P61??Can you solve the dark matter fuel riddle

It’s an incredible discovery: an ancient, abandoned alien space station filled with precursor technology. But now, every species in the galaxy is in a mad dash to get there first and claim it for themselves. And right away, you’ve got a problem. Your ship’s faster-than-light jump drive consumes 1 unit of fuel for every parsec of distance it takes you, and your ship holds only 15 units of fuel. But the space station is 23 parsecs away, and there’s only empty space between there and here. There’s one thing that can help you, though: dark matter fuel is stable in deep space. That means you can vent a cache of it from your fuel chamber, and then come back to pick it up again later. Even though your ship holds only 15 units of fuel, you’ve been granted use of all 45 units in your current location. With some strategic fuel caching along the way, you might be able to make it all 23 parsecs. So how can you reach the alien space station? Answer in 3, 2, 1. It’s possible to solve this riddle using as few as two cache points, and there are also valid solutions that use more. No matter how you go about it though, the key is determining exactly where to cache fuel along your route. Let’s work backwards from the alien space station. To reach 23 parsecs, you’ll have to leave the 8-parsec mark with a full tank of fuel. The 8-parsec point is too far from the start to use as a cache right away; you could jump there, but wouldn’t have enough fuel to return to the start, let alone store any for later. So that means you’ll need to find a cache somewhere between the start and 8. But where? There’s an interesting pattern that can help. At the start you have exactly 3 tanks’ worth of fuel. At 8 parsecs you need exactly 1. Is there a point, which we can call point X, where you could have exactly 2? That would be useful, because then you could refuel there exactly twice, making full use of your storage capacity without any waste. Wherever point X is, you’ll jump forward from it twice: once to deposit some fuel at the 8-parsec cache point, and a second time for good. So you’ll jump the distance between X and 8-parsecs 3 times in all. You’d have 2 tanks of fuel at point X, and need 1 left at the 8 parsec cache point, so you can spend one tank-- or 15 units-- going back and forth. Since 15 units divided by 3 trips is 5, we can place these two cache points 5 parsecs apart. Any farther, and you wouldn’t have enough fuel to reach the alien space station. So it looks like the earliest we can place point X is at the 3-parsec mark. Is it possible to transport 30 units of fuel there? Let’s try. You set out with a full tank of 15 units. You jump 3 parsecs, drop 9 units off at the cache point, and then jump the 3 units home, arriving with an empty tank. Repeating this process gets you 18 units of fuel at the cache point, and one more jump puts you at the 3-parsec cache with 30 total units of fuel. So far so good! Next, you jump to the 8-parsec mark, drop off 5 units of fuel, and jump back to the 3-parsec mark. You fill up your tank and jump forward again, arriving with 10 units of fuel in your tank. And now the end is in sight. You beam the 5 units of fuel in from deep space to fill your tank to capacity, and type in the coordinates of your final destination. A 15-parsec jump leaves you running on fumes, but ready to dock with the precursor space station. Time to put this alien tech to work and make life better for everyone in the galaxy.

