Prove that there is $n\in\mathbb{N}$ such that $n!$ starts with 1996

Solution 1:

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.

Solution 2:

Examples

Looking at a few examples gives a clue as to why there will be an infinite number of such numbers. $$ \begin{align} 10003183!&=1.9957478783343368038\times10^{65679340}\\ 10003184!&=\color{#C00}{1.996}3833244587984566\times10^{65679347}\\ 10003185!&=1.9970191725476385839\times10^{65679354}\\ 100006487!&=1.9959743653614076660\times10^{756622452}\\ 100006488!&=\color{#C00}{1.996}1038641782323141\times10^{756622460}\\ 100006489!&=\color{#C00}{1.996}2333913579788396\times10^{756622468}\\ 100006490!&=\color{#C00}{1.996}3629469050779724\times10^{756622476}\\ 100006491!&=\color{#C00}{1.996}4925308239615811\times10^{756622484}\\ 100006492!&=\color{#C00}{1.996}6221431190626726\times10^{756622492}\\ 100006493!&=\color{#C00}{1.996}7517837948153934\times10^{756622500}\\ 100006494!&=\color{#C00}{1.996}8814528556550287\times10^{756622508}\\ 100006495!&=1.9970111503060180035\times10^{756622516} \end{align} $$ In the case of $10003183!$, whose mantissa is a bit small, when we multiply the factorial by $10003184$, we multiply the mantissa by $1.0003184$. Since the mantissa is approximately $2$, it is increased by approximately $0.0006368$, and that brings $1.9957479$ to $1.9963833$.

In the case of $100006487!$, whose mantissa is a bit small, when we multiply the factorial by $100006488$, we multiply the mantissa by $1.00006488$. Since the mantissa is approximately $2$, it is increased by approximately $0.00012976$, and that brings $1.9959743$ to $1.9961039$.

Theoretical Considerations

$\newcommand{\bigO}[1]{\operatorname{O}\left(#1\right)}$ For $n\ge8$ and $m^2\le2\log(10)\,10^n$, we have $$ \left\{\log_{10}\left(\frac{\left(10^n+m\right)!}{10^n!}\right)\right\}=\frac{m^2}{2\log(10)\,10^n}+\bigO{\frac1{10^{n/2}}}\tag1 $$ where $\{x\}=x-\lfloor x\rfloor$ is the positive fractional part of $x$. So for $$ m\approx\left(2\log(10)\,10^n\left\{\log_{10}(1996)-\log_{10}\left(10^n!\right)\right\}\right)^{1/2}\tag2 $$ we should have $$ \left\{\log_{10}\left(\left(10^n+m\right)!\right)\right\}\approx\left\{\log_{10}(1996)\right\}\tag3 $$ If the approximation in $(3)$ is a bit off, we can adjust things by noting that $$ \begin{align} \left\{\log_{10}\left(10^n+m\right)\right\} &\approx\frac{m}{\log(10)\,10^n}\\ &\le\left(\frac2{\log(10)\,10^n}\right)^{1/2}\tag4 \end{align} $$ which is small enough to adjust within an interval of width $\log_{10}\left(\frac{1997}{1996}\right)\approx\frac1{1996\,\log(10)}$ when $n\ge8$.