JBF_Logic Puzzles Flashcards
How far can a bear walk into the woods?
A) Half-Way (after that, the bear is walking out of the woods!).
How can you flip the gummy bear pyramid upside down by moving only 3 gummy bears?
On Bert’s 14th Birthday, his younger brother Chip was half his age. If today is Bert’s 31st birthday, how old is Chip?
Chip is 24 years old (Bert is 7 years older than Chip). Happy birthday Bert!
G, H, J, M, [?], V
King Nupe of the kingdom Catan dotes on his two daughters so much that he decides the kingdom would be better off with more girls than boys, and he makes the following decree: All child-bearing couples must continue to bear children until they have a daughter!
But to avoid overpopulation, he makes an additional decree: All child-bearing couples will stop having children once they have a daughter! His subjects immediately begin following his orders.
After many years, what’s the expected ratio of girls to boys in Catan?
Don’t overthink this. Each baby born is as equally likely to be a boy as a girl. Therefore, the ratio of girls to boys must be 1:1. It’s as simple as that—honestly.
A hen and a half lays an egg and a half in a day and a half. How many eggs does one hen lay in one day?
If 1.5 hens lay 1.5 eggs in 1.5 days, it must be that one hen would lay one egg in the same time period: 1.5 days. Now, if one hen lays one egg in 1.5 days, it means that one hen would lay only ⅔ of an egg in one day. So the answer is ⅔ of an egg.
You might argue that hens can’t lay just a portion of an egg. Fair enough. But then, of course, the whole premise of the problem, which involves 1.5 hens and 1.5 eggs, doesn’t make sense from the start. And that invites pondering whether you can answer a problem that isn’t well-posed in the first place. (Is the statement “The king of France is bald” false if there is no king of France?)
Since this is a math puzzle column and not a philosophy column, I’ll stick with the answer of ⅔ of an egg.
You’re rummaging around your great grandmother’s attic when you find five short chains each made of four gold links. It occurs to you that if you combined them all into one big loop of 20 links, you’d have an incredible necklace. So you bring it into a jeweler, who tells you the cost of making the necklace will be $10 for each gold link that she has to break and then reseal.
How much will it cost?
The most straightforward approach would be to break a link on the end of each of the five chains, and then reattach the link to the back of the next chain in the loop. This would cost you $50 for the five links that were broken and resealed.
But you can actually do it for $40! Instead of breaking a link in each chain, break all four links in one of the chains and then use those four links to attach the remaining four chains together. Now you’ve saved $10. Use it on something nice.
Kenny, Abby, and Ned got together for a round-robin pickleball tournament, where, as usual, the winner stays on after each game to play the person who sat out that game. At the end of their pickleball afternoon, Abby is exhausted, having played the last seven straight games. Kenny, who is less winded, tallies up the games played:
Kenny played eight games
Abby played 12 games
Ned played 14 games
Who won the fourth game against whom?
First, we need to figure out how many total games were played. We get 34 when we add up the game totals, but since two players play each game, there were 17 games played.
Now, with a brief insight, the solution quickly unravels. The insight is that in a three-person round-robin, every player plays at least every other game. Since 17 total games were played, every player must have played at least eight games. And, in fact, since Kenny played exactly eight games, we can infer which ones, as highlighted below:
riddle of the week
Laura Feiveson
Now, had Kenny won any of his games, we would have seen him play two games in a row. Instead, it must be the case that he lost every game he played. Thus, we can conclude that Kenny lost the fourth game. But against whom?
Well, we know that Abby and Ned played against each other in every game that Kenny didn’t play, which are all the unshaded squares above. In addition, we know Abby played the last seven straight games.
This means we know for sure that Abby played games 1, 3, 5, 7, 9, 11, 12, 13, 14, 15, 16, and 17, which amounts to 12 games, accounting for all of her games. So Ned must have played all other games against Kenny, such that Ned beat Kenny in the fourth game.
The circuit breaker box in your new house is in an inconvenient corner of your basement. To your chagrin, you discover none of the 100 circuit breakers is labeled, and you face the daunting prospect of matching each circuit breaker to its respective light. (Suppose each circuit breaker maps to only one light.)
To start with, you switch all 100 lights in the house to “on,” and then you head down to your basement to begin the onerous mapping process. On every trip to your basement, you can switch any number of circuit breakers on or off. You can then roam the hallways of your house to discover which lights are on and which are off.
What is the minimum number of trips you need to make to the basement to map every circuit breaker to every light?
The solution here is amazing if you pick the right strategy. Just to set the scene, the simplest strategy would be to just switch each circuit breaker off one at a time. But this would take 99 trips to the basement—the 100th circuit breaker would be mapped by the process of the elimination). You can do much, much better.
Believe it or not, you can map all 100 circuit breakers to their respective lights in just seven trips to the basement!
Here’s the strategy:
For the ease of keeping track of things, put a piece of masking tape on each circuit breaker and on each light. On the first trip to the basement, flip 50 circuit breakers to off, mark these circuit breakers with a “0,” and mark the 50 circuit breakers that are on with a “1.” Accordingly, as you roam around the house to tally the lights, mark the 50 lights that are off with a “0” and mark the other 50 lights with a “1.”
On your second trip to the basement, keep off half of the circuit breakers that are marked with a “0,” turn off half of the circuit breakers that are marked with a “1,” and mark all of these circuit breakers with a second number of “0.” Flip on all other circuit breakers if they’re not already on, and mark their second number as “1.” Now go around the house, and again mark the lights that are off with a “0” and those lights that are on with a “1.”
Jesse’s two grandmothers want to see him every weekend, but they live on opposite sides of town. As a compromise, he tells them that every Sunday, he’ll head to the subway station nearest to his apartment at a random time of the day and will hop on the next train that arrives.
If it happens to be the train traveling north, he’ll visit his Grandma Erica uptown, and if it happens to be the train traveling south, he’ll visit his Grandma Cara downtown. Both of his grandmothers are okay with this plan, since they know both the northbound and southbound trains run every 20 minutes.
But after a few months of doing this, Grandma Cara complains that she sees him only one out of five Sundays. Jesse promises he’s indeed heading to the station at a random time each day. How can this be?
Although both trains come exactly every 20 minutes, the timing still matters: Suppose the northbound train comes on the hour, on the 20, and on the 40, so at 9:00, 9:20, 9:40, 10:00, etc., but the southbound train comes on the 4, 24, and 44, so 9:04, 9:24, 9:44, 10:04, etc.
This means in any hour, there will be only 12 minutes in which the southbound train will be the next train to arrive. That is, if Jesse arrives between 9:00-9:04, between 9:20-9:24, or between 9:40-9:44, he’ll get on the southbound train. Otherwise, he’ll get on the northbound train.
