I'm posting this answer to my own question because I think I found a solution, but I would appreciate feedback on whether the solution below is complete.
The only detail that is missing to complete part d is that I need to show that the statement is true if 24 in the original statement is replaced by 30; that is, I will try to show that there is a period of consecutive hours during which the arm wrestler had exactly 30 matches (since there seems to be no counter-example for this case).
I think I found a way to prove it, based on the suggestion given by Lopsy in the comments to the question. This attempt is very similar to the accepted answer here: Prove that 2 students live exactly five houses apart if.
We build the following sets: the 5-element sets $\{1, 31, 61, 91, 121\}$, $\{2, 32, 62, 92, 122\}$, ..., $\{5, 35, 65, 95, 125\}$, and the 4-element sets $\{6, 36, 66, 96\}$, $\{7, 37, 67, 97\}$, ..., $\{30, 60, 90, 120\}$. The purpose of these sets is to represent every possibility of number of matches played so far. They encompass all numbers from 1 to 125 and, inside each set, the difference between adjacent numbers is 30; however, there are no numbers from two different sets whose difference is 30. There are 5 five-element sets, and 25 four-element sets.
Now, if I want to choose 75 numbers from all sets so that there are no two numbers whose difference is 30, I have to choose at most 3 elements from the sets with 5 elements (for example, I can choose 1, 61 and 121 from the set $\{1, 31, 61, 91, 121\}$; if I choose one more, then clearly two of them will have difference equal to 30), and at most 2 elements from the sets with 4 elements (for example, I can choose 30 and 120 from $\{30, 60, 90, 120\}$, but, if I choose one more, then clearly two of them will have difference 30). There are 30 sets in total; since there are 5 sets with 5 elements and 25 sets with 4 elements, the maximum number of elements that can be chosen so that no two of them have 30 as difference is $3\times 5 + 2\times 25 = 15 + 50 = 65$ numbers. But I need to choose 75 numbers. So, there must be some numbers whose difference is 30.
Does this make sense, or is there some mistake?
Are you familiar with differential equations? Solving linear recurrences is the same as solving linear differential equations, essentially. Linear recurrences (ie., difference equations) are discrete differential equations.
So for $a_{n} = 4a_{n-1} - 3a_{n-2}$, we start by setting up our characteristic polynomial: $\lambda^{2} - 4\lambda + 3 = 0$. You then solve for $\lambda$, getting solutions $\lambda = 1, 3$. And so your general form equation $a_{n} = c_{1}(\lambda_{1})^{n} + c_{2}(\lambda_{2})^{n}$.
To set up the characteristic polynomial, you look at the difference of the indices. So for $a_{n}$, we see indices $n, n-1, n-2$. Clearly there is a difference of two between the largest and smallest index. So we have a quadratic. The largest indexed term has the largest power. So $a_{n} \to \lambda^{2}$, $4a_{n-1} \to 4\lambda$, and $-3a_{n-2} \to -3$. Do you see how I got that?
So $a_{n} = c_{1} + c_{2}3^{n}$. Then plug in your constraints and solve for $c_{1}$ and $c_{2}$.
Now for your example with Fibonacci, we have $a_{n} = a_{n-1} + a_{n-2}$, we have another quadratic. Do you see why? So $a_{n} \to \lambda^{2}$, $a_{n-1} \to \lambda$, $a_{n-2} \to 1$. And so our characteristic polynomial is $\lambda^{2} - \lambda - 1 = 0$. Solve for $\lambda$, which gives you two distinct roots $\lambda_{1}, \lambda_{2}$, then plug in: $a_{n} = c_{1} \lambda_{1}^{n} + c_{2}\lambda_{2}^{n}$ and solve the system of equations from your initial conditions.
Now for a first order non-homogenous recurrence of the form $a_{n} = ra_{n-1} + c$, we get a solution of the form $a_{n} = kr^{n} + c \frac{r^{n}-1}{r -1 }$ for $r \neq 1$. If $r = 1$, we get $a_{n} = k + cn$.
The constant $k$ is what we solve for based on the initial conditions.
Best Answer
If the current time is $t$, and the time $n$ hours into the future is $f$, using a $24$ hour clock, we have:
$t$ is the current time. $t+n$ is the future time. $f$ is also the future time. Thus:
$$t+n\equiv f\pmod{24}$$
using a twelve hour clock, this is instead simply $t+n\equiv f\pmod{12}$
If the current time is 2:00, and we want to know what the future time is 100 hours into the future, we have:
$f\equiv t+n\equiv 2+100\equiv 102\equiv 4\cdot 24+6\equiv 6\pmod{24}$
so the future time will read 06:00
If the current time is $t$ (currently unknown), and $100$ hours into the future the clock reads $2:00$ (the future time), we have:
$t+n\equiv f\pmod{24}$ implying
$t\equiv f-n\equiv 2 - 100\equiv -98\equiv -5\cdot 24 + 22\equiv 22\pmod{24}$
so the current time reads 22:00
Similarly, we could talk about time into the past.
If the current time is $t$ and the time $n$ hours before it into the past is $p$, using a $24$ hour clock we have:
$t$ is the current time. $t-n$ is the past time. $p$ is the past time. Thus:
$$t-n\equiv p\pmod{24}$$
If the current time is 12:00 and we are curious what time it was 45 hours into the past, we have:
$p \equiv t-n\equiv 12 - 45\equiv -33\equiv -2\cdot 24 + 15\equiv 15\pmod{24}$
thus, the time 45 hours in the past was 15:00
If the current time is unknown, $t$, but we know that 45 hours into the past the time had at that point read as 12:00, we have:
$t\equiv n+p\equiv 45+12\equiv 57\equiv 2\cdot 24 + 9\equiv 9\pmod{24}$
so, the current time is 9:00.
Again, in all of these examples, if you were using a twelve hour clock instead, you would use modulo twelve. In each case, you are asking "how much larger than the closest smaller multiple of twenty-four (or twelve as the case may be) is my number?"