# A Barrage of Brain Teasers!

[Image courtesy of SharpBrains.com.]

One of our favorite pastimes here on PuzzleNation Blog is cracking brain teasers. From riddles to logic problems, we accept challenges from all comers, be they TV detectives or fellow PuzzleNationers!

I got an email a few days ago from a reader who needed help unraveling a few brain teasers from a list she found online. She was proud to have solved most of them, but a few had eluded her.

We’re always happy to assist a fellow puzzler, so let’s take a look at those brain teasers!

In case you want to try them for yourself before we reveal the answers and how to solve each puzzle, I’ll list the original puzzles here and put a nice spoiler-safe break between the questions and the answers.

QUESTION 1: If you have a 7-minute hourglass and an 11-minute hourlass, how can you boil an egg for exactly 15 minutes?

QUESTION 2: Name the next number in the following sequence: 1, 11, 21, 1211, 111221, 312211, _____.

QUESTION 3: Four people want to cross a river, but the only option is a narrow bridge. The bridge can only support two people at a time. It’s nighttime, and the group has one torch, which they’ll need to use every time they cross the bridge. Person A can cross the bridge in 1 minute, Person B in 2 minutes, Person C in 5 minutes, and Person D in 8 minutes. When two people cross the bridge together, they must move at the slower person’s pace. Can all four get across the bridge in 15 minutes or less?

QUESTION 4: During a recent census, a man told a census taker that he had three children. When asked their ages, he replied, “The product of their ages is 72. The sum of their ages is the same as my house number.” The census taker ran to the man’s front door and looked at the house number. “I still can’t tell,” she complained. The man replied, “Oh, that’s right, I forgot to tell you that the oldest one likes chocolate pudding.” The census taker then promptly wrote down the ages of all three children. How old are they?

QUESTION 5: There are five bags of gold that all look identical, and each contains ten gold pieces. One of the five bags has fake gold, though. All five bags are identical, and the real gold and fake gold are identical in every way, except the pieces of fake gold each weigh 1.1 grams and the pieces of real gold each weigh 1 gram. You have a perfectly accurate digital scale available to you, but you can only use it once. How do you determine which bag has the fake gold?

[Image courtesy of AwesomeJelly.com.]

Okay, here’s your spoiler alert warning before we start unraveling these brain teasers.

So if you don’t want to see them, turn away now!

Last chance!

Ready? Okay, here we go!

[Image courtesy of Just Hourglasses.com.]

QUESTION 1: If you have a 7-minute hourglass and an 11-minute hourlass, how can you boil an egg for exactly 15 minutes?

A variation on the two jugs of water puzzle we’ve covered before, this puzzle is basically some simple math, though you need to be a little abstract with it.

• Step 1: Start boiling the egg and flip over both hourglasses.
• Step 2: When the 7-minute hourglass runs out, flip it over to start it again. (That’s 7 minutes boiling.)
• Step 3: When the 11-minute hourglass runs out, the 7-minute hourglass has been running for 4 minutes. Flip it over again. (That’s 11 minutes boiling.)
• Step 4: When the 7-minute hourglass runs out, another 4 minutes has passed, and you’ve got your 15 minutes of egg-boiling time.

QUESTION 2: Name the next number in the following sequence: 1, 11, 21, 1211, 111221, 312211, _____.

The answer is 13112221. This looks like a math or a pattern-matching puzzle, but it’s far more literal than that.

Each subsequent number describes the number before it. 11, for instance, isn’t eleven, it’s one one, meaning a single one, representing the number before it, 1.

The third number, 21, isn’t twenty-one, it’s “two one,” meaning the previous number consists of two ones, aka 11.

The fourth number, 1211, translates to “one two, one one,” or 21. The fifth number, 111221, becomes “one one, one two, two one.” And the sixth, 312211, becomes “three one, two two, one one.”

So, the number we supplied, 13112221, is “one three, one one, two two, two one.”

[Image courtesy of Do Puzzles.]

QUESTION 3: Four people want to cross a river, but the only option is a narrow bridge. The bridge can only support two people at a time. It’s nighttime, and the group has one torch, which they’ll need to use every time they cross the bridge. Person A can cross the bridge in 1 minute, Person B in 2 minutes, Person C in 5 minutes, and Person D in 8 minutes. When two people cross the bridge together, they must move at the slower person’s pace. Can all four get across the bridge in 15 minutes or less?

