Solve $\frac{(2n)!}{n!}=30{,}240$ Problem:$$\frac{(2n)!}{n!}=30{,}240$$
Attempt:
I know that $$\frac{(2n)!}{n!}=\prod_{k=n+1}^{2n}k=(n+1)(n+2)...(2n)$$
I've considered factoring $30240=5 \times 6 \times  7\times 8  \times 9  \times 2$
To fit the pattern, $2\times 5=10$, so I have $30240=6 \times  7\times 8  \times 9  \times 10 \implies n=5$.
Question
I'm wondering if there's a better/more general method rather than what I did.
 A: I would do it in a similar manner as you. Here is another approach (assuming there is an exact integer solution):
$$\dfrac{(2n)!}{n!} = \left(\dfrac{2n}{n}\cdot \dfrac{2(n-1)}{n-1}\cdots \dfrac{2(1)}{(1)}\right)\cdot (2n-1)(2n-3)\cdots1$$
$$=2^n\cdot \text{[stuff]}$$
Where $\text{[stuff]}$ has no factors of 2. So you want the largest $n$ such that $2^n$ is a factor of $30,240$.
A: Using algebra and making the problem more general
$$\frac{(2n)!}{n!}=k$$
Take logarithms and using Stirling approximation, we have
$$\log \left(\frac{(2 n)!}{n!}\right)=n (\log (4 n)-1)+\frac{\log (2)}{2}-\frac{1}{24 n}+O\left(\frac{1}{n^3}\right)$$ Let $m=4n$ and neglect the $\frac{1}{24 n}$ to end with
$$(m-1)\log(m)\simeq\log \left(\frac{k^4}{4}\right)$$ Then, an approximate solution
$$n=\frac {\log \left(\frac{k}{\sqrt{2}}\right)} {W\left(\frac 4 e\log \left(\frac{k}{\sqrt{2}}\right)\right) }$$ where $W(.)$ is Lambert function.
Since we neglected the term $-\frac{1}{24 n}$, we know that this is an underestimate of the solution.
For $k=30240$, this gives,as a real $n=4.99722$.
For $k=28158588057600$, this gives,as a real $n=10.99900$.
Now, small numbers
For $k=12$, this gives,as a real $n=1.99011$.
For $k=2$, this gives,as a real $n=0.97104$.
