7.5. Find the largest natural number $n$ for which $3^{n}$ divides the number $a=1 \cdot 2 \cdot 3 \cdot \ldots \cdot 2020$.
## Solution:

Consider the sequence of numbers 1, 2, 3, ..., 2019, 2020. Every third number in the given sequence is divisible by 3. Since $2020 = 3 \cdot 673 + 1$, it follows that there are 673 numbers in the sequence that are divisible by 3. Among these 673 numbers, every third one is divisible by $3^2$. Since $673 = 3 \cdot 224 + 1$, it follows that there are 224 such numbers. Proceeding similarly, we find how many numbers in the given sequence are divisible by $3^3, 3^4, 3^5$, etc. In the divisions performed, we only needed the quotients, which are the integer parts of the numbers $2020 \div 3, 2020 \div 3^2$, etc. Since $3^6 = 729$ and $3^7 = 2181$, we get $n = \left[\frac{2020}{3}\right] + \left[\frac{2020}{3^2}\right] + \left[\frac{2020}{3^3}\right] + \left[\frac{2020}{3^4}\right] + \left[\frac{2020}{3^5}\right] + \left[\frac{2020}{3^6}\right] = 673 + 224 + 74 + 24 + 8 + 2 = 1005$.

Answer: $n = 1005$.