}x + \dfrac{1}{4}(x + 1) + \dfrac{1}{5}(x + 2) \le 20 \\[1em] \Rightarrow \dfrac{x}{3} + \dfrac{x + 1}{4} + \dfrac{x + 2}{5} \le 20 \\[1em] \Rightarrow \dfrac{20x + 15(x + 1) + 12(x + 2)}{60} \le 20 \\[1em] \Rightarrow \dfrac{20x + 15x + 15 + 12x + 24}{60} \le 20 \\[1em] \Rightarrow 47x + 39 \le 1200 \\[1em] \Rightarrow 47x \le 1161 \\[1em] \Rightarrow x \le \dfrac{1161}{47} \\[1em] \Rightarrow x \le 24.702$

∴ x = 24, x + 1 = 25, x + 2 = 26.

Hence, three consecutive numbers are 24, 25 and 26.