P62??Can you solve the death race riddle

The night before the Death Race across the Wastelands is set to begin, your uncle, the great inventor Slate Kanoli, got kidnapped by the ruthless No-Side gang. The only way to get him back is to race his Coil Runner against the gang yourself. Win and they’ll give back your uncle. Lose and you’ll forfeit the Coil Runner and all his other creations. As the grueling race gets underway, you find yourself falling further and further behind. Your only chance is to take a shortcut your uncle told you about–– the Flux Ravine gambit. Fortunately, the Coil Runner comes equipped with emergency turbo thrusters. Unfortunately, your uncle was a notorious tinkerer, and the system still had some kinks to work out. Just minor things like the ignition exploding, the reactor leaking, or the oxygen levels depleting— any of which would end your racing career immediately. Before his kidnapping, Uncle Slate determined that each of these critical failures was the ultimate result of a chain reaction originating in the thrusters. He was also certain that while one factor could trigger two different effects, and two factors could each independently lead to the same effect, no effect is caused by two factors in conjunction. However, Uncle Slate never got around to pinpointing which thruster was responsible for which error. All you have are the notes from his test runs: 1. When thrusters B and C are on, the Fuel gauge glows. 2. When thrusters A, B, and D are on, the Fuel gauge glows and the Helium tank rattles. 3. When thrusters C, D, and E are on, the Fuel gauge glows and the Gravitometer spins. 4. When thrusters A, D, and E are on, the Gravitometer spins and the Helium tank rattles. 5. Shortly after the Helium tank rattles and the Gravitometer spins, the Ignition explodes and the Oxygen levels deplete. 6. Shortly after the Fuel gauge glows and the Gravitometer spins, the Reactor leaks. You need to use as many thrusters as possible to give yourself the best chance at clearing the gap, without triggering any of the three catastrophic failures. Which thrusters should you activate? Answer in 3 Answer in 2 Answer in 1 The most important thing to remember here is that even if we know that one thing causes another, the converse is not necessarily true. For example, this panic switch shuts off the coil runner’s engine. But the engine being off doesn’t necessarily mean the panic switch was engaged— the coil runner could be out of fuel, or damaged— or turned off normally. We can, however, conclude that if the engine is running, the panic switch hasn’t been engaged. With that in mind, one way we can start is to work backwards from the three defects that could knock you out of the race. So let’s look at Slate’s last two notes, since they give direct information about those. The Gravitometer spins in both cases, but the results are different. That means the spinning Gravitometer can’t be the cause of any particular malfunction. If it were, the same thing would happen each time. So we can conclude that a glowing Fuel gauge makes the reactor leak, while a rattling Helium tank makes the Ignition explode and depletes the Oxygen levels. Once we know which two errors we need to avoid, we can make a table and use the logic of cause and effect to see which thrusters trigger them. Since the Helium tank is fine during the first test run when thrusters B and C are active, we can assume neither makes it rattle. And from the third run we know that D and E don’t either. That leaves thruster A, which was indeed used in the second and fourth test runs where the Helium tank rattled. Now what causes the glowing Fuel gauge? From the fourth test run we know it can’t be thrusters A, D, or E. So is the culprit, B, C, or each of them separately? The answer can be found in the second and third test runs: the fuel tank glowed in both, but B was activated in one, and C in the other. That means the B and C thrusters each independently make the Fuel tank glow. It looks like the A, B, and C thrusters are off limits. Fortunately, the other two are just enough to clear the jump. You rocket into first place and the gang begrudgingly hands over your uncle. He thanks you profusely, and decides to celebrate your victory with a cup of tea from his latest contraption...

P63? ?Can you solve the dragon jousting riddle

After centuries of war, the world’s kingdoms have come to an agreement. Every five years, teams representing the elves, goblins, and treefolk will compete in a grand tournament of dragon jousting. Every team will face each of the others once. The kingdom whose team wins the most matches will rule all of Center-Realm until the next tournament. To prevent any outside meddling, the games are to be conducted in absolute secret except for a group of wizards who will make sure nobody uses enchantments, hexes, or spells to cheat. You’ve been given the extremely important job of recording the scores for the first inaugural tournament. But the opening celebrations get a bit out of control, and when you wake up, you realize the games are already underway. Fortunately, no one has noticed your absence so far. However, you need to get up to speed quickly; if your boss, the head tournament official, finds out you’ve been sleeping on the job, you’ll lose your head. After weighing your options, you decide to offer your life’s savings to one of the regulation wizards in return for the information, giving him your blank scorecard to fill out. But before he can finish, your boss walks into the tent. You barely manage to hide the scorecard in time, and the wizard excuses himself. Your boss chuckles. “Hope you didn’t believe anything Gorbak’s been saying— he’s been cursed to tell only lies, even in writing. Anyway, can you believe how low-scoring the tournament’s been? Every team has played at least once, yet not a single match with a combined score of more than five hits! Anyhow, I’ll be back in a minute to review your scorecard.” You laugh along, and when he leaves you look at the partially completed card, now knowing every single number on it is wrong. You’ve only got one chance to save yourself, so what’s the real score of each match? Pause now to figure it out for yourself. The incredible thing about this riddle is that you can reach the solution despite an almost complete lack of correct information. And that’s possible because knowing that something is false is meaningful information in its own right. The first key is to realize that no team will play more than two matches, since there are only two other teams. So if the elves didn’t actually play one match, and the goblins didn’t actually play two, the truth must be that elves played two and goblins played one. For the elves to have played two matches, they must’ve faced each of the other teams once. And since goblins have only played one match so far— against the elves— that means the match between goblins and treefolk has not occurred yet. We know it’s false that the treefolk tied zero matches, which means their bout against the elves must’ve tied. We also know that the elves won at least one match, and since they tied against the treefolk, they must have beaten the goblins. But can we figure out the actual scores? Let’s start with the elf-treefolk tie. Because no more than five total hits were scored, the final tally must’ve been 0-0, 1-1, or 2-2. But the treefolk must’ve scored some hits, and it’s false that they only had one hit scored against them. The only option that leaves is 2-2. In the match between elves and goblins, the goblins must’ve scored at least one hit. And the elf score must be 2 or more for them to have won the match. This leaves only a few possibilities that add up to 5 or less. The elves couldn’t have scored three, so that eliminates these two. And their total hits scored across both matches can’t add up to six, so this one’s out too. So the score must’ve been 2-1. With one match remaining, you’ve managed to save your job— and your neck. Gorbak the wizard may have lied, but your deductive skills quickly evened the score.

