The number of rational terms in the binomial expansion of $$\left(4^{1/4} + 5^{1/6}\right)^{120}$$ is ___.

The general term in the expansion of $$\left(4^{1/4} + 5^{1/6}\right)^{120}$$ is:

$$T_{r+1} = \binom{120}{r} \left(4^{1/4}\right)^{120-r} \left(5^{1/6}\right)^r = \binom{120}{r} \cdot 2^{(120-r)/2} \cdot 5^{r/6}$$

For this term to be rational, both $$2^{(120-r)/2}$$ and $$5^{r/6}$$ must be rational, which requires both $$(120-r)/2$$ and $$r/6$$ to be non-negative integers.

The condition $$r/6 \in \mathbb{Z}$$ means $$r$$ is divisible by 6. Since 120 is divisible by 2, the condition $$(120-r)/2 \in \mathbb{Z}$$ means $$r$$ is even, i.e., divisible by 2. Thus $$r$$ must be divisible by $$\text{lcm}(6, 2) = 6$$.

With $$0 \leq r \leq 120$$ and $$r \equiv 0 \pmod{6}$$, the values are $$r = 0, 6, 12, 18, \ldots, 120$$, giving $$\frac{120}{6} + 1 = 21$$ rational terms.