Since Jesse arrives at a random point every day (or equivalently at a random point in each hour), his chance of getting on the southbound train will be 12 out of 60, or one-fifth. And so that’s the answer: Both the northbound and the southbound trains arrive at the station every 20 minutes, but the southbound train always arrives 4 minutes after the northbound train.
Grandma Cara may be lucky, though. If the southbound train was scheduled to arrive only 1 minute after the northbound train, Jesse would only visit two to three times per year!
Max and Rose are ant siblings. They love to race each other, but always tie, since they actually crawl at the exact same speed. So they decide to create a race where one of them (hopefully) will win.
For this race, each of them will start at the bottom corner of a cuboid, and then crawl as fast as they can to reach a crumb at the opposite corner. The measurements of their cuboids are as pictured:
If they both take the shortest possible route to reach their crumb, who will reach their crumb first? (Don’t forget they’re ants, so of course they can climb anywhere on the edges or surface of the cuboid.)
The key to solving this problem is to come up with the length of each of their shortest routes. The simplest way to find this shortest route is to flatten the box. Once the box is flattened, it’s very easy to find the shortest route between the ant and their crumb: The shortest route between two points is a straight line!
In Max’s case, flattening the box is straightforward. Since it’s a cube, it doesn’t matter which way you flatten it. If you flatten the top front fold, you’ll see the following rectangle:
Clearly, Max’s fastest route will be a straight line to the crumb. Using the pythagorean theorem (or graph paper and a good ruler), you can determine that his shortest path is √45 inches, or 6.71 inches.
Rose’s is slightly trickier to figure out, as there are three possible ways that you could “unfold” her cuboid:
ant riddle
Laura Feiveson
Again, the pythagorean theorem—or a very good ruler—will tell us the diagonal of the second rectangle is the shortest at √41 inches, or 6.40 inches. So Rose will achieve this route if she crawls on the left most (unseen surface) and then onto the top:
ant riddle
Laura Feiveson
Since 6.40 is less than 6.71, Rose will get to her crumb first!
You’re facing your friend, Caryn, in a “candy-off,” which works as follows: There’s a pile of 100 caramels and one peppermint patty. You and Caryn will go back and forth taking at least one and no more than five caramels from the candy pile in each turn. The person who removes the last caramel will also get the peppermint patty. And you love peppermint patties.
Suppose Caryn lets you decide who goes first. Who should you choose in order to make sure you win the peppermint patty?
The best way to approach this problem is to think backwards.
Let’s suppose Caryn faces five or fewer caramels in the pile on her turn. Then she can take them all, and will win the peppermint patty. So you need to make sure Caryn never faces a pile with five or fewer caramels. But how about six?
If Caryn starts a turn with six caramels in front of her, she’ll have to take one to five caramels, leaving you with the win. Alternatively, if she starts a turn with seven, eight, nine, 10, or 11 caramels, she can take just enough to leave you with six, which would ensure a win for her.
So you want Caryn to face six caramels in front of her. You’ll definitely be able to do that if she faces 12 caramels in her prior turn. From that turn, she’ll leave you with seven, eight, nine, 10, or 11 caramels, from which you can take away just enough to leave her with six.
You can be sure to leave Caryn with 12 caramels on a turn if and only if you leave her with 18 caramels on her previous turn. And you can leave her with 18 caramels, if and only if you leave her with 24 caramels on the turn before that. You may see the pattern now: You need Caryn to face a multiple of six caramels in each turn.
Now, 100 is not a multiple of six, which means you shouldn’t let Caryn go first. In fact, the closest multiple of six that’s under 100 is 96. Therefore, you should go first and take four caramels on your first turn, leaving Caryn facing a pile with 96 caramels.
No matter what Caryn takes on her first turn, make sure to leave her with 90 caramels for her second turn, 84 caramels for her third turn, 78 for her fourth turn, and … six for her 16th and last turn. She’ll then take some amount of caramels, leaving you with the ability to take the last caramel and the peppermint patty.
There’s one interesting side effect of your strategy: Caryn can do quite well in terms of the number of caramels she receives by taking the maximum number (five) in each turn. Since she has 16 turns, she’ll end up with 16 x 5 = 80 caramels, whereas you’ll end up with only 20. You better really like that peppermint patty!
Finally, the Great American Rail-Trail across the whole country is complete! Go ahead, pat yourself on the back—you’ve just installed the longest handrail in the history of the world, with 4,000 miles from beginning to end. But just after the opening ceremony, your assistant reminds you that the metal you used for the handrail expands slightly in summer, so that its length will increase by one inch in total.
“Ha!” you say, “One inch in a 4,000 mile handrail? That’s nothing!” But … are you right?
Let’s suppose when the handrail expands, it buckles upward at its weakest point, which is in the center. How much higher will pedestrians in the middle of the country have to reach in summer to grab the handrail? That is, in the figure below, what is h? (For the purposes of this question, ignore the curvature of the Earth and assume the trail is a straight line.)
- Half of 4,000 miles is 2,000 miles, which is about 126.7 million inches.
- The handrail expands by 1 inch in total, so each of the top sides of the triangle will be 126.7 million and ½ inches.
Now, we can use Pythagorean’s Theorem (a2 + b2 = c2) to solve for h:
127,000,0002 + h2 = 127,000,000.52
Solving for h, we get: h = 11,260 inches, or 938 feet. Let’s just say that pedestrians would have to have extremely long arms to still make use of the handrail in summer.
Ungar gives some context for the real-world implications of this problem:
“It takes a surprisingly large perpendicular displacement to accommodate the extra length, which is a general geometric feature of buckling, and the simple triangle model makes a great illustration. Something like a long rail will be more likely to push in and out sideways over shorter baselines, making for smaller but still impressive displacements, as seen in these real-life buckled railways.
Two ways that engineers deal with situations like these rails are by using sliding joints and by optimizing the temperature where a rail held fixed by its supports is stress-free. Sliding joints work by allowing connected rails to expand and contract, while the second technique reduces temperature-related stresses so the rail does not break off its supports.”
Amanda lives with her teenage son, Matt, in the countryside—a car ride away from Matt’s school. Every afternoon, Amanda leaves the house at the same time, drives to the school at a constant speed, picks Matt up exactly when his chess club ends at 5 p.m., and then they immediately return home together at the same constant speed. But one day, Matt isn’t feeling well, so he leaves chess practice early and starts to head home on his portable scooter.
After Matt has been scooting for an hour, Amanda comes across him in her car (on her usual route to pick him up), and they return together, arriving home 40 minutes earlier than they usually do. How much chess practice did Matt miss?
