$x$ is an integer satisfying:
$$\left[\frac x{1!}\right]+\left[\frac x{2!}\right]+\left[\frac x{3!}\right] + \cdots +\left[\frac x{10!}\right]=1001$$
Find largest prime divisor of $x$, where $[~]$ denotes the greatest integer function.
I have no idea of solving questions of these format. Any solutions are highly requested.
I know basic Gif operations, identities like $[x] =[x/2]+[(x+1)/2]$ which can be generalised for any natural number $n$.
Best Answer
A lot of people gave great hints in the comment, but I'll answer just so this post has one
First, the left side is strickly increasing over the integer. So, if we find a value that work, it is the only one. Also, if a value of $x$ give something smaller than $1001$, then the answer is bigger. The reciprocal is also true. We could narrow our possibilities by doing a binary search.
Since $x$ is an integer, $$\left[\frac{x}{1!}\right] = [x]$$ Then $x\leq1001$
Follow up $6!=720$, then $$\frac{x}{6!} <1 \Longrightarrow \left[\frac{x}{6!}\right] = 0$$ We are left with $$\left[\frac{x}{1!}\right]+\left[\frac{x}{2!}\right]+\left[\frac{x}{3!}\right]+\left[\frac{x}{4!}\right]+\left[\frac{x}{5!}\right]=1001$$ Adding the term up to further reduce the possibilities. If we just have $$\left[\frac{x}{1!}\right]+\left[\frac{x}{2!}\right]=1001$$ Then $x$ should be around two third of $1001$. Let's try $x=667$ $$\left[\frac{667}{1!}\right]+\left[\frac{667}{2!}\right]=667+333=1000$$ $x<668$. Here it doesn't help but if the numbers were bigger, it could have remove more terms from the sum.
Finally, if we remove the $[\cdot]$ function, then $x$ is $$\frac{x}{1!}+\frac{x}{2!}+\frac{x}{3!}+\frac{x}{4!}+\frac{x}{5!} = \frac{120x+60x+20x+5x+1x}{120}$$ $$\frac{206x}{120}=1001 \Longrightarrow x = 583.106\ldots$$ Let's try $x=584$ $$\left[\frac{584}{1!}\right]+\left[\frac{584}{2!}\right]+\left[\frac{584}{3!}\right]+\left[\frac{584}{4!}\right]+\left[\frac{584}{5!}\right]=584+292+97+24+4=1001$$
Edit to include Henry comment
If we continue the sum, the equation could become $$\left[\frac{x}{1!}\right]+\left[\frac{x}{2!}\right]+\left[\frac{x}{3!}\right]+\dots+\left[\frac{x}{11!}+\dots\right]=1001$$ The left hand part is becomes $x(e-1)$ and $$x=\frac{1001}{e-1}\approx 582.56$$ From there, we test $583$ and, since it is too low, $584$.
Edit 2 to answer the question Thank's to Mark who point out that I forgot to answer to the question which is "Find the largest divisor of $x$". Since $584 = 2^3\times73$, the answer is $73$.