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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

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$a(i)$ being the number of problems solved until the $i$th day,we have for any consecutive $11$ weeks $1≤a(1)<a(2)<...<a(77)≤132$ and $21≤a(1)+20<a(2)+20<...<a(77)+20≤152$ .Then we have a total of 154 positive distinct 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....$(i)$th day a total of exactly $20$ problems.

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