Let’s call the spot at which Amanda and Matt meet on the road, point M. In this problem, Amanda drives from their home to point M, where she picks up Matt, and then drives back to their home. Let’s call the time it takes her to do this “T”.
We don’t know T, but we do know the time it took Amanda to do this is 40 minutes less than the time it usually takes her to drive back and forth from school.
From this, we can infer the back-and-forth trip she did not drive (from M to school and back to M) must take 40 minutes. Since she drives at a constant speed, the one-way trip from M to school must therefore take 20 minutes.
Since we know Amanda times her day to arrive at school for pickup at exactly 5 p.m., she must have reached M at 20 minutes before pickup, or at 4:40 pm.
Now, we know from the problem that Matt left chess club one hour before he met Amanda at point M. Thus, he must have left at 3:40 pm. Since chess typically ends at 5 pm, we have our answer: Matt missed 1 hour and 20 minutes of chess practice.
Three movie stars, Chloe, Lexa, and Jon, are filming a movie in the Amazon. They’re very famous and very high-maintenance, so their agents are always with them. One day, after filming a scene deep in the rainforest, the three actors and their agents decide to head back to home base by foot. Suddenly, they come to a large river.
On the riverbank, they find a small rowboat, but it’s only big enough to hold two of them at one time. The catch? None of the agents are comfortable leaving their movie star with any other agents if they’re not there as well. They don’t trust that the other agents won’t try to poach their star.
For example, Chloe’s agent is okay if Chloe and Lexa are alone in the boat or on one of the riverbanks, but definitely not okay if Lexa’s agent is also with them. So how can they all get across the river?
For those of you stuck at home with your families, this is a great problem to work on together with props. Even elementary school kids can enjoy puzzling this one out. But there’s no upper age limit to river crossing problems.
While the solution to this problem isn’t unique, it can’t be completed in fewer than nine river crossings. Here’s one solution:
Crossing 1: Chloe and Chloe’s agent row to the far side of the river.
Crossing 2: Chloe’s agent gets off and Chloe rows back to the near side.
Crossing 3: Lexa’s agent and Jon’s agent row to the far side.
solutions to river crossing riddle
Laura Feiveson
Crossing 4: Jon’s agent gets off and Lexa’s agent rows back to the near side.
solutions to river crossing riddle
Laura Feiveson
Crossing 5: Lexa’s agent switches places with Jon and Chloe, who row to the far side.
solutions to river crossing riddle
Laura Feiveson
Crossing 6: Jon gets off, and Chloe rows back to the near side.
solutions to river crossing riddle
Laura Feiveson
Crossing 7: Chloe switches places with Lexa and her agent, who row to the far side.
solutions to river crossing riddle
Laura Feiveson
Crossing 8: Chloe’s agent gets back in the boat and rows back to the near side.
solutions to river crossing riddle
Laura Feiveson
Crossing 9: Chloe and her agent row to the far side. They’ve all made it!
Carol was creating a family tree, but had trouble tracking down her mother’s birthdate. The only clue she found was a letter written from her grandfather to her grandmother on the day her mother was born. Unfortunately, some of the characters were smudged out, represented here with a “___”. (The length of the line does not reflect the number of smudged characters.)
“Dear Virginia,
Little did I know when I headed to work this Monday morning, that by evening we would have a beautiful baby girl. And on our wedding anniversary, no less! It makes me think back to that incredible weekend day, J___ 27th, 19___, when we first shared our vow to create a family together, and, well, here we are! Happy eighth anniversary, my love.
Love, Edwin”
John Conway’s Doomsday Rule is his ingenious method to quickly map any date in history with its day of the week. He practiced it daily for years, and got so good that he could compute the day of the week for 15 dates in his head in under 10 seconds.
To solve this problem, you need to know some calendar rules:
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The Most Important Discovery of 2022
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There are 365 days in each non-Leap Year.
There are 366 days in each Leap Year.
Leap Years occur every four years except in years that are divisible by 100, and except-except in years that are divisible by 400.
Thus, 1900 is the only year in the 20th century that is divisible by 4, but not a leap year.
Since 365 divided by seven days of the week has a remainder of 1, any date will move forward one day of the week after every non-leap year. So if July 31st is a Friday in 2020, we know it will be a Saturday in 2021. And since 366 divided by 7 has a remainder of 2, any date will move forward two days of the week when it passes a leap day.
For example, since July 31st was a Wednesday in 2019, we know it must be a Friday in 2020. (The Leap Day occurred in February 2020.)
From Edwin’s letter to Virginia, we can surmise the following three things:
The month in which they were married was January, June, or July.
They were married on a weekend day.
Eight years later, their anniversary was on a Monday.
Almost all eight-year periods involve passing over six non-Leap Years and two Leap Years. This means you’d advance by one day of the week for each of the non-Leap Years, and two days of the week for each of the two Leap Years. In total, you’d advance by 10 days of the week, which is the same as advancing by 3 days of the week.
This means if Edwin and Virginia’s wedding date was on a Saturday, eight years later should be a Tuesday, and if their wedding date was a Sunday, eight years later should be a Wednesday. But we know eight years after Edwin and Virginia’s wedding was a Monday.
This can only occur if the eight-year period includes only one Leap Year, which means the eight-year period in question must have started in 1900! Furthermore, they must have been married before February 28th, as dates after February 28th in 1900 would still pass two Leap Days over an eight-year period.
So, since the month started with J, we now have the complete answer: Edwin and Virginia must have been married on January 27th, 1900, which means Carol’s mother must have been born on January 27th, 1908!
Imagine you have a very long belt. Well, extremely long, really … in fact, it’s just long enough that it can wrap snugly around the circumference of our entire planet. (For the sake of simplicity, let’s suppose Earth is perfectly round, with no mountains, oceans, or other barriers in the way of the belt.)
Naturally, you’re very proud of your belt. But then your brother, Peter, shows up—and to your disgruntlement, he produces a belt that’s just a bit longer than yours. He brags his belt is longer by exactly his height: 6 feet.
If Peter were also to wrap his belt around the circumference of Earth, how far above the surface could he suspend the belt if he pulled it tautly and uniformly?
If you add a mere 6 feet to a belt that’s 130 million feet long, you’re adding a minuscule 0.000005 percent to its length. Accordingly, my intuition says you wouldn’t even be able to notice the difference between the two belts when they’re wrapped around the earth. But my intuition is completely wrong. Perhaps yours is better!
Let’s call Earth’s radius “R,” measured in feet. (We don’t actually need to know the radius is about 4,000 miles, or 20 million feet.) This means your belt is exactly the length of Earth’s circumference:
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The Soviet Union’s Secret Space Cannon
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Your belt = 2𝛑R
Peter’s belt is 6 feet longer than yours:
Peter’s belt = 2𝛑R+6
Imagine pulling up on Peter’s belt so it’s truly suspended above Earth’s surface. The question at hand is how high above Earth’s surface is it? Or, in other words, how much longer is the radius of the circle that it encompasses?
