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Can You Solve the Virus Riddle?

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In this video riddle, you're a researcher who needs to solve a math problem. The fate of humanity is at stake!

Here's the setup. Your research group has isolated a lethal virus and is studying it in a lab. But one night after you leave the lab, an earthquake strikes and breaks the virus vials. This means that 15 of the 16 rooms in the lab are contaminated, and you have to get past the lab's security system in order to destroy the virus. (There is time pressure, as eventually the virus will escape the lab and kill us all!)

The lab is built as a 4x4 grid, containing a total of 16 rooms, with an entrance at the northwest corner and an exit at the southeast corner. Each room is connected to the adjacent rooms by an airlock. Only the entrance and exit rooms are connected to the outside. The virus has been released in every room except the entrance room.

To destroy the virus samples, you must enter each room and pull its self-destruct switch, destroying the room and the virus within it. But there's a problem—because the lab is in lockdown mode, once you enter a contaminated room, you can't exit without activating the self-destruct switch. Furthermore, once the self-destruct switch has been activated, you cannot re-enter a contaminated room.

Your job is to enter through the entrance room, exit through the exit room, and destroy the virus in every contaminated room. How can you do it?

From the video (at the 1:41 mark), here are the official rules and restrictions:

1. You must enter the building through the entrance and leave through the exit.

2. Every room except the entrance is contaminated.

3. Once you enter a contaminated room, you must pull the switch.

4. After pulling the switch, you must immediately leave the room.

5. You cannot return to a room after its switch has been activated.

Watch the video below for a visual explanation of the problem. This one's a bit of a forehead-smacker when you see the solution.

For more on this puzzle (and its solution), check out this TED-Ed page.

Note: If you're interested in math (without puzzle spoilers), this problem is related to Hamiltonian Paths, or paths that visit each point exactly once.

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See If You Can Solve This Tricky Coin-Flipping Riddle
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Make sure your head is in working order before trying to solve this riddle from TED-Ed, because it's a stumper.

Here's the scenario: You're an explorer who's just stumbled upon a trove of valuable coins in a remote dungeon. Each coin has a gold side and a silver side, each with an identical scorpion seal. The wizard who guards the coins agrees to let you have them, but he won't let you leave the room unless you separate the hoard into two piles with an equal number of coins with the silver side facing up in each. You've just counted the total number of silver-side-up coins—20—when the lights go out. In the dark, you have no way of knowing which half of a coin is silver and which half is gold. How do you divide the pile without looking at it?

As TED-Ed explains, the task is fairly easy to complete, no psychic powers required. All you need to do is remove any 20 coins from the pile at random and flip them over. No matter what combination of coins you choose, you will suddenly have a number of silver-side-up coins that's equal to whatever is left in the pile. If every coin you pulled was originally gold-side-up, flipping them would give you 20 more silver-side-up coins. If you chose 13 gold-side-up coins and seven of the silver-side coins, you'd be left with 13 silver coins in the first pile and 13 silver ones in your new stack after flipping it over.

The solution is simple, but the algebra behind it may take a little more effort to comprehend. For the full explanation and a bonus riddle, check out the video from TED-Ed below.

[h/t TED-Ed]

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No One Can Figure Out This Second Grade Math Problem
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Angie Werner got a lot more than she bargained for on January 24, when she sat down to help her 8-year-old daughter, Ayla, with her math homework. As Pop Sugar reports, the confusion began when they got to the following word problem:

“There are 49 dogs signed up to compete in the dog show. There are 36 more small dogs than large dogs signed up to compete. How many small dogs are signed up to compete?”

Many people misread the problem and thought it was a trick question: if there are 36 more small dogs and the question is how many small dogs are competing, then maybe the answer is 36?

Wrong!

Frustrated by the confusing problem, Angie took to a private Facebook group to ask fellow moms to weigh in on the question, which led to even more confusion, including whether medium-sized dogs should somehow be accounted for. (No, they shouldn’t.) Another mom chimed in with an answer that she thought settled the debate:

"Y'all. A mom above figured it out. We were all wrong. If there is a total of 49 dogs and 36 of them are small dogs then there are 13 large dogs. That means 36 small dogs subtracted by 13 large dogs then there are 23 more small dogs than large dogs. 36-13=23. BOOM!!! WOW! Anyone saying there's half and medium dogs tho just no!"

It was a nice try, but incorrect. A few others came up with 42.5 dogs as the answer, with one woman explaining her method as follows: "49-36=13. 13/2=6.5. 36+6.5=42.5. That's how I did it in my head. Is that the right way to do it? Lol I haven't done math like this since I was in school!"

Though commenters understandably took issue with the .5 part of the answer—an 8-year-old is expected to calculate for a half-dog? What kind of dog show is this?—when Ayla’s teacher heard about the growing debate, she chimed in to confirm that 42.5 is indeed the answer, but that the blame in the confusion rested with the school. "The district worded it wrong,” said Angie. “The answer would be 42.5, though, if done at an age appropriate grade."

Want to try another internet-baffling riddle?


Here's the answer.

[h/t: Pop Sugar]

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