# Pigeonhole Principle Emergency Room problem

An emergency room physician was on duty for 36 hours. During this period there was at least one emergency case each hour, but no more than 50 emergencies total. Show that there was some period of consecutive hours during which there were exactly 21 emergency cases.

I tried making 36 boxes, and having at least 1 emergency in each box, so there's 14 leftover cases (objects) for each box. I'm not sure how to put these objects in the boxes to show that there was some period of consecutive hours with 21 emergency cases.

• Are you sure your question is right? – user2277550 Oct 14 '16 at 2:50
• I finished the other problems, I don't know why this question is wrong. – Jared Y Oct 14 '16 at 2:55

Define $a_i =$ total no. of surgeries till the $i$th hour. Then $1\le a_i\le50$. Now define $b_i$ to be $a_i + 21$. Then $1 \le b_i \le 71$.
Now the total number of terms in both the sequences is $72$, and each term takes one of only $71$ possible values. Applying the pigeonhole principle, there are $2$ terms which are equal. As each sequence is strictly increasing, we can't have $a_i=a_j$ or $b_i=b_j$. So $a_i = b_j$ for some $i$ and $j$, which implies $a_i = a_j + 21$. Which means between $i$th hour and $j$th hour , $21$ surgeries were performed.