Originally Posted by

**Fox**
I don't know how to prove this mathematically, so I put the data in a grid. Along the top of the grid I arranged all the possible combination of pairs of problems.

1,2 1,3 1,4 1,5 1,6 2,3 2,4 2,5 2,6 3,4 3,5 3,6 4,5 4,6 5,6

Down the left side of the grid I arranged the contestants with the problems they got wrong. I chose 15 contestants because it is divisible by 5 and because they could represent every possible combination of wrong problems. I made the assumption that attempting to give each contestant 4 "rights" and 2 "wrongs" and distributing the wrong problems as evenly as possible would maximize the number of contestants who got each problem pair right while minimizing the number of contestants who got 5 problems right with no perfect scores.

What I ended up with was a symmetrical grid with the same numbers across the top and down the left. I either put an X in each box of the grid or left it blank depending on whether a given contestant with 2 wrong problems had that problem pair right. Each row and, more importantly, each column had exactly 6 X's. 6 is 2/5 of 15. But since the question stipulated that each problem pair was soved by MORE than 2/5 of the contestants, they couldn't ALL have gotten only 4 right. In fact, in my example, forcing each problem pair to be represented by 7 contestants (the smallest number that is MORE than 2/5) caused ALL of the contestants to have at least 5 right answers.

I don't know whether a different combination of number of contestants or distribution of wrong answers would yield a better result.