SAUDI ARABIAN IMO Booklet 2023

Number the examiners from $1$ to $10$ and consider the results of contestants as binary sequences of the form $x_1{x}_2\dots x_{10}$, in which $x_i=0$ or $1$ if the $i$th examiner gave pass or fail response for the corresponding candidate. We can directly construct $24$ binary strings that satisfy the “separative” condition. Indeed, choose strings of the form $(x_1{x}_2{x}_3{x}_4{x}_5{x}_6)(x_7{x}_8{x}_9{x}_{10})$, where: $x_1{x}_2{x}_3{x}_4{x}_5{x}_6$ contains exactly $3$ zeros and $3$ ones; and $x_7{x}_8{x}_9{x}_{10}$ will contain all binary strings of length $4$. Since $\left(\genfrac{}{}{0ex}{}{6}{3}\right)=20$ and two strings $x_1{x}_2{x}_3{x}_4{x}_5{x}_6\ne x_1^{\prime }{x}_2^{\prime }{x}_3^{\prime }{x}_4^{\prime }{x}_5^{\prime }{x}_6^{\prime }$ would be different at at least $2$ positions, so the rest just need to differ by at least $1$ position. Since $2^4=16$, there will be $4$ repeated strings at the last $4$ positions, we just need to choose those strings that differ by at least $3$ positions out of the first $6$ positions. For detail, $$\begin{array}{lcl}(000111)(0000)& \to & (a)\\ (001011)(0001)& \to & (b)\\ (001101)(0010)& \to & (c)\\ (001110)(0011)& \to & (d)\\ (010011)(0100)& & \\ (010101)(0101)& & \\ (010110)(0110)& & \\ (011001)(0111)& & \\ (011010)(1000)& & \\ (011100)(1001)& & \\ (100011)(1010)& & \\ (100101)(1011)& & \\ (100110)(1100)& & \\ (101001)(1101)& & \\ (101010)(1110)& & \\ (101100)(1111)& & \\ (110001)(0011)& \to & (d^{\prime })\\ (110010)(0010)& \to & (c^{\prime })\\ (110100)(0001)& \to & (b^{\prime })\\ (111000)(0000)& \to & (a^{\prime }).\end{array}$$ Finally, add $4$ more strings as the following: $$\begin{array}{l}(000000)(0000),(000000)(1111),\\ (111111)(0000),(111111)(0000).\end{array}$$ It is easy to check that they are pairwise separative and also separative with the other $20$ strings. $\square$