Let’s call the height we’re looking for “H.” Then, the radius of the circle that Peter’s belt encompasses is R+H. So the length of Peter’s belt must be its circumference, or: 2𝛑(R+H). But we already know Peter’s belt is 2𝛑R+6.
Eureka! We have an equation!
2𝛑R+6 = 2𝛑(R+H)
Amazingly, the 2𝛑R drops out on both sides, and we get:
H = 3/𝛑
Since 𝛑 is about 3, H is about 1 foot! Or, to be exact, H is 0.95 feet. Incredible!
In some future time, when the shelter-in-place bans are lifted, a married couple, Florian and Julia, head over to a bar to celebrate their newfound freedom.
They find four other couples there who had the same idea.
Eager for social contact, every person in the five couples enthusiastically taps elbows (the new handshake) with each person they haven’t yet met.
It actually turns out many of the people had known each other prior, so when Julia asks everyone how many elbows they each tapped, she remarkably gets nine different answers!
The question: How many elbows did Florian tap?
First, we need to figure out which nine answers Julia heard from the other nine people in the room. Since the two people per couple already know each other, at most, any one person could have tapped elbows with eight other people (since they tapped elbows only with people they haven’t met).
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What is Sleep Paralysis?
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Thus, since the nine answers are distinct from one another, Julia must have heard the answers: 0, 1, 2, 3, 4, 5, 6, 7, 8.
Now, how can we possibly figure out which is Florian’s answer?
Let’s start by calling the person who tapped eight elbows Person 8. It must be the case that Person 8 tapped everyone’s elbow except for Person 8’s own spouse (because they already know their spouse). Thus, everyone other than Person 8’s spouse tapped at least one elbow. Therefore, Person 8’s spouse must be the person who tapped zero elbows! So we now know Person 8 must be married to Person 0.
Now let’s consider Person 7. We know they didn’t tap Person 0’s elbow because, well, Person 0 didn’t tap any elbows. This means since Person 0 and Person 7’s spouse are out, Person 7 must have tapped elbows with all the seven other people.
Now, as before, this means those seven other people tapped elbows with at least two people (Person 8 and Person 7), which means the person who tapped elbows with only one person must be Person 7’s spouse. So we conclude Person 7 must be married to Person 1.
We can use similar logic to conclude Person 6 must be partnered with Person 2, and that Person 5 must be with Person 3. Finally, we get to the last remaining couple: One of them is Person 4, and the other has tapped elbows with Person 8, Person 7, Person 6, and Person 5—or, in other words, also four people. Thus, Person 4’s spouse is another Person 4.
Since Julia heard nine different answers, and didn’t hear two people answer that they had both tapped four elbows, it must be the case that Julia herself tapped four elbows. Furthermore, it also must be the case that Florian, her husband, is Person 4.
Thus, we have our answer: Florian tapped four elbows.
Alan and Claire live by the old Scottish saying, “Never have whisky without water, nor water without whisky!” So one day, when Alan has in front of him a glass of whisky, and Claire has in front of her a same-sized glass of water, Alan takes a spoonful of his whisky and puts it in Claire’s water.
Claire stirs her whisky-tinted water, and then puts a spoonful of this mixture back into Alan’s whisky to make sure they have exactly the same amount to drink.
So: Is there more water in Alan’s whisky, or more whisky in Claire’s water? And does it matter how well Claire stirred?
One way to approach this problem quickly is by thinking in extremes.
Suppose the spoon was the same size as the entire glass. In that case, putting Alan’s “spoonful” of whisky into Claire’s water would entail mixing both glasses together, leading to a mixture that’s half water and half whisky. Then, when Claire returns a “spoonful” of this mixture to Alan’s glass, there would be exactly half water and half whisky in both glasses.
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The History of the F-35 Joint Strike Fighter
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So in this extreme, there would be the same amount of water in Alan’s whisky as there is whisky in Claire’s water. Indeed, this is the solution no matter the size of the spoon.
To answer more carefully, let’s assume each glass has 100 milliliters (mL) of each liquid to start with: Alan’s has 100 mL of whisky and Claire’s has 100 mL of water. Since the liquid transfers consist of removing and adding a spoonful to each glass, the net amount of liquid changed in each glass is zero. Thus, both glasses end with the same amount of liquid they started with: 100 mL.
This means if Alan has x mL of water in his glass at the end, then he must have exactly 100-x mL of whisky. Since we know that there’s 100 mL of whisky in total, this means there must be x mL of whisky in Claire’s glass.
So the water in Alan’s glass must have displaced whisky in Alan’s glass one-for-one, such that there is exactly the same volume of water in Alan’s glass as there is whiskey in Claire’s glass. This will be the case no matter how well Claire mixed!
The question: Can you make 100 by interspersing any number of pluses and minuses within the string of digits 9 8 7 6 5 4 3 2 1? You can’t change the order of the digits! So what’s the least number of pluses and minuses needed to make 100?
For instance, 98 - 7 - 6 + 54 - 32 shows one way of interspersing pluses and minuses, but since it equals 107, it’s not a solution.
A lot of trial and error is needed, and, although there are ways to help limit the set of possibilities, there isn’t a surefire method of arriving at a solution in a reasonable amount of time with a pen and paper.
To start with, we might notice the first two digits make the number 98, which is quite close to 100. So, if we can add and subtract the rest of the single digits to make 2, then we’ll have 100. In fact, there are eight ways to do this:
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IndyCar Driver Dalton Kellett’s Favorite Course
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98 + 7 + 6 - 5 - 4 - 3 + 2 - 1
98 + 7 - 6 + 5 - 4 + 3 - 2 - 1
98 + 7 - 6 + 5 - 4 - 3 + 2 + 1
98 + 7 - 6 - 5 + 4 + 3 - 2 + 1
98 - 7 + 6 + 5 + 4 - 3 - 2 - 1
98 - 7 + 6 + 5 - 4 + 3 - 2 + 1
98 - 7 + 6 - 5 + 4 + 3 + 2 - 1
98 - 7 - 6 + 5 + 4 + 3 + 2 + 1
But we can do better—we can make 100 with fewer than 7 pluses and minuses. Here’s one way to make sure we find all possibilities: Use a computer simulation. Each pair of digits can be connected by either nothing, a plus sign, or a minus sign. Since there are eight paired connections, there are 3^8 = 6,561 possible combinations of pluses and minuses. I simulated each one of these combinations to determine which sum to 100.
