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:
Step-by-Step Solution
Key Concept: Integral terms in $\bigl(7^{1/3}+11^{1/12}\bigr)^{n}$ require $3\mid(n-r)$ AND $12\mid r.$ Set $r=12k$, then $3\mid n$. The integer count is $\lfloor n/12\rfloor+1=183$.
General term: $\binom{n}{r}\,7^{(n-r)/3}\,11^{r/12}.$ Integral iff $3\mid(n-r)$ and $12\mid r.$
$r=12k$ ($k=0,1,\dots,\lfloor n/12\rfloor$) automatically gives $3\mid r$, so $3\mid(n-r)\Leftrightarrow 3\mid n.$
Number of integral terms $=\lfloor n/12\rfloor+1=183\Rightarrow \lfloor n/12\rfloor=182\Rightarrow n\in\{2184,2185,\dots,2195\}.$
Among these, $3\mid n$ holds for $n\in\{2184,2187,2190,2193\}$. Least: $n=2184.$
Correct Answer: 1