Prove that there is an integer $n$ such that $n^{1992}$ starts with $1992$ one's. This was taken from an old Brazilian Mathematical Olympiad (1992).
As the title says, we're supposed to prove that there is an integer $n$ such that $n^{1992}$ starts with $1992$ one's (in the decimal expansion).
In principle I'm expecting a proof using rather elementary methods, though any solution at all is appreciated and will be upvoted.
 A: In other words, you want the fractional part of
$$ \log_{10} n^{1992} $$
to lie between the fractional parts of
$$ \log_{10} \underbrace{111\cdots111}_{\text{1992 1's}} $$
$$\log_{10} \underbrace{111\cdots11}_{\text{1991 1's}}2 $$
A: Hint: Consider $n=\left\lfloor\left(\dfrac19\right)^{1/1992}\times10^{2000}\right\rfloor$
A: Let $a$ be the number fromed by $1992$ ones. Then we are looking for $n$ and $k$ such that $a\cdot 10^k\le n^{1992}<(a+1)\cdot 10^k$, that is
$$k+\log a\le 1992\log n<k+\log(a+1). $$
Let $u=\log a-1991$ and $v=\log (a+1)-1992$ and for $n\in\mathbb N$ let $f(n)=1992\log n-\lfloor (1992\log n)\rfloor$.
There are positive numbers $a,b$ (namely $\approx \frac{1992}{\ln10}$ such that 
for $n\in\mathbb N$ we have $\frac an<f(n+1)-f(n)<\frac bn$ or both $f(n)>1-bn$ and $f(n+1)<bn$. This follows by taking derivatives or by looking at the binomial expansion of $(n+1)^{1992}$.
Assume $f(n)\le u$ for almost all $n$. Then $f(n+1)>f(n)+\frac an$ for almost all $n$, which is impossible because $a\cdot\sum \frac1n$ diverges.
By the same reason, $f(n)$ cannot be $>u$ for almost all $n$. Hence there are infinitely many $n$ with $f(n-1)\le u<f(n)$. Especially, there are such $n$ with $\frac b{n-1}<v-u$, hence $u\le f(n)<v$. Each such $n$ is a solution.
A: Let $a=111...111$ Consider 
$$a\times 10^k\lt n^{1992}\lt (a+1)\times 10^k \tag{*} $$
If $k$ sufficient large, there must be an integer $n$ such that
$$\sqrt[1992]{a}\times  10^{\frac k{1992}}\lt   n\lt\sqrt[1992]{a+1}\times  10^{\frac k{1992}} $$
that is $(*)$


In general, If $f(x)$ is a polynomial, $a\in \Bbb N$, there must be $n\in \Bbb N$ such that
$f(n)$ starts with $a$

In fact, If $k$ sufficient large, there must be an integer $n$ such that
$$f(n)\lt a\times10^k\lt f(n+1)\qquad f(n+1)-f(n)\lt 10^k$$
