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This question already has an answer here:

This one comes from this week's Brooklyn Nine-Nine episode!

There is an island with 12 islanders. All of the islanders individually weigh exactly the same amount, except for one, who either weighs more or less than the other 11.

You must use a see-saw to figure out whose weight is different, and you may only use the see-saw 3 times. There are no scales or other weighing device on the island.

How can you find out which islander is the one that has a different weight?

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marked as duplicate by Joe Z., Tryth, Ivo Beckers, Aza Mar 9 '15 at 9:49

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    $\begingroup$ Made me laugh when Amy started with "Take six islan-" and Holt said, "Nope, won't work." I originally thought it would take 6 on one side, 6 on the other, and go from there, but then I realized... the islander can weigh more OR less, so you couldn't find out that way. I like Rosa's answer... squeeze until fatty confesses. :p $\endgroup$ – Josh Mar 9 '15 at 7:33
  • $\begingroup$ haha yeah loved the episode! i thought that at the end scully would come up with the answer and make rosa/gina regret rejecting his help $\endgroup$ – emdee Mar 10 '15 at 6:59
  • $\begingroup$ I wonder if this is the highest-viewed duplicate question of all time? $\endgroup$ – Rand al'Thor Mar 12 '15 at 11:47
  • $\begingroup$ sorry my bad. i checked if there was already someone that posted a riddle about the 12 islanders but didn't know it was also 12 balls and a scale $\endgroup$ – emdee Mar 13 '15 at 0:22
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    $\begingroup$ @emdee - Here is the link that you posted in the question. Feel free to provide your own comment and I can delete this one. Brooklyn 99 officially released a video of the answer with Captain Holt! youtube.com/watch?v=5K2WE9z4zL4 $\endgroup$ – Len Mar 13 '15 at 1:09
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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 wieghted one is in the other group of 4.

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

Condition 1.1

If the see saw 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 see saw is not balanced, remove one from each end. If the see-saw 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

It is easier than everyone makes it. A seesaw is binary. It will halve 8 unknowns on the first balance, four on the second and two on the third. Set it up so deduction eliminates everything else and your gold. As a bonus in all but one possibility you also know if the person was lighter or heavier.

(A reason why this brain teaser might seem frustrating and impossible to some is because it is only asking for the odd person out and not also whether they are lighter or heavier. It is impossible to know both for sure in only three steps.)

Edit: In 11/12 cases you know whether the person is lighter or heavier as the seasaw dictates it. The only case where you don’t is 1.1.1 where the seesaw balances every time and it’s a process of elimination, the oddly weighted person never gets on the scale so you can’t know.

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  • $\begingroup$ " It is impossible to know both for sure in only three steps." That's not true. You can figure out both in three steps. Example for condition 1: put three "unknowns" against three "knowns". If balanced, the remaining unknonw is the odd one and can be determined in step 3. If the unknowns go down, you know that one of them is heavy. In step 3 place one unknown on each side. If balanced, the 3rd unknown is heavy. If not balanced, the one going down is heavy. Works the other way around with "light" and "up" instead of "heavy" and "down" $\endgroup$ – Hilmar Apr 1 '15 at 14:46
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    $\begingroup$ "It is impossible to know both for sure in only three steps". Wow how did this even get 8 upvotes...mind boggling $\endgroup$ – Geralt of Rivia Dec 21 '15 at 9:09
  • $\begingroup$ @Hilmar: It's indeed not that easy to know both, but yes it's possible like you said: puzzling.stackexchange.com/a/224/1649 $\endgroup$ – justhalf Jul 19 '16 at 6:04
  • $\begingroup$ This shouldn't be the accepted answer since it isn't actually the solution... It is a good write up though $\endgroup$ – Nathan Loyer Jan 1 at 4:39
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OK, I think I have it, now the problem of explaining it, here goes:

We are going to name the islanders 1 2 3 4 5 6 7 8 9 10 11 12

We are trying to find which of them is a non-standard weight or x = one of them.

//are comments during explanation

Use 1:

1 2 3 4 against 5 6 7 8

I) 1 2 3 4 = 5 6 7 8 then Use 2: 9 against 10 //9 10 11 or 12 are x

     A) 9 > 10 or 9 < 10 then Use 3:  9 against 11    //9 or 10 are x

          i)  9 > 11 or 9 < 11 then 9 = x

         ii)  9 = 11  then 10 = x

     B) 9 = 10 then Use 3:  9 against 11      //11 or 12 are x

          i) 9 > 11 or 9 < 11 then 11 = x

         ii) 9 = 11 then 12 = x

//Ok four down, eight to go, that was the easy part

II) 1 2 3 4 > 5 6 7 8 then Use 2: 1 2 3 5 against 4 10 11 12 //10 11 12 are not x now

  A) 1 2 3 5 > 4 10 11 12 then Use 3: 1 against 2  //1 2 or 3 are x now and x is heavier than the rest

         i) 1 > 2 then 1 is x //x is heavier

        ii) 1 < 2 then 2 is x

       iii) 1 = 2 then 3 is x

  B) 1 2 3 5 < 4 10 11 12 then Use 3:  4 against 12 // 4 or 5 is x. The switched 4 and 5 caused a reversal

        i) 4 > 12 or 4 < 12 then 4 is x

       ii) 4 = 12 then 5 is x

 C) 1 2 3 5 = 4 10 11 12 then Use 3:  6 against 7   //6 7 or 8 are x and lighter than the rest 

        i) 6 > 7 then 7 is x

       ii) 6 < 7 then 6 is x

      iii) 6 = 7 then 8 is x

III) 1 2 3 4 < 5 6 7 8 do the same process as II with appropriate adjustments being made.

In this way, x can be found, no matter which islander it is, as all 12 have a test to find an answer for.

Hope that makes sense. :D

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  • $\begingroup$ If you tweak I.A you can tell whether it is lighter or heavier. Weigh 9/10 against 11/4. i) they weigh the same, weigh 12/4 to see if 12 is lighter or heavier. ii) weigh 9/10, if they are different then the one that goes the same direction as 9/10 before is that direction. If they are the same then 11 is the direction it went. For example 9/10 are lighter than 11/4, 10 is lighter than 9 means 10 is the odd one and lighter. If 10 and 9 were the same then 11 would be the odd one and it would be heavier. $\endgroup$ – Jason Goemaat Mar 12 '15 at 4:21
  • $\begingroup$ That's true, but I don't think you have to know if they are lighter or heavier, just who it is. :D $\endgroup$ – Jared Anderson Mar 12 '15 at 4:24
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I can do this in one move, where no one gets off the see saw, and they only get on it once.

It is a see saw, not a giant scale! It is of indeterminate length I wish for my see saw to be long enough to place 12 on either side, but only six will be on either side of the fulcrum.

Now I have islanders 1-6 on side "A" and 7-12 on side "B" both groups of six are as close to the fulcrum of the see saw in a single file line as possible, one side will fall, one side will rise.
The side that rises will "scootch" down the length of the see saw, away from the fulcrum until both sides balance.
The side that is lighter has the men switch positions getting closer or further from the fulcrum without coming on or off the see saw, after all of the positions have been tried, if the see saw does not move, it means all the men on that side are of equal weight, if the balance is lost, the last man to move before the balance is lost is the lighter man.
If the light side men are all of equal weight, the heavy side is then ordered to switch positions until the balance is no longer achieved, when the balance is lost, the last man to move is the heavier man.

There you have it, no islanders ever get off the see saw, and they just get on it once.

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