There are 12 people on an island and one is heavier than the others. You have to figure out which Islander is the heaviest using a seesaw. But you can only use the seesaw 3 times. How do you figure out which one is heavier?

Divide them into 3 groups of 4 people.

Put any two groups on each side of the see-saw. (First Use)

Condition 1

If the see-saw balances, we are sure that the oddly weighted one is in the other group of 4.

In that case, take two people from that group and place them on one end of the seesaw and two of the balanced eight on the other. (Second Use)

Condition 1.1

If the seesaw balances, remove all but one from the seesaw and put one of the remaining two opposite them. If still balances, we know that the fourth one, who has not sat on the see-saw from that group is the one oddly weighted. (Third Use)

Condition 1.2

If the seesaw is not balanced, remove one from each end. If the seesaw is balanced, the one of the unknown four just removed was the oddly weighted one. Otherwise, the one who stayed is the oddly weighted one. (Third Use)

Condition 2

If the two groups of 4 don’t balance remember which side was lighter, have three get off one end, and the remaining person swap places with one of the other four. Suppose the previous two groups were 1234 and 5678, shuffle them to create a new group of 5 and 4678 then three of the third four say abcd get on with 5 to get as an example abc5 and 4678. (Second Use)

Condition 2.1.1

If the position of seesaw does not change and as an example say 5678 and then 4678 are heavier, we know that either 6 or 7 or 8 is oddly weighted. Now put 7 on one end and 8 on the other. If one is heavier they are the odd one otherwise it is 6. (Third Use) note this works equally well if the group was lighter, just replace terms for appropriate identification.

Condition 2.1.2

If the seesaw reverses, ether 4 or 5 is the oddly weighted one. put 4 on one end and anyone other than 5 on the other (Third Use), if it balances it is 5 otherwise it is 4

Condition 2.1.3

If the seesaw balances we know that either 1 or 2 or 3 is oddly weighted. Say as example 1234 were lighter. Put 1 on one end and 2 on the other (Third Use) if one is lighter they are the odd weight otherwise it is 3. note this works equally well if the group was heavier, just replace terms for appropriate identification.

Done – easy peasy. Right?