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
$a_i$ being the number of problems solved until the $i^{th}$ day, we have for any consecutive $11$ weeks $$1\leq a_1<a_2<\ldots<a_{77}\leq 132$$ and $$21\leq a_1+20<a_2+20<\ldots<a_{77}+20\leq 152$$ Then we have a total of $154$ positive integers all of which are less than or equal to $152$.Then by pigeon hole principle there exist $i$ and $j$ such that $a_i=a_j+20$. Then $a_i-a_j=20$. Then the problem solver has solved on $(j+1)^{th}$ day,$(j+2)^{th}$ day $\ldots$ $i^{th}$ day a total of exactly $20$ problems.