How many factors does the product $1 \cdot 2 \cdot 3 \cdot \ldots \cdot n$ have, if we know that it ends with 1981 zeros?
I. solution. A number ends with $N=1981$ zeros if in its prime factorization, the smaller of the exponents of 2 and 5 is $N$. Since among the integers, even numbers are more frequent than those divisible by 5, it is expected that only a solution exists where the exponent of 5 is the smaller one. Let's denote the exponent of 5 in the prime factorization of $n!=1 \cdot 2 \cdot 3 \cdot \ldots \cdot n$ by $K(n)$. Therefore, we are looking for an $n$ such that

$$
K(n)=N
$$

Since $nN$, only the five numbers listed can be the number of factors if the product ends with 1981 zeros.

II. solution. We continue to use the notation introduced in the first solution and the result that it is sufficient to solve equation (1). Let's write the sought $n$ number in the base 5 number system! Since $n2$ and $K(n) \geq 3 \cdot 781=2343$, and for $P2$ the value on the left side is at least 468, for $Q<2$ the value is at most 308. If $P=Q=2$, (1) is equivalent to the equation

$$
31 R+6 S+T=107
$$

Here, the value of $R$ can only be 3, and in this case $S=2, T=2$. Therefore, the form of the sought $n$ number in the base 5 number system is 223220, 223221, 223222, 223223, or 223224, so the suitable values of $n$ in the base 10 number system are 7935, 7936, 7937, 7938, and 7939.