The simulation unearthed that there are seven other ways of making 100:
98 - 7 - 6 - 5 - 4 + 3 + 21
9 + 8 + 76 + 5 + 4 - 3 + 2 - 1
9 + 8 + 76 + 5 - 4 + 3 + 2 + 1
9 - 8 + 76 + 54 - 32 + 1
9 - 8 + 76 - 5 + 4 + 3 + 21
9 - 8 + 7 + 65 - 4 + 32 - 1
98 - 76 + 54 + 3 + 21
The bolded solution is the winner. It uses only four pluses and minuses!
The computer simulation also revealed that it’s possible to make every number from 1 to 100, which could keep you doodling for many meetings. (In fact, it’s possible to make every number in more ways than one with one notable exception: 9 + 87 - 65 + 4 - 32 - 1 is the unique way to make 2.)
Cecilia loves testing the logic of her very logical friends Jaya, Julian, and Levi, so she announces:
“I’ll write a positive number on each of your foreheads. None of the numbers are the same, and two of the numbers add up to the third.”
She scribbles the numbers on their heads, then turns to Jaya and asks her what her number is. Jaya sees Julian has 20 on his forehead, and Levi has 30 on his. She thinks for a moment and then says, “I don’t know what my number is.” Julian pipes in, “I also don’t know my number,” and then Levi exclaims, “Me neither!” Cecilia gleefully says, “I’ve finally stumped you guys!”
“Not so fast!” Jaya says. “Now I know my number!”
What is Jaya’s number?
Jaya’s number is 50.
To figure this out, let’s go back to what Jaya initially observes: She sees Julian has 20 on his forehead and Levi has 30 on his. That means she can either be 50 (their sum), or 10 (their difference).
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Let’s suppose Jaya were 10. Then Julian would have seen 10 (on Jaya) and 30 (on Levi), thus thinking he was either 20 or 40, and would say he doesn’t know what number he is. Now it comes to Levi, who would see 10 on Jaya and 20 on Julian. He would think, then, he’s either 10 (the difference) or 30 (the sum).
But wait! Levi can’t be 10, because Cecilia told everyone all three numbers are different from one another, and Jaya is already 10. So Levi would know he was 30, and would say so. Since he said he didn’t know his number, Jaya can’t be 10. Thus, she knows she’s 50.
For completeness, we need to confirm if Jaya were 50, Julian and Levi would respond as they did: If she were 50, Julian would have seen 50 (on Jaya) and 30 (on Levi), such that we wouldn’t know if he were 20 or 80. Then it would come to Levi, who would see 50 on Jaya and 20 on Julian. He could therefore be either 30 or 70, and he wouldn’t know which one. (For extra completeness, and quite tediously, we need to make sure all answers are also consistent from Julian’s and Levi’s perspectives. For instance, if Levi were 70, is it the case that Julian or Jaya couldn’t have figured out their numbers previously? It is.)
Extra credit: I love these problems in which it appears on the face that no information has been exchanged, and yet there was sneakily enough exchanged to pinpoint a solution. Often, these solutions rely on constraints that are subtly introduced in the body of the problem. In this solution, we rely on the tidbit that Cecilia shared: All the numbers are different from one another. However, given the other constraints already embedded in the problem, we don’t actually need that piece of information. Why?
Solution to extra credit: The bit of information we still need is all of the numbers are positive, and therefore can’t be zero. Once again, we consider what would happen if Jaya had a 10 on her forehead. As before, when it comes to Levi, he’d see 10 on Jaya and 20 on Julian, and think he’s either 10 or 30. This time, he eliminates the possibility he’s 10 by thinking to himself, “If I were 10, then Julian would have seen 10 on Jaya and 10 on me, and therefore would have been able to figure out he was 20, since we all know he can’t be zero. Since he said he didn’t know what his number was, I can’t be 10!” The rest of the solution follows as before.
It’s 2024, and there are five candidates running in the democratic primary: Taylor Swift, Oprah Winfrey, Mark Cuban, Keanu Reeves, and Dwayne Johnson. (Hey, it could happen.) As usual, the first primary is in Iowa.
In an effort to overcome its embarrassment after the 2020 caucus debacle, the Iowa Democratic Party has just announced a new, foolproof way of finding the best candidate: there will be four consecutive elections.
First, candidate 1 will run against candidate 2. Next, the winner of that will run against candidate 3, then that winner will run against candidate 4, and finally the winner of that election will run against the final candidate. By the transitive property, the winner of this last election must be the best candidate … so says the Iowa Democratic Party.
Candidate Keanu has been feeling pretty low, as he knows he is ranked near the bottom by most voters, and at the top by none. In fact, he knows the Iowa population is divided into five equal groups, and that their preferences are as follows:
Text, Font, Line, Organism, Document, Number, Handwriting, Calligraphy, Smile, Art,
.
Keanu is childhood friends with Bill S. Preston, Esq., the new head of the Iowa Democratic Party. Preston, confident that the order of the candidates doesn’t matter for the outcome, tells Keanu he can choose the voting order of the candidates.
So what order should Keanu choose?
Remarkably, although Reeves is, at best, ranked third out of five, he can still be the overall winner in a series of consecutive one-to-one elections if he picks the correct order. This is because the “transitive” property doesn’t necessarily hold for majority preferences: that is, if the population prefers candidate A to candidate B, and candidate B to candidate C, it’s not necessarily the case that they prefer candidate A to candidate C.
This phenomenon, called the “Condorcet paradox,” was first famously discussed by the Marquis de Condorcet, a French mathematician and philosopher in the late 1700s. In addition to his mathematical works, he also published radical works denouncing slavery and promoting gender equality, and was ultimately imprisoned due to his political beliefs.
But back to Keanu. He needs to choose the order of candidates in the following diagram:
Text, Font, Line, Number, Diagram, Parallel, Slope,
.
He wants to be the “FINAL WINNER,” and so begins by working backward. He first looks to see how he would do against every other candidate. To look at this, he needs to view the relative rankings of each Keanu voter versus the candidate in question. For instance, against Oprah, the preferences are:
Text, Font, White, Black, Line, Document, Organism, Number, Paper, Smile,
.
Oof—Keanu would get zero votes in a contest against Oprah. Similarly, he would lose against Swift (with zero votes) and Cuban (with only one out of five votes). But, wait! Against Johnson, he would win with three out of five votes!
Text, Font, White, Black, Line, Document, Organism, Paper, Number, Paper product,
.
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So Keanu knows he can only win one election, which means he needs to put himself as Candidate 5 to have any hope of winning the whole thing. Furthermore, he needs to go against Johnson, which means Johnson needs to be the winner of the previous election:
Text, Font, Line, Number, Diagram, Parallel,
.
