2. For a natural number ending not in zero, one of its digits was replaced by zero (if it was the leading digit, it was simply erased). As a result, the number decreased by 6 times. Find all numbers for which this is possible.
Answer: $12 a$ when $a=1,2,3,4$.

Solution. Represent the original number in the form $m+10^{k} a+10^{k+1} n$, where $a$ is a decimal digit, $k, m, n-$ are non-negative integers, and $m0$, otherwise $m=0$ and $a=0$. Then the equation transforms to $m=2 a \cdot 10^{k-1}$. Due to the condition, the number $m$ does not end in 0 and therefore is not divisible by 10. Thus, $k=1, m=2 a$, and the original number is $12 a$. Since $m<10$, the digit $a$ takes the values $1,2,3,4$.