The least value of n for which the number of integral terms in the Binomial expansion of $$(\sqrt[3]{7}+\sqrt[12]{11})^{n}$$ is 183, is :

We need to find the least value of $$n$$ for which the binomial expansion of $$(\sqrt[3]{7} + \sqrt[12]{11})^n$$ has 183 integral terms.

The general term is: $$T_{r+1} = \binom{n}{r}(\sqrt[3]{7})^{n-r}(\sqrt[12]{11})^r = \binom{n}{r} \cdot 7^{(n-r)/3} \cdot 11^{r/12}$$

For $$T_{r+1}$$ to be an integer, the exponents of 7 and 11 must be non-negative integers:

(i) $$\frac{n-r}{3}$$ is a non-negative integer, i.e., $$n - r \equiv 0 \pmod{3}$$, i.e., $$r \equiv n \pmod{3}$$

(ii) $$\frac{r}{12}$$ is a non-negative integer, i.e., $$r \equiv 0 \pmod{12}$$

Since $$r$$ must be a multiple of 12, and 12 is a multiple of 3, condition (i) requires $$n \equiv 0 \pmod{3}$$.

When $$n$$ is a multiple of 3, both conditions are satisfied whenever $$r = 0, 12, 24, \ldots$$ with $$r \leq n$$.

The number of valid values of $$r$$ is $$\lfloor n/12 \rfloor + 1 = 183$$.

So $$\lfloor n/12 \rfloor = 182$$, which means $$2184 \leq n < 2196$$.

The least such $$n$$ that is also a multiple of 3 is $$n = 2184$$ (since $$2184 = 3 \times 728$$).

The correct answer is Option 1: 2184.