3. The $n$th term of a sequence is the first non-zero digit of the decimal expansion of $\frac{1}{\sqrt{n}}$. How many of the first one million terms of the sequence are equal to 1 ?
Solution
In the first one million terms, $0.001 \leq \frac{1}{\sqrt{n}} \leq 1$.
If the first non-zero digit is 1 , we have four inequalities which satisfy the question.
$$
\begin{array}{c}
1 \leq \frac{1}{\sqrt{n}}<2 \\
0.1 \leq \frac{1}{\sqrt{n}}<0.2 \\
0.01 \leq \frac{1}{\sqrt{n}}<0.02 \\
0.001 \leq \frac{1}{\sqrt{n}}<0.002
\end{array}
$$

Inequality (1) gives $\frac{1}{4}<n \leq 1$.
Inequality (2) gives $25<n \leq 100$.
Inequality (3) gives $2500<n \leq 10000$.
Inequality (4) gives $250000<n \leq 1000000$.
Adding up the number of integers in each solution set gives the total number of solutions as $1+75+7500+750000=757576$