3. Find all natural numbers whose decimal representation ends with two zeros and have exactly 12 natural divisors.
Answer: 200 and 500. Solution: Since the number ends with two zeros, it is divisible by 100, that is, it has the form $n \cdot 100=n \cdot 2^{2} \cdot 5^{2}$. We will prove that if a number has exactly 12 divisors, then $n$ can only be 2 or 5. The smallest number of the form $n \cdot 2^{2} \cdot 5^{2}$ is the number 100 (the case $n=1$), which has 9 divisors. These can be found directly, but it can also be done as follows: all divisors of the number 100 have the form $2^{k} \cdot 5^{m}$, where $k$ and $m$ can be 0, 1, or 2. Therefore, the number of divisors is $3 \cdot 3=9$. If $n$ is not divisible by 2 or 5, then the number $n \cdot 100$ will have at least 18 divisors: 9 divisors of the number 100 and 9 divisors obtained by multiplying the divisors of the number 100 by $n$. If $n=2$ or $n=5$, then, as is easy to check, there will be exactly 12 divisors. If $n$ is divisible by 2 or 5 to a power higher than the first, then $n$ is divisible by 200 or 500, and in this case, it is greater than 200 or 500, respectively, so it has more than 12 divisors. Hence - the answer.