2. For a natural number ending not in zero, one of its digits (not the most significant) was erased. As a result, the number decreased by 9 times. How many numbers exist for which this is possible?
Answer: 28.

Solution. Let's represent the original number in the form $m+10^{k} a+10^{k+1} n$, where $a$ is a decimal digit, and $k, m, n$ are non-negative integers, with $m>0$. By erasing the digit $a$, we get the number $m+10^{k} n$. According to the condition,

$$
m+10^{k} a+10^{k+1} n=9\left(m+10^{k} n\right) \Longleftrightarrow 8 m=10^{k}(a+n)
$$

Note that $k>0$, otherwise $m=0$ and $a=n=0$. Then the number $8 m$ is divisible by 10 and thus ends in 0. Given the condition, the number $m$ does not end in 0. Therefore, the last digit of $m$ is 5 and the number $m$ is odd. Therefore, $8 m$ is not divisible by 16, which implies $k \leqslant 3$. Let's consider three cases.

1) Suppose $k=3$. Then $m=125(a+n)$. Since the number $m$ is odd and less than $1000, a+n$ can take the values $1,3,5,7$. Note that the pair $(a, n)$ uniquely determines the original number, and each value of $a+n$ gives $a+n$ different pairs. Thus, we get $1+3+5+7=16$ options.
2) Suppose $k=2$. Then $m=25 \cdot \frac{a+n}{2}$. Since the number $m$ is odd and less than $100, a+n$ equals 2 or 6. These values give us $2+6=8$ options.
3) Suppose $k=1$. Then $m=5 \cdot \frac{a+n}{4}$. Since the number $m$ is odd and less than 10, we get $a+n=4$, which gives us 4 options.

Note that in 1) we get four-digit numbers, in 2) — three-digit numbers, in 3) — two-digit numbers. Therefore, each number satisfying the condition of the problem falls into exactly one of the sets 1) - 3). Thus, the total number of options is $16+8+4=28$.