# Pigeonhole-principle with two choices

I am able to solve this sort of problem pretty easily.

An arm wrestler is the champion for a period of 75 hours. The arm wrestler had at least one match an hour, but no more than 125 total matches. Show that there is a period of consecutive hours during which the arm wrestler had exactly 24 matches.

However this homework problem I can't seem to figure out.

Tyler has five weeks to prepare for his driver’s test. His mother volunteers to ride with him for either 15 minutes or one half hour every day until the test but not for more than 15 hours in all. Show that during some period of consecutive days, Tyler and his mother will drive for exactly 8.75 hours.

If I try doing it the same way as the first problem. I get $70$ pigeons and $95$ pigeon holes. I honestly don't know where to start with this problem. I thought that maybe I could assume that they only drove $.25$ hours on each day and prove that there is a period of consecutive days where they drove for exactly $8.75$ hours. Then I could say that we assume there is one day that they drove for $.5$ hours and work my way up, but I can't seem to get that to work either.

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Let $F(n)$ be the number of hours driven from day 1 to day $n$ inclusive, with $F(0) = 0$. Then $F(n) - F(n-1)$ is either $1/2$ or $1/4$, for $n$ from $1$ to $35$. Say day $b$ is the first where $F(b) >= 8.75$. Then we must have $F(b) = 9$ and $F(b-1) = 8.5$. Moreover, $F(1) = .5$ since otherwise we could take days $2$ to $b$. Thus they must have driven $1/2$ hour on at least two days ($2$ and $b$), so a total of at least 9.25 hours. Therefore $b < 35$ and $F(b+1)$ is either $9.25$ or $9.5$. It can't be $9.25$, otherwise we could take days $2$ to $b+1$, so it must be $9.5$. Moreover, they must have driven $1/2$ hour on the second day, otherwise you could take days $3$ to $b+1$. So there were at least $4$ half-hour days $(1,2, b, b+1)$, which implies a total of at least 9.75 hours. Repeat...
Let $F(n)$ be the number of $15$ minute intervals Tyler practices from day $1$ to day $n\leq 35$ inclusive, with $F(0)=0$. First we assume that $F(i) \equiv 0\pmod{35}$. From this assumption there must exist an integer $k$ such that $F(i) = 35k$. Because $0<F(i)\leq 60$, $F(i) = 35$. We must assume that this is not the case, because it leads to the hours driven from the $1^{st}$ day to the $i^{th}$ day being equal to $8.75$. Applying PHP we now that there must be two positive integers $i$ and $j$ such that $F(i)$ and $F(j)$ are in the same congruence class of $35$. From this we can conclude that there exist positive integers $i<j \leq 35$ such that $F(i) \equiv F(j)\pmod{35}$. From this we know that there exists an integer $k$ such that $F(j)-F(i) = 35k$. Because $0<F(j)-F(i)\leq 60$, $F(j)-F(i) = 35$. This means that the sum of the hours driven from the $(i+1)^{th}$ day to the $j^{th}$ day is $8.75$. $\square$