Puzzles in Pop Culture: Brooklyn Nine-Nine

In previous editions of Puzzles in Pop Culture, we’ve explored opinions about crosswords, embarked on scavenger hunts with sitcom characters, and even saved New York with brain teasers alongside John McClane in Die Hard with a Vengeance. But it’s rare when a movie or TV show poses a puzzler and leaves it to the audience to solve it.

In “Captain Peralta,” a recent episode of the Fox police sitcom Brooklyn Nine-Nine, a subplot featured Captain Holt posing a brain teaser to his fellow officers.

There are 12 men on an island. 11 weigh exactly the same amount, but one of them is slightly lighter or heavier. You must figure out which.

The island has no scales, but there is a seesaw. You can only use it three times.

With Beyonce tickets going to the person who solved the puzzle, the competition was fierce. Rosa suggested using the seesaw to threaten the men into confessing. Amy and Terry suggested that the first seesaw ride involve putting six men on one side and six on the other, which Captain Holt quickly said wouldn’t work.

As it turns out, Holt was hoping one of his officers would solve the puzzle for him, since he’s been unable to crack it for years. The episode ended without providing the audience with a solution.

Thankfully, your friends here at PuzzleNation Blog would never leave you high and dry like that. Let’s get puzzling!

Now, this WOULD be a simple logic problem if you knew you were looking for someone lighter or you knew you were looking for someone heavier. In that case, a 3×3 ride, 4×4 ride, or even the 6×6 ride Amy and Terry suggested, would eliminate part of the field immediately, and the remaining two uses could determine the heavy person or the light person.

Unfortunately, in Captain Holt’s puzzle, you don’t know if the person is heavier or lighter, which makes this more difficult. For instance, if you knew you were looking for someone heavier, you could immediately eliminate anyone on the side of the seesaw higher in the air. But if you don’t know if the subject in question is lighter or heavier, then you could have a heavier person on one side or a lighter person on the other.

Diabolical.

But, with some careful deduction, you CAN solve this puzzle.

First, let’s identify our 12 castaways with the letters A through L. Now let’s divide them into three groups of four: ABCD, EFGH, and IJKL.

For the first seesaw ride, let’s weigh ABCD vs. EFGH.

There are three possible outcomes:

  • They balance, meaning we can eliminate all eight of them and our mystery person is in IJKL.
  • ABCD sinks while EFGH rises, meaning there’s a heavier person in ABCD or a lighter person in EFGH, so we can eliminate IJKL.
  • EFGH sinks while ABCD rises, meaning there’s a heavier person in EFGH or a lighter person in ABCD, so we can eliminate IJKL.

OUTCOME 1: They balance

For the second seesaw ride, we’ll take IJK and weigh them against any three of the eliminated people — let’s say ABC — because we know they weigh the same.

OUTCOME 1-1: if IJK balances against ABC, we know that L is our guy. For the third seesaw ride, weigh L against A to determine if L is lighter or heavier.

OUTCOME 1-2: if IJK sinks, one of them is heavier than ABC. For the third seesaw ride, weigh I against J. If they balance, K is the heavy one. If I or J sinks, he’s the heavy one.

OUTCOME 1-3: if IJK rises, one of them is lighter than ABC. For the third seesaw ride, weigh I against J. If they balance, K is the light one. If I or J rises, he’s the light one.


OUTCOME 2: ABCD sinks while EFGH rises

For the second seesaw ride, we have eight possible suspects — four heavy, four light — so we mix up the two previous groupings in order to eliminate some suspects. We’ll take E, F, and A and weigh them against G, B, and L. That’s two from the lighter side and one from the heavier vs. one from the lighter, one from the heavier, and one we know is standard.

Again, there will be three possible outcomes:

OUTCOME 2-1: If EFA balances with GBL, they’re all eliminated, leaving either H as a lighter person or either C or D as a heavier person. For the third seesaw ride, weigh C against D. If they balance, H is lighter. If they don’t, whichever is heavier is our guy.

