Problem 6

In a mathematical competition, in which

6

problems were posed to the participants, every two of these problems were solved by more than

\frac25

of the contestants. Moreover, no contestant solved all the

6

problems. Show that there are at least

2

contestants who solved exactly

5

problems each.

Step 5 of 6: Reduce modulo

3

for any split of

\{1,2,3,4,5\}

Previous step Next step

t_{de}\equiv 1+t_a+t_b+t_c+t_{ab}+t_{bc}+t_{ca}\pmod 3\quad(\{a,b,c,d,e\}=\{1,2,3,4,5\})

Detailed analysis

Reducing the identity of Step 4 modulo $3$ gives $t_{45}\equiv 1+t_1+t_2+t_3+t_{12}+t_{23}+t_{13}\pmod3$ . Since the setup is symmetric in the five problems $1,2,3,4,5$ , the same congruence holds for every partition of $\{1,2,3,4,5\}$ into a triple $\{a,b,c\}$ and a pair $\{d,e\}$ .