In this week’s Riddler Classic we are asked to find the highest posssible score (where high is bad) in a 3 part climbing competition that could conceivably result in a win or overall tie. There are 8 contestants and no ties in the 3 individual events. Scoring is based on the the product of the finishes in the 3 individual events.

We could easily simulate this but look for a heuristic approach instead. We note that:

The worst possible score for any of the three different contest winners is (1,7,8) = 56. Therefore our highest possible winning answer can be no higher than 56.

If we look at possible scores from 56 down we see the first few are: 56 = [(1,7,8), (2,4,7)], 50 = (2,5,5), 49= [1,7,7], 48 = [(3,4,4), (2,3,8), (2,4,6), (1,6,8)].

For 56 if we use 3 (1,7,8) triplets for the individual contest winners then the remaining possible allocation for the second place finishes is (2,5,6) = 60 which we need to use three times to stay >= 56. That leaves us with three 3’s and three 4’s to allocate among two remaining contestants, and there is no way to do that and remain above 56. If we use (2,4,7) then we are left with a maximum for one of the winners of (1,6,8) = 48 and so 56 is out.

Continuing on in similar fashion we see that constraints on the 1’s and 2’s pretty much determine the story:

.

So the answer to the Classic is 48, with four different possibilities for one player acheiving 48 and examples above (not unique) for how the other players could score to make it work.