During a common year a student solves at least $1$ problem in combinatorics, each day. But, so he would not get too tired, he solves at most $12$ problems per week. Prove that there exist several consecutive days during which student has solved, exactly, $20$ problems.
What I have tried is dividing $365$ days in $19$ $"boxes"$. Hence $365 > 19*19$ there must be a $"box"$ with $20$ consecutive days. Also, hence $19 = 2*7+5$ he can't have solved more than $2*12+5$ problems. But from this I cannot infer how I get there is no $21$ days in the $"box"$.
Is my approach even good? Thanks in advance