This is for self-study. This question is from Rosen's "Discrete Mathematics And Its Applications", 6th edition.
An arm wrestler is the champion for a period of 75 hours. (Here, by an hour, we mean a period starting from an exact hour, such as 1 P.M., until the next hour.) The arm wrestler had at least one match an hour, but no more than 125 total matches.
1 - Show that there is a period of consecutive hours during which the arm wrestler had exactly 24 matches.
2 - Is the statement in the previous exercise true if 24 is replaced by
a) 2? b) 23? c) 25? d) 30?
My solution to part 1 is the following, based on Rosen's solution to a similar problem given as example in the text:
1 - If I consider $a_i$ to be the number of competitions until the $i^{th}$ hour, then, $1\leq a_1
There are 150 numbers $a_1,\cdots,a_{75},a_1+24,\cdots,a_{75}+24$. By the inequalities above, these numbers range from 1 to 149. Then, by the pigeonhole principle, at least two of them are equal (in a list of 150 integers ranging from 1 to 149, at least two are equal).
Because all the numbers $a_1,\cdots,a_{75}$ are distinct, and all numbers $a_1+24,\cdots,a_{75}+24$ are also distinct, it follows that $a_i = a_j + 24$ for some $i > j$. Therefore, from the $(j+1)^{th}$ hour to the $i^{th}$ hour, there were exactly 24 competitions.
Now, I will show the attempt at a solution for part 2:
2 - a) The same reasoning as above can be used:
$1\leq a_1
$3\leq a_1 + 2
So, there are 150 numbers that range from 1 to 127. Therefore, by the pigeonhole principle, at least $\left \lceil \frac{150}{127} \right \rceil$ = 2 numbers must be equal. So, there is an $a_i = a_j + 2$. This guarantees that there is a period of consecutive hours during which there were exactly 2 competitions.
b) The same reasoning as above can be used:
$1\leq a_1
$24\leq a_1 + 23
So, there are 150 numbers that range from 1 to 148. Therefore, by the pigeonhole principle, at least $\left \lceil \frac{150}{148} \right \rceil$ = 2 numbers must be equal. So, there is an $a_i = a_j + 23$. This guarantees that there is a period of consecutive hours during which there were exactly 23 competitions.
c) $1\leq a_1
$26\leq a_1 + 25
In this case, there are 150 numbers that range from 1 to 150. So, there are not necessarily two equal numbers. Therefore, we can't conclude anything directly.
But it is possible to show that the statement is not true for 25, because an explicit counter-example (suggested below in the comments) can be found. Suppose the number of matches until each one of the 75 hours is, respectively: $\{1,2,\cdots,25,51,\cdots,75,101,\cdots,125\}$. Here, there are no pairs of numbers whose difference is 25.
d) $1\leq a_1
$31\leq a_1 + 30
In this case, there are 150 numbers that range from 1 to 155. So, we can't apply the pigeonhole principle here, similarly to the above situation. It is not easy to find a counter-example in this case; I think that, for 30, the statement is always true (that is, there is always a period of consecutive hours during which there were exactly 30 matches). But I'm not sure how to prove it.
Edit: I think I found a way to prove it, based on the suggestion given by Lopsy; I've included it as an answer to this question.
Thank you in advance.