Now, how about Johnson? Johnson wins only against Cuban. Thus for Johnson to survive to the last round, he must be Candidate 4, and run against Cuban in the third election:
Text, Font, Line, Number, Diagram, Parallel,
.
Cuban beats Oprah, but loses to Swift. Thus, Cuban must be Candidate 3, and Oprah must be the winner of the first election. Finally, we can see this is possible because Oprah does indeed beat Swift in a one-to-one election! The final order, which leads to Keanu winning, is:
Text, Font, Line, Number, Diagram, Design, Parallel,
There are 100 lockers that line the main hallway of Chelm High School. Every night, the school principal makes sure all the lockers are closed so that there will be an orderly start to the next day. One day, 100 mischievous students decide that they will play a prank.
The students all meet before school starts and line up. The first student then walks down the hallway, and opens every locker. The next student follows by closing every other locker (starting at the second locker). Student 3 then goes to every third locker (starting with the third) and opens it if it’s closed, and closes it if it’s open. Student 4 follows by opening every fourth locker if it’s closed and closing it if it’s open. This goes on and on until Student 100 finally goes to the hundredth locker. When the principal arrives later in the morning, which lockers does she find open?
The key here is to consider which students will open or shut any particular locker. Let’s take an example of locker 24. Student 1 will open it, since student 1 opens every locker. Student 2 will then close it, since student 2 closes every even locker and 24 is even. Student 3 will open it and then student 4 will shut it, since 24 is a multiple of both 3 and 4. Student 5 will pass it by since 24 is not a multiple of 5.
This example makes it clear that each student will change the status of all lockers that have a number that is a multiple of the student’s number. Conversely, every locker will have its status changed by the students that are numbered by one of the locker’s factors. Thus, locker 24 will have its status changed by students 1, 2, 3, 4, 6, 8, 12, and 24.
Now, how does this lead us to figure out which lockers are opened at the end? Locker 1, which has one factor, will clearly be open at the end, since the only student who touches it is the first student, who opens it. Locker 2, with two factors, will be closed, since the only two students to touch it are student 1, who opens it, and then student 2, who closes it.
Locker 3, also with two factors, will also be closed at the end. On the other hand, locker 4, which has three factors (1, 2, and 4), will be open, shut, and then open again. This line of thinking leads us to the conclusion: Only those lockers with an odd number of factors will be left open at the end of the prank.
So which numbers have an odd number of factors?
Consider that a factor is an integer that, when multiplied by another integer, produces the number of interest. Thus, factors are always paired with “another integer.” For instance, the factors of 24 are paired in the following way: (1, 24), (2, 12), (3, 8), (4, 6). Since factors come in pairs, most numbers have an even number of factors. The only exception occurs when factors are paired with themselves.
Let’s look at the factor “pairs” for 16: (1, 16), (2, 8), (4, 4). In this case, 4 is paired with itself to produce the number 16. Thus, 16 has an odd number of factors: 1, 2, 4, 8, 16. Factors are only ever paired with themselves in the case of perfect squares, which means that perfect squares are the only numbers with an odd number of factors! Therefore, the lockers that are left open are: 1, 4, 9, 16, 25, 36, 49, 64, 81, and 100.
How many triangles?
Unlike some viral math problems that are purposely vague and open for interpretation, this one actually does have a slam-dunk, no-doubt-about-it solution, and it’s 18. Let’s hear from some of the geometry experts as to why.
“I would approach this just like one approaches any mathematical problem: reduce it and find structure,” says Sylvester Eriksson-Bique, Ph.D., a postdoctoral fellow with the University of California Los Angeles’s math department.
The only way to form triangles in the figure I drew, Erikkson-Bisque says, is if the top vertex (corner) is part of the triangle. The base of the triangle will then have to be one of the three levels below. “There are three levels, and on each you can choose a base among six different ways. This gives 18, or 3 times 6 triangles.”
Let’s look at the master triangle again.
line, triangle, parallel,
Andrew Daniels
“It’s convenient to generalize to the case where there are n lines passing through the top vertex, and p horizontal lines,” says Francis Bonahon, Ph.D., a professor of mathematics at the University of Southern California.
In our case, n = 4, and p = 3. Any triangle we find in the drawing should have one top vertex and two others on the same horizontal line, so for each horizontal line, the number of triangles with two vertices on that line is equal to the number of ways we can choose these vertices, Bonahon says—namely the number of ways we can choose two distinct points out of n, or “n choose 2.”
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Remember high school math? That’s n(n-1)/2. And since there are p horizontal lines, says Bonahan, this gives p n(n-1)/2 possible triangles. In our case, that’s 3x4(4-1)/ 2=18.
Here’s a handy breakdown of how to find each possible triangle:
Flavius Josephus, a Jewish-Roman historian from the first century, tells the story like this: A company of 40 soldiers, along with Josephus himself, were trapped in a cave by Roman soldiers during the Siege of Yodfat in 67 A.D. The Jewish soldiers chose to die rather than surrender, so they devised a system to kill off each other until only one person remained. (That last person would be the only one required to die by their own hand.)
All 41 people stood in a circle. The first soldier killed the man to his left, the next surviving soldier killed the man to his left, and so on. Josephus was among the last two men standing, “whether we must say it happened so by chance, or whether by the providence of God,” and he convinced the other survivor to surrender rather than die.
This tale may be apocryphal and fantastic, but it gives rise to a fascinating math problem. That is: If you’re in a similar situation to Josephus, how do you know where to stand so you will be the last man standing?
If you start running through this sequence with different numbers of people in the starting circle, you will see a few patterns emerge. First of all, the final survivor is never someone in an even-numbered position because all of the people standing in even-numbered positions are killed first (1 kills 2, 3 kills 4, and so on). Do enough trial and error and you might notice that any time the starting number of people is a power of 2, the final person standing is the same as the person who started the sequence (position number 1). This is the key to figuring out where you should stand. When the number of people left standing is equal to a power of 2, then you want it to be your turn to kill your neighbor.
As Numberphile demonstrates, you can use math to determine the winning spot beforehand. You just need to figure out what the highest power of 2 is that is smaller than the starting number of people. For Josephus, the starting number is 41, and the highest power of 2 that is fewer than 41 is 32 (2 to the power of 5). You want it to be your turn when there are exactly 32 people left. Because of the way the problem works, with every other person dying, the position you want to stand in is 2 times the difference between 41 and 32 (41 - 32 = 9), plus 1. So, 2 x 9 + 1 = 19. There’s the magic number: Josephus must have been standing in position 19 of the circle (or his fellow survivor was, in which case he was in position 35, second-to-last standing). The video does a bang-up job of explaining this part visually, so give it a watch.
