find n, if the coefficient of $x^2$ in $(3 + 2x)^n$ is 20412. Find $n$, if the coefficient of $x^2$ in $(3+2x)^n$ is 20412
I solved it using binomial theorem and got 
$n(n-1)3^{n-2} = 10206$
The term like $n(n-1)3^{n-2}$, I am not sure how to solve it.

The coefficient should be $2n(n-1)3^{n-2}$.
To obtain $n$ use prime factorisation: $\;20412=2^2\cdot3^6\cdot 7$. Maybe $n$ or $n-1$ is divisible by $3$, so the equality
$$2n(n-1)3^{n-2}=2^2\cdot3^6\cdot 7$$
implies $\;n\le 8$. 
By Euclid's lemma, it also implies $7$ divides $n$ or $n-1$. Combining with $n\le 8$, we have that $n=7$ or $n=8$. However $n=8$ is impossible, as the expression is divisible only by $2^2$. Thus there is only one solution: $\;n=\color{red}7$.
