# Understandable questions which are hard for non-mathematicians but easy for mathematicians [closed]

A friend of mine has set me the challenge of finding an example of the following:

Is there a question, that everyone (both mathematicians and non-mathematicians) can understand, that most mathematicians would answer correctly, instantly, but that most non-mathematician would struggle to solve/take longer to arrive at a solution?

When I say mathematician, I mean a person who has studied maths at degree level. The key feature of such a question is that it should be understandable to an average person. I realise, of course, that this is a subjective question - what does 'most mathematicians' mean, what would 'most mathematicians' be able to answer? Nonetheless, I would be interested to hear peoples' opinions and ideas:

Can you think of a question, that in your opinion, is an example of the above?

An example of such a question, I think, would provide a good way of explaining to people how mathematicians think. It could also be a good teaching tool (i.e to show how mathematicians approach problem solving).

My friend suggested the following question:

Does there exist a completed Sudoku grid with top row $1,2,3,4,5,6,7,8,9$?

I won't give the answer (so that you can see for yourself if it works!). When we asked this to fellow maths researchers, nearly all were able to give the correct answer immediately. When I asked the undergraduates that I teach, most of them (but not all) could answer correctly and pretty quickly. I like this question but I'm sure there's a better one.

Thanks!

Edit: As for the Sudoku question, most of the researchers I asked answered in 10 seconds with the solution:

Yes. You can relabel any Sudoku (i.e swaps sets of numbers - change all $1$s for $2$s for example) and still have a valid Sudoku solution. So, in a sense all Sudoku are equivalent to a Sudoku with one to nine in the first row.

• My conjecture: "for all Math questions posted below, I can find at least one "mathematicians" that do not know how to solve the questions" – user99914 Feb 28 '18 at 9:43
• Indeed! This question is just for a bit of fun. It is clearly subjective as I pointed out in my question. – Zestylemonzi Feb 28 '18 at 9:45
• The 36 officer problem is easily stated and understood, but I doubt many mathematicians would be able to prove it. (You have 6 regiments and each regiment has 6 ranks of officer. Can you arrange these officers in a $6\times 6$ square such that each row and column contains an officer of each rank and of each regiment? Its a bit like lying two SuDoKu puzzles, one on top of the other. This was stated in 1782 by Euler, solved in 1901 by Tarry.) – user1729 Feb 28 '18 at 9:57
• The traditional example is "What is the sum of every number from 1 to 100?" Few non-mathematicians get this the first time they hear it, but almost all mathematicians will figure it out very quickly. And the non-mathematicians who do get it are usually in "math-adjacent" fields like programmers, engineers and physicists. – RBarryYoung Feb 28 '18 at 18:29
• @DannyPflughoeft Take any completed sudoku grid. Say the top row happens to be 967213584. Then change all the 9's in the grid to 1's, all the 6's to 2's, all the 7's to 3's, etc... and now you still have a valid grid but the top row is 123456789. – user253751 Feb 28 '18 at 23:27

The standard test for distinguishing mathematicians from normal people is a two-part test. In Part One, there is a kettle full of water on the floor, and a stove with one burner lit: how do you heat the water in the kettle? Everyone answers, pick the kettle up off the floor, and put it on the lit burner.

In Part Two, there is a kettle of water on a table, and a stove with one burner lit: how do you heat the water in the kettle? If you answered, take the kettle off the table, and put it on the lit burner, well, there's nothing wrong with that answer, but it does prove you're not a mathematician. The mathematician's answer is,

take the kettle off the table, and put it on the floor. This reduces it to a problem already solved.

