A teacher wants to divide the $2010$ questions she asked in the exams during the school year into three folders of $670$ questions and give each folder to a student who solved all $670$ questions in that folder. Determine the minimum number of students in the class that makes this possible for all possible situations in which there are at most two students who did not solve any given question.
Problem
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Tags: combinatorics proposed, combinatorics
AmisitRegina
25.12.2020 21:49
Let's say there are $x$ students in the class. If $x=4$, it is not possible to have three students who can take a folder in every situation. For example; Say first student's correct questions are $1, 2, 3, ..., 669$ second student's are $670, 671, ..., 1338$ third student's are $1339, 1340, ..., 2007$ and last student made all questions correct. It is clear that there is no question that three students made incorrect in common. So $4$ doesn't provide conditions. for $x=5$ we could say the same, the only difference is fifth student, which could solve every problem correctly and in this way $5$ doesn't provide conditions. Say we have $6$ students in this class. We have $2\times 2010$ rights to make incorrect questions. A student should have at least $1341$ incorrect, if this student doesn't get any folder. We need $4$ students who can't get a folder but $1341\times 4 = 5364$ and $2\times 2010 =4020$. So we can't have 4 students out of 6, who can't get a folder. $6$ provides the condition.