2. Find the smallest natural number that has exactly 12 different natural divisors, the largest prime divisor of which is the number 101, and the last digit is zero.
Solution. Let $n$ be the desired number. According to the condition, $n: 101, n: 2, n: 5$. Consider the number $m=2 \cdot 5 \cdot 101$, note that it has exactly 8 different natural divisors $(1$, $2,5,101,2 \cdot 5, \ldots, 2 \cdot 5 \cdot 101$ ), so $n>m$. Since $n$ is the smallest natural number satisfying the condition of the problem, we will further consider the number $2 \cdot m=2 \cdot 2 \cdot 5 \cdot 101$, obtained from $m$ by multiplying by the smallest of the prime numbers. Note that the number $2 m=2^{2} \cdot 5 \cdot 101$ has exactly 12 different natural divisors $(1,2,4,5,101$, $2 \cdot 5, \ldots, 2 \cdot 2 \cdot 5 \cdot 101$ ), so $n=2^{2} \cdot 5 \cdot 101=2020$. Answer: 2020.

Comment. In the solution, the well-known function $\tau(n)$ - the number of natural divisors of the number $n$ - can be used. Moreover, if the canonical decomposition into prime factors of the number $n$ has the form $n=p_{1}^{\alpha_{1}} \cdot p_{2}^{\alpha_{2}} \cdot p_{3}^{\alpha_{3}} \cdot \ldots \cdot p_{k}^{\alpha_{k}}$, then $\tau(n)=$ $\left(\alpha_{1}+1\right) \cdot\left(\alpha_{2}+1\right) \cdot \ldots \cdot\left(\alpha_{k}+1\right)$. If an olympiad participant uses the last formula without proof, no points are deducted.