Hum maybe i miss something but if a student has $0$ il all the tests. He has no chance for being a team member?

mamouaz1 wrote: Hum maybe i miss something but if a student has $0$ il all the tests. He has no chance for being a team member? I think it asks if there exists a configuration of rankings so that for every student, there is a permutation of the tests in which that student qualifies. Edit: I live in Sugar Land

rafayaashary1 wrote: I live in Sugar Land That's pretty intresting

What is the maximum number of students there can be so that there is still a chance of being a team member?

MathPanda1 wrote: What is the maximum number of students there can be so that there is still a chance of being a team member? I think there was an example for $14$ students. But don't have an idea about the maximum number.

TEST: 1 2 3 4 5 6 A A A A A A B B B B B B C C C D D D D D D C C C E E F F G G H I J K L M Satisfies the conditions

MathPanda1 wrote: What is the maximum number of students there can be so that there is still a chance of being a team member? Here is an example for $14$ students: \begin{tabular}{l*{5}{c}r} Test & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline 1st place& A & A & A & A & A & A \\ 2nd place & B & B & B & B & B & B \\ 3rd place & C & C & D & D & E & E \\ 4th place& F & G & G & H & H & F \\ 5th place& I & J & K & L & M & N \\ \end{tabular} For example, $I$ can get into the team with the order $(4,3,5,2,6,1)\rightarrow(A,B,E,C,F,I)$.

"That's an awfully tricky yes-no question to put on a TST..." You should know about pre-history of a problem: https://artofproblemsolving.com/community/c6h46293p292638

$\textbf{Yes.}$ This is in fact possible, with the following setup, where each column represents the top 6 scorers on each problem, and the contestants are numbered $01,02,\ldots,13$. 01 01 01 01 01 01 02 02 02 02 02 02 03 03 03 03 03 03 04 04 04 04 04 04 05 05 06 06 07 07 08 09 10 11 12 13

It is indeed possible. Number the students 01 to 13 and suppose that the top 6 on each problem is as follows 01 01 01 01 01 01 02 02 02 02 02 02 03 03 03 03 03 03 04 04 04 04 04 04 05 05 06 06 07 07 08 09 10 11 12 13 And assign the rest of the rankings arbitrarily. Then in any IMO team, students 01 through 04 must be included. Students 05 and 09 can be included through the permutation 654312, and something similar can be sued to include any of the other students. $\blacksquare$

\begin{tabular}{l*{5}{c}r} Test & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline 1st place& 1 & 1 & 1 & 2 & 2 & 2 \\ 2nd place & 3 & 4 & 5 & 3 & 4 & 5 \\ 3rd place & 6 & 7 & 8 & 8 & 6 & 7 \\ 4th place& 9 & 10 & 11 & 12 & 13 & 14 \\ \end{tabular} is a shorter example for $14$. You work backwards to verify that this works, notice that showing any $4$th place student can get into the team is sufficient. Let me demonstrate an example for any student in $5$th place say $11$: To get $11$ you must get students $8,5,1$ into the team first, you have to get $8$ from Test $4$ and $5$ from Test $6$ which are always different by construction. And you should also take student $3$ from Test $1$ before you start taking $8$, notice that Test $1$, $4$, $6$, $3$ are all different again. So you first take the left out tests, $2$, $5$ in this case, then take $1$, $4$, $6$, $3$ in order.

