The squares of the natural numbers from 1 to 99 were written one after another, forming the number 14916253649... What is the digit that occupies the 100th position? (The positions are counted from left to right, so the $1^{\underline{a}}$ position is the 1, the $2^{\underline{a}}$ is the 4, and so on.)
Separating the numbers whose squares have 1, 2, and 3 digits, we have,

$$
\begin{array}{ll}
\text { with } 1 \text { digit: } & 1,2,3 \\
\text { with } 2 \text { digits: } & 4,5,6,7,8,9 \\
\text { with } 3 \text { digits: } & 10,11,12, \ldots, 31
\end{array}
$$

Up to $31^{2}$, the number already has $3+12+66=81$ digits. Abbreviating digit by "dgs", we have

$$
\underbrace{1^{2}, 2^{2}, 3^{2}}_{1 \times 3 \text { dgs }}, \underbrace{4^{2}, \ldots, 9^{2}}_{2 \times 6=12 \text { dgs }}, \underbrace{10^{2}, \ldots, 31^{2}}_{3 \times 22=66 \text { dgs }}
$$

Thus, we are short of $100-81=19$ digits to reach 100. Since only $100^{2}$ has 5 digits, and since $19=4 \times 4+3$, we will have 4 more numbers, each with 4 digits, which are $32^{2}$, $33^{2}$, $34^{2}$, and $35^{2}$, plus the 3 digits (thousands, hundreds, tens) of the number $36^{2}=1296$, as follows.

$$
\underbrace{1^{2}, 2^{2}, 3^{2}}_{1 \times 3 \text { dgs }}, \underbrace{4^{2}, \ldots, 9^{2}}_{2 \times 6=12 \text { dgs }}, \underbrace{10^{2}, \ldots, 31^{2}}_{3 \times 22=66 \text { dgs }}, \underbrace{32^{2}, 33^{2}, 34^{2}, 35^{2}}_{4 \times 4=16 \text { dgs }}, 12 \underbrace{9}_{100^{\text{th}} \text { dgs }} 6
$$

Thus, we see that the digit 9 occupies the 100th position.