If $$n$$ is the number of irrational terms in the expansion of $$\left(3^{1/4} + 5^{1/8}\right)^{60}$$, then $$(n-1)$$ is divisible by:

The general term in the expansion of $$\left(3^{1/4} + 5^{1/8}\right)^{60}$$ is given by $$T_{r+1} = \binom{60}{r} \cdot 3^{(60-r)/4} \cdot 5^{r/8}$$, where $$r = 0, 1, 2, \ldots, 60$$.

For a term to be rational, both exponents must be integers. The exponent of 3 is $$(60-r)/4$$, which is an integer when $$r \equiv 0 \pmod{4}$$. The exponent of 5 is $$r/8$$, which is an integer when $$r \equiv 0 \pmod{8}$$.

Both conditions are satisfied simultaneously when $$r$$ is divisible by $$\text{lcm}(4,8) = 8$$. So $$r \in \{0, 8, 16, 24, 32, 40, 48, 56\}$$, giving us 8 rational terms.

The total number of terms in the expansion is $$60 + 1 = 61$$. Therefore, the number of irrational terms is $$n = 61 - 8 = 53$$.

We need to check the divisibility of $$n - 1 = 52$$. We have $$52 = 4 \times 13 = 2^2 \times 13$$. Checking each option: $$52 / 26 = 2$$ (divisible), $$52 / 30$$ is not an integer, $$52 / 8 = 6.5$$ (not divisible), and $$52 / 7$$ is not an integer. Hence $$(n-1)$$ is divisible by 26.