Prove that there is $n\in\mathbb{N}$ such that $n!$ starts with 1996 Basically the statement says that there is some number $n$ such that:
$$1996\cdot 10^k<n!<1997 \cdot 10^k$$
$$k + \log 1996<\sum_{i=1}^n \log i<k+\log 1997$$
A nice idea how to solve a similar problem can be found here, but I could not utilize it in this problem. 
Origin of this problem is a mystery - I have found it on a page where people listed their favorite math problems but this one had no solution.
 A: As the question basically states, the issue is to show there exists positive integers $k$ and $n$ such that
$$k + \log_{10} 1996 < \sum_{i=1}^n \log_{10} i < k + \log_{10} 1997 \tag{1}\label{eq1}$$
This basically means proving there exists an $n$ such that the fractional part of $\log_{10} n!$ is between $\log_{10} 1996$ and $\log_{10} 1997$, which has a difference of
$$\log_{10} 1997 - \log_{10} 1996 = 3.30037806\ldots - 3.30016053\ldots \approx 2.17 \times 10^{-4} \tag{2}\label{eq2}$$
First, consider
$$n_1 = 10^9 + 5 \times 10^4 \tag{3}\label{eq3}$$
The fractional part of the base $10$ log of this is
$$\log_{10}(n_1) - 9 \approx 2.17 \times 10^{-5} \tag{4}\label{eq4}$$
Next, consider
$$n_2 = 10^9 + 10^5 \tag{5}\label{eq5}$$
The fractional part of the base $10$ log of this is
$$\log_{10}(n_2) - 9 \approx 4.34 \times 10^{-5} \tag{6}\label{eq6}$$
Note the value in \eqref{eq6} is less than that of \eqref{eq2}. As the logarithm function is strictly increasing, the fractional parts of the logarithm, to base $10$, of values between $n_1 + 1$ and $n_2$ must be greater than of $n_1$ given in \eqref{eq4}. As there are $50000$ of them, their sum is more than $50000$ times the value in \eqref{eq4}, with this product being $\approx 1.08$. As this is $\gt 1$, it means any fractional interval greater than that in \eqref{eq6}, including of \eqref{eq2}, must have at least one value $n$ between $n_1$ and $n_2$ where the fractional part of $n!$ lies in this range (otherwise, it means the smallest value of $n$ just after the range must be $1$ more than the largest value just before range, so the size of the difference must be larger than the range, which is not possible here). Thus, there's an $n$ value (actually, there will be several) in this range which satisfies the question, i.e., $n!$ will start with the digits $1996$.
Note this proof technique can be generalized to show that any sequence of $m \ge 1$ digits, starting with a non-zero value, will have at least one positive $n$ where $n!$ starts with these digits. However, as this question only asked specifically for $1996$, I just gave the first example I found. I'm leaving the generalization to anybody who's interested, although I suggest you start with Dave L. Renfro's useful MO answer which he provided in a comment below.
