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 $100^{th}$ position? (The positions are counted from left to right: the $1^{st}$ position is the 1, the $2^{nd}$ is the 4, etc.)
Separating the numbers whose squares have 1, 2, and 3 digits, we have:

1 digit: $1, 2, 3$

2 digits: $4, 5, 6, 7, 8, 9$

3 digits: $10, 11, 12, \ldots, 31$

Up to $31^{2}$, the sequence has $3+12+66=81$ digits.

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

Thus, we are missing $100-81=19$ digits to reach the $100^{\text{th}}$ digit. 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 first 3 digits (thousands, hundreds, tens) of the number: $36^{2}=1296$.

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

Therefore, the number is 9.