The Brooklyn 99 Twelve Islanders problem.

The problem is:

1. There are eleven islanders who all weight exactly the same and one who is either heavier or lighter than those eleven.
2. You have a see-saw with accurately marked seats that you can use as weighing scale, and can use it three times.
3. Can you always solve it in a maximum of three comparisons?

After four or five hours of trial and error, I have found that the answer is “yes” come up with a bomb-proof system (one of no doubt dozens), pdf below (click to enlarge).

You start by numbering the islanders 1 to 12.
Let’s assume 6 is heavier, but you don’t know that yet.
Test 1 is always 1,2,3,4 on one side of the see-saw against 5,6,7,8 on the other.
Result 1 is 5,6,7,8 are heavier, so go to the second table.
Test 2 is 1,2,5,6 against 3,7,9,10.
Result 2 is 1,2,5,6 are heavier.
Test 3 is 3,5 against 7,9.
Result 3 is both sides weight the same
Answer is then read off in the bottom row of the second table, “6 H”

If anybody spots a mistake, please leave a comment