P64? ?Can you solve the egg drop riddle

The city has just opened its one-of-a-kind Fabergé Egg Museum with a single egg displayed on each floor of a 100-story building. And the world's most notorious jewel thief already has her eyes on the prize. Because security is tight and the eggs are so large, she'll only get the chance to steal one by dropping it out the window into her waiting truck and repelling down before the police can arrive. All eggs are identical in weight and construction, but each floor's egg is more rare and valuable than the one below it. While the thief would naturally like to take the priceless egg at the top, she suspects it won't survive a 100-story drop. Being pragmatic, she decides to settle for the most expensive egg she can get. In the museum's gift shop, she finds two souvenir eggs, perfect replicas that are perfectly worthless. The plan is to test drop them to find the highest floor at which an egg will survive the fall without breaking. Of course, the experiment can only be repeated until both replica eggs are smashed. And throwing souvenirs out the window too many times is probably going to draw the guards' attention. What's the least number of tries it would take to guarantee that she find the right floor? Pause here if you want to figure it out for yourself! Answer in: 3 Answer in: 2 Answer in: 1 If you're having trouble getting started on the solution, it might help to start with a simpler scenario. Imagine our thief only had one replica egg. She'd have a single option: To start by dropping it from the first floor and go up one by one until it breaks. Then she'd know that the floor below that is the one she needs to target for the real heist. But this could require as many as 100 tries. Having an additional replica egg gives the thief a better option. She can drop the first egg from different floors at larger intervals in order to narrow down the range where the critical floor can be found. And once the first breaks, she can use the second egg to explore that interval floor by floor. Large floor intervals don't work great. In the worst case scenario, they require many tests with the second egg. Smaller intervals work much better. For example, if she starts by dropping the first egg from every 10th floor, once it breaks, she'll only have to test the nine floors below. That means it'll take at most 19 tries to find the right floor. But can she do even better? After all, there's no reason every interval has to be the same size. Let's say there were only ten floors. The thief could test this whole building with just four total throws by dropping the first egg at floors four, seven, and nine. If it broke at floor four, it would take up to three throws of the second egg to find the exact floor. If it broke at seven, it would take up to two throws with the second egg. And if it broke at floor nine, it would take just one more throw of the second egg. Intuitively, what we're trying to do here is divide the building into sections where no matter which floor is correct, it takes up to the same number of throws to find it. We want each interval to be one floor smaller than the last. This equation can help us solve for the first floor we need to start with in the 100 floor building. There are several ways to solve this equation, including trial and error. If we plug in two for n, that equation would look like this. If we plug in three, we get this. So we can find the first n to pass 100 by adding more terms until we get to our answer, which is 14. And so our thief starts on the 14th floor, moving up to the 27th, the 39th, and so on, for a maximum of 14 drops. Like the old saying goes, you can't pull a heist without breaking a few eggs.