Yes, you can get all four across the bridge in 15 minutes.

This one’s a little tougher, because people have to cross the bridge in both directions so that the torch remains in play. Also, there’s that pesky Person D, who takes so long to get across.

So what’s the most time-efficient way to get Person D across? You’d think it would be so send D across with Person A, so that way, you lose 8 minutes with D, but only 1 minute going back with the torch with A. But that means only 6 minutes remain to get A, B, and C across. If you send A and C together, that’s 5 minutes across with C, and 1 minute back with A, and there’s your 15 minutes gone, and A and B aren’t across.

So the only logical conclusion is to send C and D across together. That’s 8 minutes down. But if you send C back down, that’s another 5 minutes gone, and there’s no time to bring A, B, and C back across in time.

So, C and D have to cross together, but someone faster has to bring the torch back. And suddenly, a plan comes together.

• Step 1: A and B cross the bridge, which takes 2 minutes. A brings the torch back across in 1 minute. Total time used so far: 3 minutes.
• Step 2: C and D cross the bridge, which takes 8 minutes. B brings the torch back across in 2 minutes. Total time used so far: 13 minutes.
• Step 3: A and B cross the bridge again, which takes 2 minutes. Total time used: 15 minutes.

(It technically doesn’t matter if A returns first and B returns second or if B returns first and A returns second, so long as they are the two returning the torch.)

QUESTION 4: During a recent census, a man told a census taker that he had three children. When asked their ages, he replied, “The product of their ages is 72. The sum of their ages is the same as my house number.” The census taker ran to the man’s front door and looked at the house number. “I still can’t tell,” she complained. The man replied, “Oh, that’s right, I forgot to tell you that the oldest one likes chocolate pudding.” The census taker then promptly wrote down the ages of all three children. How old are they?

Their ages are 3, 3, and 8.

Let’s pull the relevant information from this puzzle to get started. There are three children, and the product of their ages is 72.

So let’s make a list of all the three-digit combinations that, when multiplied, equal 72: 1-1-72, 1-2-36, 1-3-24, 1-4-18, 1-6-12, 1-8-9, 2-2-18, 2-3-12, 2-4-9, 2-6-6, 3-3-8, 3-4-6. We can’t eliminate any of them, because we don’t know how old the man is, so his children could be any age.

But remember, after being told that the sum of the children’s ages is the same as the house number, the census taker looks at the man’s house number, and says, “I still can’t tell.” That tells us that the sum is important.

Let’s make a list of all the sums of those three-digit combinations: 74, 39, 28, 23, 19, 18, 22, 17, 15, 14, 14, 13.

The census taker doesn’t know their ages at this point. Which means that the sum has multiple possible combinations. After all, if there was only one combination that formed the same number as the house number, the census taker would know.

And there is only one sum that appears on our list more than once: 14.

So the two possible combinations are 2-6-6 and 3-3-8.

The chocolate pudding clue is the deciding fact. The oldest child likes chocolate pudding. Only 3-3-8 has an oldest child, so 3-3-8 is our answer.

[Image courtesy of Indy Props.com.]

QUESTION 5: There are five bags of gold that all look identical, and each contains ten gold pieces. One of the five bags has fake gold, though. All five bags are identical, and the real gold and fake gold are identical in every way, except the pieces of fake gold each weigh 1.1 grams and the pieces of real gold each weigh 1 gram. You have a perfectly accurate digital scale available to you, but you can only use it once. How do you determine which bag has the fake gold?

With only one chance to use the scale, you need to maximize how much information you can glean from the scale. That means you need a gold sample from at lesst four bags (because if they all turn out to have real gold, then the fifth must be fake). But, for the sake of argument, let’s pull samples from all five bags.

How do we do this? If we pull one coin from each bag, there’s no way to distinguish which bag has the fake gold. But we can use the variance in weight to our advantage. That .1 difference helps us.

Since all the real gold will only show up before the decimal point, picking a different number of coins from each bag will help us differentiate which bag has the fake gold, because the number after the decimal point will vary.

For instance, if you take 1 coin from the first bag, 2 coins from the second, 3 coins from the third, 4 coins from the fourth, and 5 coins from the fifth, you’re covered. If the fake gold is in the first bag, your scale’s reading will end in .1, because only one coin is off. If the fake gold is in the second bag, your scale’s reading will end in .2, because two coins are off. And so on.

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# 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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