1. For every natural number $n$ with 3 decimal digits (so the first digit is not zero), we consider the number $n_{0}$ obtained from $n$ by removing its digits that are equal to zero. For example, if $n=205$ then $n_{0}=25$.

Determine the number of integers $n$ with three digits for which $n_{0}$ is a divisor of $n$ different from $n$.
Solution: We immediately exclude the case where all the digits of $n$ are different from zero, as in this case we have $n_{0}=n$. Therefore, there are two possibilities left:

1) $n$ has the third digit equal to zero; the decimal representation of $n$ is of the form $n=a b 0, n_{0}=a b$ and $n_{0} \mid n$ for any choice of $a \in\{1, \ldots, 9\}$ and $b \in\{0,1, \ldots, 9\}$. Relative to this case, there are thus $9 \times 10=90$ possibilities for $n$.
2) $n$ has the third digit different from zero but the second digit equal to zero; in this case $n$ is written in the form $a 0 b$ and $n_{0}$ in the form $a b$. Recalling the meaning of the decimal representation, we seek the cases where $10 a+b$ divides $100 a+b$. Since $10 a+b$ certainly divides $10(10 a+b)=100 a+10 b$, it is also equivalent to see when $10 a+b$ divides the difference $(100 a+10 b)-(10 a+b)=9 b$.

Writing $k(10 a+b)=100 a+b$ and considering the last decimal digit, we obtain that $(k-1) b$ is divisible by 10, so at least one of $k-1$ and $b$ is divisible by 5. Since obviously $1<k<10$ and $1 \leq b \leq 9$, we have that $k-1=5$ or $b=5$.

For $k-1=5$ we get $60 a+6 b=100 a+b$, from which $b=8 a$ and thus $a=1, b=8$. It is verified that $n=108$ is indeed a solution, as 18 divides 108.

For $b=5$ we use the equivalent property that $10 a+b$ divides $9 b$, which becomes $10 a+5$ divides 45; it is immediately seen that the divisors of 45 of the form $10 a+5$ with $a \geq 1$ are 15 and 45, giving the two solutions $n=105$ (15 divides 105) and $n=405$ (45 divides 405).

In conclusion, there are 90 solutions in case 1) and 3 solutions in case 2), for a total of 93 solutions.