P65? ?Can you solve the false positive riddle

Mining unobtainium is hard work. The rare mineral appears in only 1% of rocks in the mine. But your friend Tricky Joe has something up his sleeve. The unobtainium detector he’s been perfecting for months is finally ready. The device never fails to detect unobtainium if any is present. Otherwise, it’s still highly reliable, returning accurate readings 90% of the time. On his first day trying it out in the field, the device goes off, and Joe happily places the rock in his cart. As the two of you head back to camp where the ore can be examined, Joe makes you an offer: he’ll sell you the ore for just $200. You know that a piece of unobtanium that size would easily be worth $1000, but any other minerals would be effectively worthless. Should you make the trade? Pause here if you want to figure it out for yourself. Answer in: 3 Answer in: 2 Answer in: 1 Intuitively, it seems like a good deal. Since the detector is correct most of the time, shouldn’t you be able to trust its reading? Unfortunately, no. Here’s why. Imagine the mine has exactly 1,000 pieces of ore. An unobtainium rarity of 1% means that there are only 10 rocks with the precious mineral inside. All 10 would set off the detector. But what about the other 990 rocks without unobtainium? Well, 90% of them, 891 rocks, to be exact, won’t set off anything. But 10%, or 99 rocks, will set off the detector despite not having unobtanium, a result known as a false positive. Why does that matter? Because it means that all in all, 109 rocks will have triggered the detector. And Joe’s rock could be any one of them, from the 10 that contain the mineral to the 99 that don’t, which means the chances of it containing unobtainium are 10 out of 109 – about 9%. And paying $200 for a 9% chance of getting $1000 isn’t great odds. So why is this result so unexpected, and why did Joe’s rock seem like such a sure bet? The key is something called the base rate fallacy. While we’re focused on the relatively high accuracy of the detector, our intuition makes us forget to account for how rare the unobtanium was in the first place. But because the device’s error rate of 10% is still higher than the mineral’s overall occurrence, any time it goes off is still more likely to be a false positive than a real finding. This problem is an example of conditional probability. The answer lies neither in the overall chance of finding unobtainium, nor the overall chance of receiving a false positive reading. This kind of background information that we’re given before anything happens is known as unconditional, or prior probability. What we’re looking for, though, is the chance of finding unobtainium once we know that the device did return a positive reading. This is known as the conditional, or posterior probability, determined once the possibilities have been narrowed down through observation. Many people are confused by the false positive paradox because we have a bias for focusing on specific information over the more general, especially when immediate decisions come into play. And while in many cases it’s better to be safe than sorry, false positives can have real negative consequences. False positives in medical testing are preferable to false negatives, but they can still lead to stress or unnecessary treatment. And false positives in mass surveillance can cause innocent people to be wrongfully arrested, jailed, or worse. As for this case, the one thing you can be positive about is that Tricky Joe is trying to take you for a ride.

