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Assume an integer number of hours will be worked each day for $49$ consecutive days. Further assume that at least $ 1 \frac{\text{hrs}}{\text{day}}$ and at most $11 \frac{\text{hrs}}{\text{wk}}$ can be worked. Given these assumptions, we are to show that $$ \text{there exists a sequence of days such that exactly} \ 20 \ \text{hours are worked.} \tag{1}$$

It seems that $(1)$ can be shown through some application of number theory or by the pigeonhole principle. How would you approach a proof for this?

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Let $x_i$ be the number of hours worked up to (and including) the $i$th day. At least one hour is worked each day so the sequence $\{x_i\}_{i=1}^{49}$ is strictly increasing. The restriction of $11$ hours per week in $7$ weeks gives us that $x_{49} \leq 77$.

Now consider the sequence $x_1, x_2, \ldots x_{49}, x_1+20, x_2 + 20, \ldots, x_{49}+ 20$ which is increasing and ranges from $1$ to $97$. There are $96$ integers in $[1,97]$ and the sequence has $98$ distinct elements. The pigeonhole principle implies there are two elements of the sequence that are equal. These elements have to be of the form $x_i$ and $x_j+20$ since $\{x_i\}_{i=1}^{49}$ and $\{x_i+20\}_{i=1}^{49}$ are strictly increasing. It follows that from the $(j+1)$-th day up to the $i$-th day, exactly $20$ hours were worked.

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