这是一份leeds利兹大学MATH223101作业代写的成功案例
Just as in the treatment of the case $k=200$ above, we subtract the number of primes up to $10^{k-1}$ from the number of primes up to $10^{k}$. By the Prime Number Theorem, this number is about
$$
\frac{10^{k}}{k \ln 10}-\frac{10^{k-1}}{(k-1) \ln 10}=\frac{(9 k-10) 10^{k-1}}{k(k-1) \ln 10} .
$$
Since
$$
\frac{9 k-10}{k-1}=9-\frac{1}{k-1}
$$
is very close to 9 if $k$ is large, we get that the number of primes with $k$ digits is approximately
$$
\frac{9 \cdot 10^{k-1}}{k \ln 10}
$$
Comparing this with the total number of positive integers with $k$ digits, which we know is $10^{k}-10^{k-1}=9 \cdot 10^{k-1}$, we get
$$
\frac{9 \cdot 10^{k-1}}{k \ln 10 \cdot 9 \cdot 10^{k-1}}=\frac{1}{(\ln 10) k} \approx \frac{1}{2.3 k} .
$$
MATH223101 COURSE NOTES ：
Simplifying, we obtain
$$
4 k^{2}-4 n k+n^{2}-n-2<0 .
$$
Solving for $k$, we get that the left-hand side is nonpositive between the two roots:
$$
\frac{n}{2}-\frac{1}{2} \sqrt{n+2} \leq k \leq \frac{n}{2}+\frac{1}{2} \sqrt{n+2}
$$
So the first integer $k$ for which this is nonpositive is
$$
k=\left\lceil\frac{n}{2}-\frac{1}{2} \sqrt{n+2}\right]
$$