P66? ?Can you solve the famously difficult green

Imagine an island where 100 people, all perfect logicians, are imprisoned by a mad dictator. There's no escape, except for one strange rule. Any prisoner can approach the guards at night and ask to leave. If they have green eyes, they'll be released. If not, they'll be tossed into the volcano. As it happens, all 100 prisoners have green eyes, but they've lived there since birth, and the dictator has ensured they can't learn their own eye color. There are no reflective surfaces, all water is in opaque containers, and most importantly, they're not allowed to communicate among themselves. Though they do see each other during each morning's head count. Nevertheless, they all know no one would ever risk trying to leave without absolute certainty of success. After much pressure from human rights groups, the dictator reluctantly agrees to let you visit the island and speak to the prisoners under the following conditions: you may only make one statement, and you cannot tell them any new information. What can you say to help free the prisoners without incurring the dictator's wrath? After thinking long and hard, you tell the crowd, "At least one of you has green eyes." The dictator is suspicious but reassures himself that your statement couldn't have changed anything. You leave, and life on the island seems to go on as before. But on the hundredth morning after your visit, all the prisoners are gone, each having asked to leave the previous night. So how did you outsmart the dictator? It might help to realize that the amount of prisoners is arbitrary. Let's simplify things by imagining just two, Adria and Bill. Each sees one person with green eyes, and for all they know, that could be the only one. For the first night, each stays put. But when they see each other still there in the morning, they gain new information. Adria realizes that if Bill had seen a non-green-eyed person next to him, he would have left the first night after concluding the statement could only refer to himself. Bill simultaneously realizes the same thing about Adria. The fact that the other person waited tells each prisoner his or her own eyes must be green. And on the second morning, they're both gone. Now imagine a third prisoner. Adria, Bill and Carl each see two green-eyed people, but aren't sure if each of the others is also seeing two green-eyed people, or just one. They wait out the first night as before, but the next morning, they still can't be sure. Carl thinks, "If I have non-green eyes, Adria and Bill were just watching each other, and will now both leave on the second night." But when he sees both of them the third morning, he realizes they must have been watching him, too. Adria and Bill have each been going through the same process, and they all leave on the third night. Using this sort of inductive reasoning, we can see that the pattern will repeat no matter how many prisoners you add. The key is the concept of common knowledge, coined by philosopher David Lewis. The new information was not contained in your statement itself, but in telling it to everyone simultaneously. Now, besides knowing at least one of them has green eyes, each prisoner also knows that everyone else is keeping track of all the green-eyed people they can see, and that each of them also knows this, and so on. What any given prisoner doesn't know is whether they themselves are one of the green-eyed people the others are keeping track of until as many nights have passed as the number of prisoners on the island. Of course, you could have spared the prisoners 98 days on the island by telling them at least 99 of you have green eyes, but when mad dictators are involved, you're best off with a good headstart.

P67? ?Can you solve the fish riddle

You are the cargo director on the maiden voyage of the S.S. Buoyant, and you've agreed to transport several tanks containing the last specimens of a critically endangered fish species to their new aquarium. Unfortunately, as you're passing through shark-infested waters, the boat is battered by a fierce storm, throwing your precious cargo overboard. And to make matters worse, no one seems certain just how many fish tanks are missing. Fortunately, you have a rescue sub at your disposal, but only enough fuel for one trip to the ocean floor. You need to know where the tanks are so you can gather them all in one quick pass. Not a single fish can be lost. You decide to scan the three sectors of the ocean floor where the cargo could have landed. Thermal imaging shows 50 organisms in the area, and you quickly realize that that number includes both your fish and some ravenous sharks. You flip on the sonar to get a better look. The image for Sector Alpha shows four tanks and two sharks, the image for Sector Beta shows two tanks and four sharks, and the image for Sector Gamma is blank. Your sonar has malfunctioned, and you're going to have to go with the info you have. You check the shipping notes, but all you learn is that each tank had the same number of fish inside. The cargo hold had space for anywhere from 1 to 13 total tanks. And finally, the old captain tells you that this area has the odd property that no two sectors can have the same number of sharks, but every sector will have at least one, and no more than seven. There's no time to waste. The tanks won't withstand the pressure much longer. As you descend in the sub, you review everything you know. How many fish tanks do you need to find in Sector Gamma? Hurry, the fate of an entire species depends on you. Pause here if you want to figure it out for yourself. Answer in: 3 Answer in: 2 Answer in: 1 At first, it seems like there are just too many missing pieces of information. After all, you don't know how many fish or how many tanks there are, let alone how many fish are in each one. But then you remember the best way to compare multiple pieces of partial information - a table. Since we know there are thirteen tanks at most, and we already see six tanks in Sectors Alpha and Beta, we know the total number of tanks must be between 6 and 13. We also know that each sector has a different amount of sharks with no more than seven in each one. Since there are two in Sector Alpha and four in Sector Beta, Sector Gamma can have 1, 3, 5, 6, or 7 sharks. What about the number of endangered fish? Out of the 50 total organisms in all three sectors, we know at least seven are sharks, leaving a maximum of 43 fish inside all the tanks. And the more sharks we find in Sector 3, the fewer fish there are to save. Now, remember that the fish are equally distributed across all the tanks. Why is that important? Because it means that one of the possible values for the total amount of fish must be divisible by one of the possible values for the total amount of tanks. And looking at the table, we can see that the only combination that works is 39 fish divided between 13 tanks with three fish in each. With sharks swarming around, you quickly pilot the sub through the first two sectors before retrieving the remaining seven tanks in Sector Gamma. You've saved the species and taken an impromptu dive. All in all, not a bad day, unless you happen to be a hungry shark.

