2. For a natural number ending not in zero, one of its digits was replaced by zero (if it is the leading digit, it was simply erased). As a result, the number decreased by 9 times. How many numbers exist for which this is possible?
Answer: 7.

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 $m>0$, otherwise $m=a=0$. Then the number $8 m$ is a multiple of 10 and therefore ends in 0. By the condition, the number $m$ does not end in 0. Thus, 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 $m=125 a$. Since the number $m$ is odd and less than 1000, the digit $a$ can take the values $1,3,5,7$, which gives us 4 options.
2) Suppose $k=2$. Then $m=\frac{25 a}{2}$. Since the number $m$ is odd and less than 100, the digit $a$ is 2 or 6. These values give us another 2 options.
3) Suppose $k=1$. Then $m=\frac{5 a}{4}$. Since the number $m$ is odd and less than 10, we get $a=4$.

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 $4+2+1=7$.