1 2 3 5 8 1 2 3 6 9 1 2 3 7 A 1 2 4 5 B 1 2 4 6 C 1 2 4 7 D $\blacksquare$

Quite difficult but very satisfying to solve. Consider the construction $$\begin{tabular}{c|c|c|c|c|c|c} & Problem 1 & Problem 2 & Problem 3 & Problem 4 & Problem 5 & Problem 6 \\ \hline Student 1 & 1 & 1 & & 2 & & \\\hline Student 2 & 2 & & 1 & 1 & & \\\hline Student 3 & & 2 & & & 1 & \\\hline Student 4 & & & 2 & & & 1 \\\hline Student 5 & 3 & 3 & & & & \\\hline Student 6 & 4 & & & & & \\\hline Student 7 & & 4 & & & & \\\hline Student 8 & & & 3 & 3 & & \\\hline Student 9 & & & 4 & & & \\\hline Student 10 & & & & 4 & & \\\hline Student 11 & & & & & 2 & 2 \\\hline Student 12 & & & & & 3 & \\\hline Student 13 & & & & & & 3 \end{tabular}$$ where the columns denote problems, rows denote students, and the entry in $(r, c)$ corresponds to the rank of student $r$ on problem $c$. The places left blank can be chosen arbitrarily. Now, observe that: Students $1, 2, 3, 4$ can get on the team as if the problem they are ranked first on is chosen first; Student $5$ can get on the team if the problems are chosen in the order $3, 4, 5, 1$; Student $6$ can get on the team if the problems are chosen in the order $3, 4, 5, 2, 1$; Student $7$ can get on the team if the problems are chosen in the order $3, 4, 5, 1, 2$; Students $8$ through $10$ can get on the team by symmetry (with $3, 4$ replaced with $2, 1$ and so on). Student $11$ can get on the team if the problems are chosen in the order $4, 1, 2, 3, 5$; Student $12$ can get on the team if the problems are chosen in the order $4, 1, 2, 3, 6, 5$; Student $13$ can get on the team if the problems are chosen in the order $4, 1, 2, 3, 5, 6$. So all $13$ students can get on the team, which concludes the proof. Remark: The underlying idea behind the construction was to use the top $2 \times 4$ square to easily be able to eliminate top ranks; then, for each pair, we would be able to uplift three students who were ranked lower, which suffices for the $13$ students.

Very tricky question. We can $choose$ This $ configuration$ satisfies 1 2 5 9 1 2 6 10 1 3 7 11 1 3 5 12 1 4 6 13 1 4 7 8

What a great problem! Label the students with the numbers the letters $A$ through $M$. Motivation: Notice that the following 'pyramid' structure is very natural to consider. If the tests were chosen in numerical order, then $A$ would be admitted first, in the second test $B$ would be admitted as $A$ has already been chosen, and so on until $F$ is admitted. \begin{tabular}{l*{5}{c}r} Test & 1 & 2 & 3 & 4 & 5& 6 \\ \hline 1st& A & A& A & A& A& A\\ 2nd& {} & B & B & B & B& B \\ 3rd & {} & {} & C & C & C & C \\ 4th & {} & {} & {} & D & D & D \\ 5th & {} & {} & {} & {} & E & E \\ 6th & {} & {} & {} & {} & {} & F\\ \end{tabular}This is not necessary but it is sufficient in that every student that appears at the bottom of a pyramid (size $1$ to $6$) in some permutation can be possibly chosen. This idea leads naturally to the following construction. Construction: \begin{tabular}{l*{5}{c}r} Test & 1 & 2 & 3 & 4 & 5& 6 \\ \hline 1st& A & A& A & A& A& A\\ 2nd& B & B & B & B & B& B \\ 3rd & C & C & C & C & C & C \\ 4th & G & G & G & D & D & D \\ 5th & H & H & H & E & E & E \\ 6th & I & J & K & L & M & F\\ \end{tabular}

Great problem! (i love these yes no problems, specially this kind) Constriction: \begin{tabular}{l*{5}{c}r} Test & 1 & 2 & 3 & 4 & 5& 6 \\ \hline 1st& A & A& A & A& A& A\\ 2nd& B & B & B & B & B& B \\ 3rd & C & C & C & C & C & C \\ 4th & D & D & D & D & D & D \\ 5th & E & E & F & F & G & G \\ 6th & M & L & K & J & I & H\\ \end{tabular}It is not difficult to verify how this works.

Yes, consider \[\begin{array}{|c|c|c|c|c|c|} \hline 1 & 2 & 3 & 4 & 5 & 6 \\ \hline A & A & A & A & A & A \\ \hline B & B & B & B & B & B \\ \hline C & C & C & C & C & C \\ \hline D & D & D & D & D & D \\ \hline E & E & F & F & G & G \\ \hline H & I & J & K & L & M \\ \hline \end{array}\]