P68? ?Can you solve the frog riddle

So you're stranded in a huge rainforest, and you've eaten a poisonous mushroom. To save your life, you need the antidote excreted by a certain species of frog. Unfortunately, only the female of the species produces the antidote, and to make matters worse, the male and female occur in equal numbers and look identical, with no way for you to tell them apart, except that the male has a distinctive croak. And it may just be your lucky day. To your left, you've spotted a frog on a tree stump, but before you start running to it, you're startled by the croak of a male frog coming from a clearing in the opposite direction. There, you see two frogs, but you can't tell which one made the sound. You feel yourself starting to lose consciousness, and realize you only have time to go in one direction before you collapse. What are your chances of survival if you head for the clearing and lick both of the frogs there? What about if you go to the tree stump? Which way should you go? Press pause now to calculate odds yourself. 3 2 1 If you chose to go to the clearing, you're right, but the hard part is correctly calculating your odds. There are two common incorrect ways of solving this problem. Wrong answer number one: Assuming there's a roughly equal number of males and females, the probability of any one frog being either sex is one in two, which is 0.5, or 50%. And since all frogs are independent of each other, the chance of any one of them being female should still be 50% each time you choose. This logic actually is correct for the tree stump, but not for the clearing. Wrong answer two: First, you saw two frogs in the clearing. Now you've learned that at least one of them is male, but what are the chances that both are? If the probability of each individual frog being male is 0.5, then multiplying the two together will give you 0.25, which is one in four, or 25%. So, you have a 75% chance of getting at least one female and receiving the antidote. So here's the right answer. Going for the clearing gives you a two in three chance of survival, or about 67%. If you're wondering how this could possibly be right, it's because of something called conditional probability. Let's see how it unfolds. When we first see the two frogs, there are several possible combinations of male and female. If we write out the full list, we have what mathematicians call the sample space, and as we can see, out of the four possible combinations, only one has two males. So why was the answer of 75% wrong? Because the croak gives us additional information. As soon as we know that one of the frogs is male, that tells us there can't be a pair of females, which means we can eliminate that possibility from the sample space, leaving us with three possible combinations. Of them, one still has two males, giving us our two in three, or 67% chance of getting a female. This is how conditional probability works. You start off with a large sample space that includes every possibility. But every additional piece of information allows you to eliminate possibilities, shrinking the sample space and increasing the probability of getting a particular combination. The point is that information affects probability. And conditional probability isn't just the stuff of abstract mathematical games. It pops up in the real world, as well. Computers and other devices use conditional probability to detect likely errors in the strings of 1's and 0's that all our data consists of. And in many of our own life decisions, we use information gained from past experience and our surroundings to narrow down our choices to the best options so that maybe next time, we can avoid eating that poisonous mushroom in the first place.

