1. If we subtract the unit digit from any natural number with at least two digits, we get a number that is one digit "shorter." Find all original numbers that are equal to the absolute value of the difference between the square of the "shorter" number and the square of the removed digit.
1. Let the sought number be $10a + b$, where $a, b$ are integers, $a \geq 1, 0 \leq b \leq 9$. According to the problem, we have

$$
10a + b = \left|a^2 - b^2\right|.
$$

First, assume that $a \geq b$. In this case, simple manipulations yield

$$
\begin{aligned}
10a + b & = a^2 - b^2, \\
a^2 - 10a + 25 & = b^2 + b + 25, \\
(a-5)^2 & = b^2 + b + 25.
\end{aligned}
$$

We then substitute $b = 0, b = 1, \ldots, b = 9$ into the last equation to check if $b^2 + b + 25$ is a perfect square of some non-negative integer. The equation is satisfied by the pairs $(b=0, a=0); (b=0, a=10); (b=7, a=14)$.

In the case where $a < b$, similar manipulations yield

$$
\begin{aligned}
10a + b & = b^2 - a^2, \\
a^2 + 10a + 25 & = b^2 - b + 25, \\
(a+5)^2 & = b^2 - b + 25.
\end{aligned}
$$

Similarly, as in the first case, we obtain the pairs $(b=0, a=0); (b=1, a=0); (b=8, a=4)$.

Conclusion: Considering the conditions of the problem, the solutions are the three numbers 48, 100, 147.

For a complete solution, award 6 points, 1 point for randomly finding one or two numbers, 2 points for randomly finding all three numbers. For an incomplete discussion, deduct 2 to 4 points.