Problem: In the following 30 days you will get 46 homework sets out of which you will do at least one every day and - of course - all during the 30 days. Show that there must be a period of consecutive days during which you will do exactly 10 homework sets!
Solution: Let $f_n$ denote the number of homeworks from day $1$ to day $n$, where $n\le 30$. So, let us consider from $f_1$ up to $f_{11}$. There are ten possibilities for the remainder when each is divided by $10$. By the pigeonhole principle, there must exist two that have the same remainder, call these $f_i$ and $f_j$, for some $i,j\in [1,11]$. Therefore $$f_i - f_j \equiv 0 \pmod{10}.$$ But also $f_i - f_j \not = 20$. Hence $f_i - f_j = 10$.
I think this is on the right track. However, I have not convinced myself that $f_i - f_j \not = 20$
from Hot Weekly Questions - Mathematics Stack Exchange
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