Al has 75 days to master discrete mathematics. He decides to study at least one hour every day, but no more than a total of 125 hours. Assume Al always studies in one hour units. Show there must be a sequence of consecutive days during which he studies exactly 24 hours.

Step 1
Let $a_k$ be the number of hours of work Al has done after k days. Then $\{a_1,a_2,a_3,\dots ,a_{75}\}$ is an increasing sequence of distinct positive integers since Al does at least one hour of work each day. Observe that $1\le a_k\le 125$ since Al does at most 125 hours of work over the 75 days.
Let $b_k=a_k+24$. Then the sequence $\{b_1,b_2,b_3,\dots ,b_{75}\}$ is also an increasing sequence of distinct positive integers. Observe that $1+24=25\le b_k\le 149=125+24$.
Step 2
Now consider the union of the two sequences. It consists of 150 numbers that are at least 1 and at most 149. Thus, two of them must be the same. Hence, $b_k=a_k+24=a_j$ for some j,k. Thus, $a_j-a_k=24$, so Al does exactly 24 hours of work from day $a_{k+1}$ to day $a_j$.
This is a clever application of the Pigeonhole Principle that forced me to consult my combinatorics notes.