OUTCOME 2-2: If EFA sinks, either A is heavy (because E and F were on the lighter side before) or G is light (because B was on the heavier side and L has already been eliminated), and we can eliminate C, D, and H. For the third seesaw ride, weigh G against L. If they balance, A is heavy. If they don’t, then G is light.

OUTCOME 2-3: If EFA rises, either B is heavy (because G was on the lighter side and L has already been eliminated) or either E or F is light (because A was on the heavier side), and we can eliminate C, D, and H. For the third seesaw ride, weigh E against F. If they balance, then B is heavy. If they don’t, whichever is lighter is our guy.


OUTCOME 3: EFGH sinks while ABCD rises

For the second seesaw ride, we have eight possible suspects — four heavy, four light — so we mix up the two previous groupings in order to eliminate some suspects. We’ll take A, B, and E and weigh them against C, F, and L. That’s two from the lighter side and one from the heavier vs. one from the lighter, one from the heavier, and one we know is standard.

Again, there will be three possible outcomes:

OUTCOME 3-1: If ABE balances with CFL, they’re all eliminated, leaving either D as a lighter person or either G or H as a heavier person. For the third seesaw ride, weigh G against H. If they balance, D is lighter. If they don’t, whichever is heavier is our guy.

OUTCOME 3-2: If ABE sinks, either E is heavy (because A and B were on the lighter side) or C is light (because F was on the heavier side and L has already been eliminated), and we can eliminate D, G, and H. For the third seesaw ride, weigh C against L. If they balance, E is heavy. If they don’t, then C is light.

OUTCOME 3-3: If ABE rises, either F is heavy (because B was on the lighter side and L has already been eliminated) or either A or B is light (because E was on the heavier side), and we can eliminate D, G, and H. For the third seesaw ride, weigh A against B. If they balance, then F is heavy. If they don’t, whichever is lighter is our guy.


This was one whopper of a brain teaser, to be sure, and I’m not surprised it stumped even the likes of the impressive Captain Holt. But, as a special treat, if you’d like to see the Captain himself explain the solution, go here and check out the embedded video. Enjoy.

Of course, it doesn’t answer the real question: who cares about weight? Why aren’t they building a boat to escape the island?

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19 thoughts on “Puzzles in Pop Culture: Brooklyn Nine-Nine

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  3. That’s wrong you have them all get on the seesaw and get them to come off two at a time, at different ends till the seesaw is balanced, then you only have two suspects the ones that got off before it was balanced. You put one of your two suspects on the seesaw with one of the ten you have eliminated, if it’s unbalanced he’s your guy and if it’s balanced it’s the other guy

    • Removing them two-at-a-time wouldn’t work with the 3-use requirement unless you were lucky enough to have that person in one of the first three groups.

      • It would’nt matter it’s all the one go, Your just changing how they get off it

      • You need to rember I said you put all of them on the seesaw there is no groups. There is six on one side and six on the other

    • If it was a mathematical equation taking two off would change the dynamics of the equation and would count but it’s not a equation. It says you can only use it three time the key words are use it. That’s why getting two off at a time only counts as one time you used it

      • I am definitely enjoying the image of people trying to fall off a seesaw in tandem (since both sides would have to remain up in the air in your scenario, and any poor timing would likely topple everyone).

      • It’s you that gets to use it three time. I must say your last post had me in stitches lol. It’s word play not Mathematics. I do understand how you see it, but I’m sorry it’s wrong. I do under stand you put a lot of work into your answer and at first it’s the way I went as well but it’s still wrong

      • My post isn’t wrong, it’s a logistical approach to solving the problem. You finding another solution — one that depends on both a seesaw that fits 12 people AND on those people having exceptional balance — doesn’t invalidate my solution.

  4. They don’t need exceptional balance, I’ve went over that already and the puzzle already has a restriction, you can only use it three time. Now do you think they would put that in and forget that you could only fit some of them on it. The answer to this is (no). As far as a seesaw that can fit 12 people there are lots of they. I take it you’ve never been to a adventure park, the seesaws there can fit a lot more than 12 people on them

  5. Anyway you have answered it as if it’s a problem in a math test, witch isn’t thinking outside the box and you alway need too think outside the box with this type of puzzle

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