Look here for a very similar problem: http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2439466&sid=cacae6af292a61637bc1f88b8bfe0bbe#p2439466

hmm,I cannot see any relevance between the link above and pro itself. By the way,solution: Represent everyone's points with littices such that $(x,y)$ represents one's points in the 1st and the 2nd problem.Partition the $7*7$ points set into several groups. $A_1=\{ (x,y)|0\le x\le 7,y=0,or,x=7,1\le y\le 7 \},A_2=\{ (x,y)|0\le x\le 6,y=2,or,x=6,2\le y\le 7 \},A_3=\{ (x,y)|0\le x\le 5,y=2,or,x=5,3\le y\le 7 \},A_4=\{ (x,y)|0\le x\le 4,y=3,or,x=4,4\le y\le 7 \},A_5=\{ (x,y)|x=2,3,4\le y\le 7 \},A_6=\{ (x,y)|x=0,1,4\le y\le 7 \}$. By pigeonhole,there must be $\lfloor \frac {49-1}{5} \rfloor+1=9$ points in some group.If $9$ points in $A_5$ or $A_6$,there are $2$ points coincide,take the higher points on 3rd pro be $A$ is ok.If $A_i,i=1,2,3,4$ has $9$ points,observe there is one students whose $1$ and $2$ pro is better than the other one,also the 3rd pro has only $8$ possible points,so by pigeonhole,there are $2$ has same points on 3rd pro.Done.

mathmaths wrote: hmm,I cannot see any relevance between the link above and pro itself. Change the bounds to $0\leq a,b,c\leq 7$ and let $a$ be the score on the first problem, $b$ the score on the second, etc...

Phrasing mathematically, we have $49$ triples $(x, y, z)$ of integers from $0$ to $7$. Thus, it suffices to show that there exist two triples $(x, y, z)$ and $(x', y', z')$ such that $x \ge x'$, $y \ge y'$, and $z \ge z'$. To make this a divisors condition, we can reformulate the condition as \[2^{x'} \cdot 3^{y'} \cdot 5^{z'} | 2^x \cdot 3^y \cdot 5^z.\] Suppose for the sake of contradiction that among any subset of the positive divisors of $N = 2^7 \cdot 3^7 \cdot 5^7$ with $49$ elements, there exist no two divisors such that $d_1 | d_2$. This is equivalent to there existing an antichain of length $49$ of $N$. However, by the Maximal Antichain Lemma, the maximal size of an antichain is the number of solutions to the equation $x + y + z = \left\lfloor\frac{7 + 7 + 7}{2}\right\rfloor = 10$ with $0 \le x,y,z \le 7$. By Stars and Bars with PIE or pure bash, we get that the number of solutions is $\binom{12}{2} - 3\binom{4}{2} = 66 - 18 = 48$. This is a contradiction, so there must always exist two divisors such that $d_1 | d_2$.

Would you be willing to give a citation for this lemma?

aryabhata000 wrote: Would you be willing to give a citation for this lemma? It’s essentially a generalization of Sperner’s Theorem, where we consider antichains of positive divisors instead of sets. Rough sketch: First, prove the BTK Theorem, which says that the set of positive divisors of $n$ can be partitioned into symmetric chains. A simple induction on the number of distinct prime factors of $n$ will suffice. Then, construct a bijection between the number of symmetric chains and the number of solutions to the particular equation, by considering the middle term in every symmetric chain. Then, note that any antichain can pick only one element from each symmetric chain by definition, so we get the conclusion.