The number of natural numbers $$n \leq 2019$$ such that $$\sqrt[4]{48n}$$ is an integer is

Since $$48=2^4 \times 3$$, for $$48n$$ to be a perfect fourth power, $$n$$ must supply a factor of $$3^3$$ times a perfect fourth power, so $$n=27s^4$$ for a positive integer $$s$$. Checking $$n \leq 2019$$ gives $$s=1$$ ($$n=27$$) and $$s=2$$ ($$n=432$$) as the only solutions, so there are 2 such values.