Find three smallest consecutive natural numbers such that the difference between one-third of the largest and one-fifth of the smallest is at least 3.

Let first least natural number = x
then, second number = x + 1
and third number = x + 2

Given,

$\dfrac{1}{3}(x+2) - \dfrac{1}{5}(x) \ge 3 \\[0.5em] \Rightarrow \dfrac{x}{3} - \dfrac{x}{5} \ge 3 - \dfrac{2}{3} \\[0.5em] \Rightarrow \dfrac{5x- 3x}{15} \ge \dfrac{9-2}{3} \\[0.5em] \Rightarrow \dfrac{2x}{15} \ge \dfrac{7}{3} \\[0.5em] \Rightarrow 2x \ge \dfrac{7 \times 15}{3} \\[0.5em] \Rightarrow 2x \ge 35 \\[0.5em] \Rightarrow x \ge \dfrac{35}{2} \\[0.5em] \Rightarrow x \ge 17\dfrac{1}{2}$

Since the three consecutive numbers should be natural numbers
∴ x = 18
    x + 1 = 19
    x + 2 = 20

Hence, the three smallest consecutive natural numbers are 18, 19, 20