JEE Main 2025 Jan 29 Shift 1, Mathematics Q8: Introduction to Binomial Expansion

General term: $T_{r+1}=\left(\genfrac{}{}{0ex}{}{n}{r}\right)7^{(n-r)/3}11^{r/12}.$

For $T_{r+1}$ to be an integer (for arbitrary $n$ here) we need $r/12\in \mathbb{Z}$, i.e. $r$ is a multiple of $12$. With $r\in \{0,1,\dots ,n\}$ the count of multiples of $12$ is $\lfloor n/12\rfloor +1.$

Setting $\lfloor n/12\rfloor +1=183$ gives $\lfloor n/12\rfloor =182$, so $n\ge 12\cdot 182=2184.$

(One also needs $(n-r)/3$ to be an integer for those $r$; with $r=12k$ a multiple of $3$, this holds whenever $3\mid n$, which is satisfied by $n=2184.$)

$\therefore$ Least $n=2184.$