3. Find the smallest natural number $\mathrm{n}$ such that the number of zeros at the end of the number ( $\mathrm{n}+20$ )! is exactly 2020 more than the number of zeros at the end of the number n!.
3. Answer: $5^{2017}-20$. From the condition, it follows that in the product $(\mathrm{n}+1)^{*} \ldots *(\mathrm{n}+20)$, the power of the prime factor 5 in the prime factorization is 2020. Among 20 consecutive numbers, exactly 4 are multiples of 5, of which only one can be a multiple of 25 or higher powers of 5. That is, it must be at least $5^{2017}$. The smallest $\mathrm{n}$ for which this is possible is $5^{2017}-20$.