I'm going to argue that $F(k) = F(\gcd(m,k))$ where we have a clock that's capable of displaying numbers up to $m$ without overflowing. (So in this problem, $m = 10^4$ ).

Now, suppose that $m$ is prime (say 3 for the sake of brevity), then if $x = m$ , the only sequence it can ever generate is the constant sequence $0$ , but if we let $x$ be any other number less than $m$ , then after enough cycles, we're going to generate every number between $0,\cdots, m-1$ (proof: the sequence $\{kx \mod m \mid k \in {1,\cdots,m-1}\}$ must be all disjoint because none of $kx$ divides $m$ , and furthermore they must all be less than $m$ and positive; there are exactly $m-1$ disjoint positive number less than $m$ , so $\{kx \mod m \mid k \in {1,\cdots,m-1}\} = \{1,\cdots,m-1\}$ ).

Next, suppose $m = p^aq^b$ where both $p$ and $q$ are primes, then using the same argument as above, we can see that if $x \not\mid m$ , then $\{kx \mod m \mid k \in {0,\cdots,m-1}\} = \{0,\cdots,m-1\}$ ). Now, suppose that $x=\gamma \times p^cq^d, c<a,d<b$ , then it can only generate the sequence $0,\gamma p^cq^d,2\gamma p^cq^d,\cdots, \gamma(p^{a-c}q^{b-d}-1)p^cq^d, 0,\cdots$ , so it's length is $\frac{m}{p^cq^d}$ . It doesn't take much to also see that $p^cq^d = \gcd(p^aq^b,\gamma \times p^cq^d)$ , which is why $F(k) = F(\gcd(m,k))$ .

Here is another proof: Because there are finitely many combinations for the last 4 digits, at least one number must repeat. Clearly, once one number repeats, all the numbers after that number repeat. The goal is now to find the first number where they repeat.

Suppose that there were two four-digit numbers (that may start with 0), a and b, that would produce the same next number, c. In other words, a+x $\equiv$ b+x(mod 10000). We subtract x from both sides to get a $\equiv$ b (mod 10000). However, this means that the last 4 digits of a and b are the same. Therefore, a = b because they are both 4 digits long.

Suppose that the first number were not 0000 but some other integer m. Then we take the term before m. Because this is unique, this is the term before m both times that it repeats. This contradicts the assumption that it was the first number that repeats. However, if we take the first term that repeats to be 0000, there is no contradiction because there is no term before 0000 because that is what we started with.

We now must solve $x*k \equiv 0$ (mod 10000). In other words, we must find the minimum k such that 10000 | $x*k$ . In other words, we want to find the minimum k such that $10000*d=x*k$ for some integer d.

We rewrite x as c $*$ gcd(x, 10000). We now want to find the minimum k such that 10000 $*$ d=k $*$ c $*$ gcd(x, 10000). Because gcd(x, 10000) is a divisor of 10000, we can dividie both sides by gcd(x, 10000) and leave both sides as integers. We now have $\frac{10000}{gcd(x, 10000}*d=k*c$ . If we divide both sides by c, we get $\frac{10000}{gcd(x, 10000}*\frac{d}{c}=k$ . Therefore, d is a multiple of c. To minimize k, we let d = c. Therefore, $\frac{10000}{gcd(x, 10000}*\frac=k$ .