Is there a positive integer $n$ such that $1 \cdot 2 \cdot 3 \cdots \cdots(n-1) \cdot n$, that is, $n$! ends exactly in 100 zeros?
The condition means that $n!$ should be divisible by exactly the 100th power of 10. Since the prime factorization of 10 is $2 \cdot 5$, this is precisely when the prime factorization of $n!$ contains 2 and 5, one of which is exactly the 100th power, and the other is at least the 100th power. It is expected that 5 will appear with a smaller exponent, so we first check when the exponent of 5 will be 100.

Since every fifth natural number is divisible by 5, the product of the numbers up to $n$ can only be exactly divisible by $5^{100}$ if there are at most 100 numbers divisible by 5 up to $n$, i.e., if $n \leq 500$.

Now let's write $n$ in "base 5," i.e., in the form $n=a+5b+25c+125d+625e+\ldots$, where the coefficients are non-negative integers less than 5 (such a representation exists for every positive $n$ by the division algorithm). Since $n \leq 500$, only the first four terms in this representation can be non-zero: $n=a+5b+25c+125d$.

Let's check how many numbers up to $n$ are divisible by 5, 25, and 125. Every fifth number is divisible by 5, every twenty-fifth by 25, and every one hundred and twenty-fifth by 125. Therefore, the number of such numbers is respectively $b+5c+25d$, $c+5d$, and $d$.

Thus, in the product, the factor 5 appears $(b+5c+25d)$ times, an additional 5 appears $(c+5d)$ times, and another 5 appears $d$ times. Therefore, the total number of 5 factors is:

$$
(b+5c+25d) + (c+5d) + d = b + 6c + 31d = 100
$$

$d > 3$ is impossible because $31 \cdot 4 > 100$. Since $b, c \leq 4$, we have:

- If $d=3$, then $b + 6c = 7$. The only solution is $b=1$ and $c=1$, giving $n = 1 + 5 \cdot 1 + 25 \cdot 1 + 125 \cdot 3 = 406$.
- If $d=2$, then $b + 6c = 38$. The only solution is $b=2$ and $c=6$, but $c \leq 4$, so this is not possible.
- If $d=1$, then $b + 6c = 69$. The only solution is $b=3$ and $c=11$, but $c \leq 4$, so this is not possible.
- If $d=0$, then $b + 6c = 100$. The only solution is $b=4$ and $c=16$, but $c \leq 4$, so this is not possible.

Thus, the only solution is $n = 406$.

Remark. The above procedure can be performed for any natural number $k$ instead of 100. However, we do not always get a solution.

For example, can $n!$ end in 5 zeros? Here only $n=a+5b$ is possible. Then the number of zeros is $(a+5b) + b = a + 6b = 5$. This is only possible if $b=0$, which implies $a=5$, contradicting $a \leq 4$. Similarly, it is impossible for $5 + 6b$ zeros, where $0 \leq b < 5$. For $k=36$, for example, only $c=1$ is possible. This would imply $a + 6b = 5$, which, as we have seen, is impossible. But it is also impossible for $k=31+30$. This would only be possible if $c=1$. The resulting $a + 6b = 30$ equation would only hold if $b=5$, which contradicts $b \leq 4$. For $31 \leq k < 62$, only $c=1$ is possible, which implies $0 \leq a + 6b < 31$. Given $0 \leq a, b \leq 4$, we must exclude numbers of the form $5 + 6b$ and $a + 30$. These are:

$$
k=5+31; 5+6+31; 5+12+31; 5+18+31; 5+24+31; 30+31
$$

(The next number to be written does not satisfy $k < 62$.) Therefore, $n!$ cannot end in 36, 42, 48, 54, 60, or 61 zeros.

Similarly, it is impossible for the number of zeros to be:

$$
k=5+31c; 5+6+31c; 5+12+31c; 5+18+31c; 5+24+31c; 30+31c
$$

where $1 < c < 5$.

For $n \geq 625$, finding the cases to exclude becomes more complex, but the result can be stated.