Some quick mental math could help you figure this out in a pinch, but a computer scientist would have an even easier time determining where to stand. When you express your starting number in binary, a quick, simple pattern exists to let you know where you need to stand to live. (You’ll just have to watch the video to see the binary shortcut to a quick answer.) So, if you didn’t already have enough reason to study computer science, then being able to act quickly when confronted with a problem like Josephus’s should be plenty incentive to crack open a textbook on coding.
Headshot of Jay Bennett
JAY BENNETT
Solving Polynomials by Radicals
Remember the quadratic formula? Given ax²+bx+c=0, the solution is x=(-b±√(b^2-4ac))/(2a), which may have felt arduous to memorize in high school, but you have to admit is a conveniently closed-form solution.
Now, if we go up to ax³+bx²+cx+d=0, a closed form for “x=” is possible to find, although it’s much bulkier than the quadratic version. It’s also possible, yet ugly, to do this for degree 4 polynomials ax⁴+bx³+cx²+dx+f=0.
The goal of doing this for polynomials of any degree was noted as early as the 15th century. But from degree 5 on, a closed form is not possible. Writing the forms when they’re possible is one thing, but how did mathematicians prove it’s not possible from 5 up?
The world was only starting to comprehend the brilliance of French mathematician Evariste Galois when he died at the age of 20 in 1832. His life included months spent in prison, where he was punished for his political activism, writing ingenious, yet unrefined mathematics to scholars, and it ended in a fatal duel.
Galois’ ideas took decades after his death to be fully understood, but eventually they developed into an entire theory now called Galois Theory. A major theorem in this theory gives exact conditions for when a polynomial can be “solved by radicals,” meaning it has a closed form like the quadratic formula. All polynomials up to degree 4 satisfy these conditions, but starting at degree 5, some don’t, and so there’s no general form for a solution for any degree higher than 4.
Trisecting an Angle
The Ancient Greeks wondered about constructing lines and shapes in various ratios, using the tools of an unmarked compass and straightedge. If someone draws an angle on some paper in front of you, and gives you an unmarked ruler, a basic compass, and a pen, it’s possible for you to draw the line that cuts that angle exactly in half. It’s a quick four steps, nicely illustrated like this, and the Greeks knew it two millennia ago.
What eluded them was cutting an angle in thirds. It stayed elusive for literally 15 centuries, with hundreds of attempts in vain to find a construction. It turns out such a construction is impossible.
Modern math students learn the angle trisection problem—and how to prove it’s not possible—in their Galois Theory classes. But, given the aforementioned period of time it took the math world to process Galois’ work, the first proof of the problem was due to another French mathematician, Pierre Wantzel. He published his work in 1837, 16 years after the death of Galois, but nine years before most of Galois’ work was published.
Either way, their insights are similar, casting the construction question into one about properties of certain representative polynomials. Many other ancient construction questions became approachable with these methods, closing off some of the oldest open math questions in history.
So if you ever time-travel to ancient Greece, you can tell them their attempts at the angle trisection problem are futile.
The Classification of Finite Simple Groups
From solving Rubik’s Cube to proving a fact about body-swapping on Futurama, abstract algebra has a wide range of applications. Algebraic groups are sets that follow a few basic properties, like having an “identity element,” which works like adding 0.
Groups can be finite or infinite, and if you want to know what groups of a particular size n look like, it can get very complicated depending on your choice of n.
If n is 2 or 3, there’s only one way that group can look. When n hits 4, there are two possibilities. Naturally, mathematicians wanted a comprehensive list of all possible groups for any given size.
The complete list took decades to finish conclusively, because of the difficulties in being sure that it was indeed complete. It’s one thing to describe what infinitely many groups look like, but it’s even harder to be sure the list covers everything. Arguably the greatest mathematical project of the 20th century, the classification of finite simple groups was orchestrated by Harvard mathematician Daniel Gorenstein, who in 1972 laid out the immensely complicated plan.
By 1985, the work was nearly done, but spanned so many pages and publications that it was unthinkable for one person to peer review. Part by part, the many facets of the proof were eventually checked and the completeness of the classification was confirmed.
By the 1990s, the proof was widely accepted. Subsequent efforts were made to streamline the titanic proof to more manageable levels, and that project is still ongoing today.
The Four Color Theorem
From solving Rubik’s Cube to proving a fact about body-swapping on Futurama, abstract algebra has a wide range of applications. Algebraic groups are sets that follow a few basic properties, like having an “identity element,” which works like adding 0.
Groups can be finite or infinite, and if you want to know what groups of a particular size n look like, it can get very complicated depending on your choice of n.
If n is 2 or 3, there’s only one way that group can look. When n hits 4, there are two possibilities. Naturally, mathematicians wanted a comprehensive list of all possible groups for any given size.
The complete list took decades to finish conclusively, because of the difficulties in being sure that it was indeed complete. It’s one thing to describe what infinitely many groups look like, but it’s even harder to be sure the list covers everything. Arguably the greatest mathematical project of the 20th century, the classification of finite simple groups was orchestrated by Harvard mathematician Daniel Gorenstein, who in 1972 laid out the immensely complicated plan.
By 1985, the work was nearly done, but spanned so many pages and publications that it was unthinkable for one person to peer review. Part by part, the many facets of the proof were eventually checked and the completeness of the classification was confirmed.
By the 1990s, the proof was widely accepted. Subsequent efforts were made to streamline the titanic proof to more manageable levels, and that project is still ongoing today.
The problem: For a goat to be able to eat grass in a circle with an area of exactly one half acre, how much rope does it need?
The answer has been approximate for centuries, providing food for thought for mathematicians who enjoyed pondering an idle headscratcher. Now, there’s finally a concrete equation.
The goat problem is a living example of what it means to round off your answer. Steve Nadis at Quanta explains the distinction:
“To illustrate the difference, consider the equation x2 − 2 = 0. One could derive an approximate numerical answer, x = 1.4142, but that’s not as accurate or satisfying as the exact solution, x = √2.”
Or consider Zeno’s paradox, the famous thought experiment in which a frog halves its distance across a pond—and literally never gets to the other side. How does Zeno’s chicken cross the road? (It doesn’t.)
With a few moments of thought, the goat problem quickly turns into an exercise in many intersecting approximations. This is why every answer offered since the 1700s has been an approximation as well.
In the 1980s, mathematicians made big progress by blowing out a very hypothetical two dimensions—easy in pure math, impractical in reality—into a 3D space with different mathematics. If this sounds counterintuitive, think about how much of calculus is enabled by switching from x to x2.
