1 A certain exam consists of 5 multiple-choice questions, each with 4 different options, and each person selects exactly one option per question. Among 2000 answer sheets, there exists a number $n$ such that in any $n$ answer sheets, there are 4 sheets where any two of them have at most 3 answers in common. Find the minimum possible value of $n$. (2000 China Mathematical Olympiad Problem)
1. The minimum possible value of $n$ is 25. Denote the 4 possible answers to each question as $1, 2, 3, 4$, and each test paper's answers as $(g, h, i, j, k)$, where $g, h, i, j, k \in \{1,2,3,4\}$. For all 2000 answer sets $(g, h, i, j, k)$, group those with the last 4 components identical into one class, resulting in $4^4 = 256$ classes. Since $2000 = 256 \times 7 + 208$, there must be 8 test papers in the same class $A$. Remove these 8 test papers, leaving 1992 test papers, of which 8 still belong to the same class $B$. Remove these 8 test papers again, leaving 1984 test papers, of which 8 still belong to the same class $C$. The 24 test papers removed belong to 3 classes $A, B, C$. Therefore, when $n \leqslant 24$, any $n$ test papers taken from these 24 will have any 4 test papers with at least 2 belonging to the same class, which does not meet the problem's requirements, so $n \geqslant 25$. Now, construct such 2000 test papers, which have 250 different answers, with 8 test papers for each answer. These 250 different answers are any 250 of the $4^4 = 256$ answers $(g, h, i, j, k)$ that satisfy $g + h + i + j + k \equiv 0 \pmod{4}$. Clearly, for any 2 different answers, they can have at most 3 components the same; otherwise, if 4 components are the same, then by the congruence, the 5th component would also be the same, which is a contradiction. In such 2000 test papers, any 25 taken will have at least 4 test papers that are pairwise different, and they can have at most 3 components the same, so $n = 25$ meets the problem's requirements.