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

Step 1: Identify the general term

The general term in the binomial expansion of $(a+b)^n$ is given by $T_{r+1} = \binom{n}{r} a^{n-r} b^r$. In this problem, $a = 7^{1/3}$ and $b = 11^{1/3}$. We substitute these values into the general term formula.

Step 2: Determine conditions for integral terms

For the term $T_{r+1}$ to be an integer, the powers of 7 and 11 must be integers. This means that $(n-r)$ must be a multiple of 3, and $r$ must also be a multiple of 3. Since $r$ is a multiple of 3, $n-r$ being a multiple of 3 implies that $n$ must also be a multiple of 3.

Step 3: Express r in terms of k

Since $r$ must be a multiple of 3, we can write $r = 3k$ for some integer $k$. The possible values for $r$ range from $0$ to $n$. Therefore, $0 \le 3k \le n$, which implies $0 \le k \le n/3$.

Step 4: Calculate the number of integral terms

The number of integral terms is equal to the number of possible values for $k$. Since $k$ ranges from $0$ to $\lfloor n/3 \rfloor$, the total number of values for $k$ is $\lfloor n/3 \rfloor + 1$. We are given that the number of integral terms is 183.

Step 5: Solve for n

We set the expression for the number of integral terms equal to 183 and solve for $n$. From the definition of the floor function, if $\lfloor x \rfloor = m$, then $m \le x < m+1$.

Step 6: Determine the range for n and find the least value

Multiplying the inequality by 3, we find the range for $n$. Since $n$ must be a multiple of 3 (as established in Step 2), the possible values for $n$ are 546. The least value of $n$ in this range that is a multiple of 3 is 546.

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