Jessica is studying combinatorics during a $7$-week period. She will
study a positive integer number of hours every day during the $7$
weeks (so, for example, she won't study for $0$ or $1.5$ hours), but
she won't study more than $11$ hours in any $7$-day period. Prove that
there must exist some period of consecutive days during which Jessica
studies exactly $20$ hours.
Here are my thoughts so far: Let $f(n)$ represent the total number of hours Jessica has studied after day $n$. Clearly, there are $49$ days in total, and the domain of $f$ is integers in the interval $[0,49]$. Proving that there must exist some period of consecutive days during which Jessica studies exactly $20$ hours is equivalent to proving that there must exist $i$ and $j$ such that $f(i)-f(j)=20$.
This is a really interesting question, but I don't see a clear path forward. How do you solve this question? Please try not to use extremely advanced math or I won't understand 😛
Best Answer
HINT: Let $S=\{f(k):k=0,\ldots,49\}$; you want to show that there is some $k$ such that $f(k)+20\in S$. (You want to allow $k=0$ to cover the possibility that some $f(k)=20$; $f(0)$ is of course $0$.)
You know that $f(49)\le 7\cdot11=77$, so all of the $100$ integers $f(k)$ and $f(k)+20$ for $k=0,\ldots,49$ are in the range $[0,97]$. However, there are only $98$ integers in that range. Can you see how to apply the pigeonhole principle to get the desired result? I’ve finished the argument in the spoiler-protected block below.