And now, finally, there’s an exact solution for the first time. Mathematician Ingo Ullisch took a cue from the previous researchers who made progress on the problem. He introduced complex analysis, which is kind of like algebra with an optional imaginary-number add-on.
By multiplying out a series of values expressed as the telltale a+bi of complex numbers, he was able to reduce the problem to a still-bewildering, but exact expression. Quanta explains the catch:
“Ullisch’s solution is not something simple like the square root of 2. It’s a bit more abstruse—the ratio of two so-called contour integral expressions, with numerous trigonometric terms thrown into the mix—and it can’t tell you, in a practical sense, how long to make the goat’s leash. Approximations are still required to get a number that’s useful to anyone in animal husbandry.”
What’s fun about the goat problem, which mathematicians admit has no relationship to other questions or even mathematical fields, is that it acts as a kind of mathematical Rosetta stone. Whatever your field is, there’s probably a way to approach the problem and model it using your own modeling and analysis.
Something cool about any exact equation is that, technically speaking, it can be set equal to any other exact equation and studied for commonalities.
Flying Around the World
You’ve designed an incredibly advanced aircraft, a true marvel of aeronautics, the X-100. You want to fly all the way around the world in it without stopping. The only problem is that the plane can carry just enough fuel to make it halfway around the Earth.So you build a total of three planes, one for you to fly and two more for your assistants John and Jane. You equip these planes with some pretty incredible and futuristic features to help you along the way. For one, they can instantaneously transfer any amount of fuel to each other. One plane can even pass fuel to the other two simultaneously. Secondly, the aircraft can turn on a dime, literally reversing direction instantaneously, flying at the exact same velocity as soon as they about-face.
Now for the numbers: Each plane can carry 180 gallons of fuel. You plan to fly all the way around the world along the equator, where your plane can fly at the blistering speed of 1º of longitude per minute, meaning it’ll take 360 minutes to make the whole 360-degree journey. (This is a speed of about Mach 5.4, for those who are curious). The three aircraft burn 1 gallon of fuel per 1º longitude traveled.
One more thing: There is only one airport along the equator route that you plan to take. You must start at this airport and finish there. John and Jane can return to the airport to refuel, filling their tanks to the full 180 gallons. The one airport is the only place that anyone can land.As you can see, with 180 gallons of fuel, one of the planes by itself would only make it 180º around the world, half of the required 360º. Fortunately, the planes have their nifty instantaneous refueling and about-facing features, and you have John and Jane to help you make the trip.How do you fly all the way around the world without stopping or turning around, and without any of the three pilots running out of fuel and crashing?
Hint
As mentioned, you are probably going to want pencil and paper for this. Draw out a circle representing the world and mark 90º, 180º, 270º and 360º. Imagine you are looking at the south pole and the equator runs along the circumference of the circle. It might help to make a table to keep track of all three planes’ fuel levels as well. Where do the refuelings need to occur for you to make it all the way around in one continuous trip?
Note: If a plane hits 0 gallons of fuel, and at that exact moment another plane is there to refuel it, the plane does not crash. The fuel transfers instantaneously.
So, you need to make it all the way around with out stopping or turning around. Your assistants John and Jane can turn around instantaneously and they can land at the airport to refuel. In the diagrams below, you are flying in plane A, John is in plane B, and Jane is in plane C.
All three of you take off from the airport heading due west (counter-clockwise around the diagram). Once the three planes reach the 45º mark, 45 minutes have passed, and you take the first refueling. Here’s what happens then:
All three planes hit the 45º mark with 135 gallons of fuel, having burned 45 gallons of their 180.
Plane C passes 45 gallons of fuel apiece to plane A and plane B, losing a total of 90 gallons. This fills up A and B to the full 180 gallons, but takes plane C down to 45.
Planes A and B press on, while plane C instantaneously turns around and heads east (clockwise) back to the airport.
The second refueling occurs when planes A and B reach the 90º mark and 90 minutes have passed. Plane C reaches the airport at the same time and lands just as her aircraft uses up the last bits of fuel. Here’s what happens then:
Planes A and B hit the 90º mark with 135 gallons of fuel each, having flown 45º and burned 45 gallons since being topped off in the first refueling.
Plane B passes 45 gallons of fuel to plane A. Plane A goes from 135 gallons of fuel up to the full 180. Plane B goes from 135 gallons down to 90.
Plane A presses on, and plane B turns around to head east back to the airport. Meanwhile, plane C is on the ground at the airport refueling up to the full 180 gallons.
Now you are heading around the far side of the globe, making the longest part of the journey solo. When 180 minutes have passed and you hit the 180º mark in plane A, plane C—which, remember, has been refueling at the airport—takes off heading east (clockwise). At the same time, plane B arrives at the airport and lands to refuel just as he runs out of gas.
At this point, planes A and C are headed toward each other. They meet at the 270º mark.
At this moment, plane A has 0 gallons of fuel, having used all 180 gallons in the tank since its refueling at the 90º point. Plane C has 90 gallons.
Plane C pulls an incredible aerobatics maneuver, passing 45 gallons of fuel to plane A and turning around to head west (counter-clockwise) at the same time.
Planes A and C are now heading west toward the airport from the 270º with 45 gallons of fuel each.
At the same moment that plane C passes fuel to plane A, plane B takes off from the airport where it has been refueling to the full 180 gallons, heading east.
At the 315º mark, when 315 minutes have passed, you are reunited with both your fellow pilots John and Jane again. Now it’s up to John, in plane B, to pass fuel to you and Jane so you don’t crash.
Plane A and C are at 0 gallons of fuel when they hit the 315º mark. Fortunately, at that exact moment, plane B is there with 135 gallons, just enough fuel for all three aircraft.
Plane B passes 45 gallons of fuel to both planes A and C, and it about-faces to head west at the same moment. Each plane now has 45 gallons of fuel, enough to make it back to the airport at the 360º mark.
Congratulations! You successfully flew all the way around the world without stopping or letting either of your loyal assistants perish in a fiery X-100 plane crash.
A father said to his son, “Two years ago I was three times as old as you; but in fourteen years I shall be only twice as old as you. What were the ages of each?”
As Talwalker goes on to explain, the solution is a relatively simple one, represented by the equation f-2 = 3 (s-2), if “f” and “s” represent the father and son’s ages, respectively.
Now for the second part: If the father will only be twice as old as his son in 14 years, their ages at that time can be represented by the equation f+14 = 2(s+14), with “f” and “s” once again standing in for the father and son.
math equation
The Answer is 4.
Write down each of the number given on the left side of the equation in English e.g. TWELVE, SIX etc. The number on the right side gives in the number of letters of the English word written on the left side of equation.
So TWELVE = 6
SIX = 3 etc and so on and hence FOUR = 4
