Everyone enjoys working out a riddle, but it can get really frustrating when you can’t figure out the answer. This one went viral after featuring on the hit show Brooklyn 99, but now it’s had a resurgence on the internet. We explain the puzzle and tell you how to solve the 12 people on an island riddle.
Riddles are always going viral on social media, but when a TV show features them then it gives it a helping hand. A few years ago the 12 people on an island riddle was used on Brooklyn 99, said by the much-loved Raymond Holt.
Many took to Google to find the answer as a result. However, it’s gone viral once again as people are still left confused by the tricky riddle.
What Is The Brooklyn 99 Riddle?
The riddle goes like this…
There are 12 people on an island. 11 weigh exactly the same, but one of them is slightly heavier or slightly lighter. Figure out which of the people it is.
There is no way any of the people can escape the island, however, there is a see-saw which can hold a number of people at a time. Unfortunately, you can only use the see-saw three times to work it out.
Unfortunately, there isn’t one definitive right answer to this riddle, and solving it takes a little bit of mental maths.
How To Sole The 12 People On An Island Riddle
There is a reason it’s taken the internet by storm and that’s because it’s more complicated than more. Now, we can explain how to solve it, but we recommend getting a piece of paper to help you out.
There are a number of different outcomes, so we have split it into conditions to try and make it simpler.
First, number all the people from 1-12. We’re then going to split them into three groups of four, so we have:
- Group 1 = 1,2,3,4
- Group 2 = 5,6,7,8
- Group 3 = 9,10,11,12.
Condition 1
- Weigh 1
For our first weigh, we will put Group 1 on one side and Group 2 on the see-saw. In this instance, the see-saw remains balanced. That means that all of Group 1 and Group 1 weigh the same, meaning the odd one out is in Group 3.
- Weigh 2
The next weigh requires putting any three members of Group 3 on one side of the see-saw with three members of either Group 1 or 2. For this example, let’s say this is 9, 10 and 11.
We do this because we know both Group 1 and 2 weigh the same and this will now determine if the man is heavier or lighter.
There are a couple of possible outcomes here. If the see-saw remains balanced, then you know that 12, the member of Group 3 left off the see-saw, does not weigh the same.
- Weigh 3
You weigh 12 against any other person on the island to determine whether they are lighter on heavier.
Condition 2
For condition 2, everything is the same as condition 2, apart from the weigh 2 reveals an imbalance. This means that one of 9, 10 or 11 of Group 3 does not weigh the same.
However, we know that the other group weighs the same. So weigh 2 has determined whether the odd person out is heavier or lighter, depending on how the see-saw moved.
So, we know know that 1-8 and 12 all weigh the same.
- Weigh 3
At this point, we know the person is heavier or lighter. For weigh 3 we put 9 and 10 on other sides of the see-saw. If it is balanced, then we know that 11 is the odd one out. If it doesn’t, then, depending on how the see-saw moved in weigh 2, we can figure out which of 9 and 10 weighs differently.
Condition 3 Of The 12 Person On An Island Riddle
- Weigh 1
In condition 3, when we put Groups 1 and 2 on opposite sides of the see-saw, there is an imbalance. This is where it gets a little more complicated.
However, we now know that all of Group 3 weigh the same. Now, we swap one member from Groups 1 and 2 at a time into Group 3
- Weigh 2
Our next weigh involves weighing the amended groups against Group 3 as we know they all weigh the same. This means that we have:
1,2,3,5 vs 4, 9,10, 11
At this point, we know that 9, 10, and 11 all weigh the same. So, if weigh 2 leaves the see-saw balanced, the offending member must be either 6, 7 or 8.
- Weigh 3
Like Condition 2, we put 6 and 7 on opposite sides of the see-saw. If it is equal, then 8 is our odd one out. If not, then we know which one isn’t the same because of weigh 1.
Depending on how Group 2 moved in the weigh 1, we know if they are heavier or lighter. If their side went up, then the odd one out has to be lighter. The opposite, if it was heavier.
Condition 4
In the penultimate condition, weigh 2 of condition 3 reveals an imbalance. This means that either 1,2,3,4 is the odd person on the island out.
If the imbalance is different to that of weigh 1, then we know that 4 or 5 is the problem because they are the only things to have changed.
- Weigh 3
For the final weigh here, we put 4 on one side and any other person apart from 5 on the other. If it is imbalanced, we know that 4 weighs differently and if they are heavier or lighter.
If it is equal, we know that 5 weighs differently. We then use how the see-saw was imbalanced in weigh 1 to figure out if they are heavier or lighter.
Condition 5: The Last Of The 12 People On An Island Riddle
Well done if you have made it this far into the 12 people on an island riddle, but we promise this is your last stop!
Okay, so condition 5 means that weigh 2 of condition 3 reveals an imbalance, but it is the same as that of weigh 1. This tells us that changing 4 and 5 made no difference, so the odd one out has to be 1,2 or 3.
Based on weigh 1, we know if they are heavier or lighter, so it’s a case of putting 1 and 2 on opposite ends of the see-saw. If it’s balanced, we know it’s 3. If it’s not, then we know it’s 1 or 2 depending on the imbalance of the see-saw.
Phew! That was a lot, but we hope you understand the riddle a little better now!