P69? ?Can you solve the giant cat army riddle

The villainous Dr. Schr?dinger has developed a growth ray and intends to create an army of giant cats to terrorize the city. Your team of secret agents has tracked him to his underground lab. You burst in to find… that it’s a trap! Dr. Schr?dinger has slipped into the next room to activate his device and disabled the control panel on the way out. Fortunately, your teammates are masters of spy-craft. Agent Delta has hacked into the control panel and managed to reactivate some of its functionality. Meanwhile, Agent Epsilon has searched through surveillance to find the code for the door: 2, 10, 14. All you have to do is enter those numbers and you’ll be free. But there’s a problem. The control panel has only three buttons: one which adds 5 to the display number, one which adds 7, and one which takes the square root. You need to make the display output the numbers 2, 10, and 14, in that order. It’s okay if it outputs different numbers in between, but there’s no way to reset the display, so once you get to 2, you’ll have to continue on to 10 and 14 from there. Not only that, Agent Delta explains that there are other traps built into the panel. If it ever shows the same number more than once, a number greater than 60, or a non-whole number, the room will explode. Right now, the display reads zero, and time is running out. There’s only one way to solve the puzzle, with a few small variations. How will you input the code to escape from Dr. Schr?dinger’s lair and save the day? Pause the video now if you want to figure it out for yourself! Answer in: 3 2 1. You look over your options. Adding 5 or 7 increases the number, and the square root button will make it smaller. But there are only a few options where you can use that button: 4 9 16 25, 36, and 49. You’d love to make 4 or 16. Then you could hit the square root button once or twice to get 2. But you can’t make either with just the 5 and 7 buttons. What will you do? You look at the other possible options for numbers you could take the square root of. Nine you can’t reach. Twenty-five and 49 would take you back to 5 or 7, and you can already get to each of those. Thirty-six is your only option. You add 5, 7, 5, 7, 5, 7, and then hit the square root button. Why that series of 5s and 7s? It’s somewhat arbitrary, but you know that you want to avoid 10, 14, and perfect squares, since you’ll need them later. This gets you to 6. Does that help? Looking at your options, you see that 16 is now in your sights. You add 5 twice more to reach it. Then hit square root twice. That gets you to 2. You’re on your way! Now to 10. You can’t get straight there through addition alone, so you’re going to have to reach another square. Taking the square root of 9 or 25 would get you to a good place, but it turns out that 25 is unreachable from 2. So you add 7 to get to 9, then take the square root again. That gets you to 3. Adding 7 again makes 10. Finally, you need to reach 14. Thinking backwards, you imagine where you could be before 14: 7 or 9. But 9 won’t work because you’ve already used 9. However, you could get to 7 by reaching 49 first. You add your way towards it, being careful not to hit any of the numbers you’ve hit so far. You thread your way carefully, adding five 5s and two 7s. Then, square root to 7, and add 7 more. The door opens, and you’re out of the trap. Thanks to your problem-solving skills, your team gets Schr?dinger’s cats out of the box in the nick of time. As for Schr?dinger, you can be certain of one thing: he’ll be spending quite some time in a box of his own.

P70? ?Can you solve the giant iron riddle

The family of giants you work for is throwing a fancy dinner party, and they all want to look their best. But there’s a problem – the elder giant’s favorite shirt is wrinkled! To fix it, you’ll need to power up… the Giant Iron. The iron needs two giant batteries to work. You just had 4 working ones and 4 dead ones in separate piles, but it looks like the baby giant mixed them all up. You need to get the iron working and press the giant shirt, fast – or you’ll end up being the main course tonight! How can you test the batteries so that you’re guaranteed to get a working pair in 7 tries or less? Pause the video now if you want to figure it out for yourself Answer in 3 Answer in 2 Answer in 1 You could, of course, take all eight batteries and begin testing the 28 possible combinations. You might get lucky within the first few tries. But if you don’t, moving the giant batteries that many times will take way too long. You can’t rely on luck – you need to assume the worst possibility and plan accordingly. However, you don’t actually need to test every possible combination. Remember – there are four good batteries in total, meaning that any pile of six you choose will have at least two good batteries in it. That doesn’t help you right away, since testing all six batteries could still take as many as 15 tries. But it does give you a clue to the solution – dividing the batteries into smaller subsets narrows down the possible results. So instead of six batteries, let’s take any three. This group has a total of three possible combinations. Since both batteries have to be working for the iron to power up, a single failure can’t tell you whether both batteries are dead, or just one. But if all three combinations fail, then you’ll know this group has either one good battery, or none at all. Now you can set those three aside and repeat the process for another three batteries. You might get a match, but if every combination fails again, you’ll know this set can have no more than one good battery. That would leave only two batteries untried. Since there are four good batteries in total and you’ve only accounted for two so far, both of these remaining ones must be good. Dividing the batteries into sets of 3, 3, and 2 is guaranteed to get a working result in 7 tries or less, no matter what order you test the piles in. With no time to spare, the iron comes to life, and you manage to get the shirt flawlessly ironed. The pleased elder and his family show up to the party dressed to the nines … well, almost.