• The physicist answer would be to say that the most likely sequence of event is that someone heated the kettle and forgot to turn off the heater. All other paths are negligible,hence the water is already warm. – Three Diag Feb 28 '18 at 12:45
• I don't get it. Interacting with the kettle in anyway will heat the water inside of it eventually. How does it being on the floor solve the problem of heating it? – Mazura Feb 28 '18 at 17:50
• putting the kettle on the floor means that the second scenario is identical to the first one. Since there is a solution for the (known) first problem you can use the same solution for the second problem. Basically, if you can get the kettle on the floor you can get it on the burner too. – BlueWizard Feb 28 '18 at 18:05
• @Mazura : it's a joke, one of many based on the idea that when proving something if you are trying to get from A to B, but have already shown how to get from C to B then getting from A to C gets you to B without any additional effort. So even if A to C to B is longer than A to B, a mathematician will choose it anyways as long as A to C is shorter than A to B. – Arcanist Lupus Feb 28 '18 at 20:32
• Does the experimental physicist light the table on fire? – rrauenza Feb 28 '18 at 23:12

A good example is the Bridges of Königsberg puzzle. An important city in 18th century Prussia was the city of Königsberg (modern day Kaliningrad, a Russian enclave) which had seven bridges. The residents played a game: try to cross every bridge precisely once. No one could solve this puzzle for a long time.

Euler proved this was impossible. Every mathematician knows the solution and how to solve similar problems, although this is by training rather than their own mental guile :-)

• Thanks for your answer - as you say though a mathematician would answer quickly due to their training opposed to their problem solving ability. The same would be true if you set them the Monty Hall problem for example! – Zestylemonzi Feb 28 '18 at 9:44
• I think that in the 18th century Konigsberg was in Prussia, not Russia. – Gerry Myerson Feb 28 '18 at 11:26
• (According to wikipedia, Königsberg was the capital of Prussia in the 1730s. It was briefly part of Russia from 1758 to 1763. It was then part of a Prussia/Germany until 1946.) – user1729 Feb 28 '18 at 11:46
• Wikipedia sez, "On 31 December 1757, Empress Elizabeth I of Russia issued an ukase about the incorporation of Königsberg into Russia. On 24 January 1758, the leading burghers of Königsberg submitted to Elizabeth. Five Imperial Russian general-governors administered the city during the war from 1758–62; they included William Fermor and Nikolaus Friedrich von Korff. With the end of the Seven Years' War the Russian army abandoned the town in 1763." Before and after those few years, it was part of Prussia. – Gerry Myerson Feb 28 '18 at 11:46
• @KonradRudolph Wikipedia gets confusing here - the article on Berlin claims that Berlin was the capital post 1701, while the article on Königsberg claims that Königsberg was the capital. I'm just going to change this to "an important city in Prussia". – user1729 Feb 28 '18 at 15:50

Here's a famous one:

Cover, if you can, a chess $8\times 8$ with $31$ pieces $2\times 1$ leaving empty the upper-right and bottom-left corners.

I saw many people wasting several hours on possible coverings. On the other hand, a mathematician (or a good high school student) would solve it quite quickly (which is: a piece $2\times 1$ cover a black and a white small square; but the upper-right and bottom-left corners have the same colour).

Here's one: is $$111^3-58^3$$ a prime number? Most non-mathematicians would compute that number, get $$1\,172\,519$$ and then waste some time (or perhaps a lot of time) in search of prime factors.

A mathematician would say that that number is equal to $$(111-58)\times(111^2+111\times58+58^2)$$ and therefore it can not possibly be a prime number.

• I think a smart high schooler would answer as quickly as a PhD this particular question though... – Glougloubarbaki Feb 28 '18 at 9:42
• it's enough to notice that 111-58 must divide the number. No need to waste time on the second factor. – Ittay Weiss Feb 28 '18 at 10:18
• That's not really a math problem, but a computation problem. – Clearer Feb 28 '18 at 13:10
• @Clearer The point is, it's not. – Kimball Feb 28 '18 at 16:06
• $a^3-b^3=(a-b)(a^2+ab+b^2)$ – VortexYT Feb 28 '18 at 17:33

If most mathematicians shall be able to solve it instantly and most non-mathematicians shall be able to solve it but use longer time, I would go for something real simple like:

What is the sum of all numbers from 1 to 100

The more the upper limit is increased (e.g. 100000 instead of 100), the more difficult it will be for non-mathematicians while the upper limit wouldn't make much difference for mathematicians.

• To increase the time difference, perhaps increase the upper limit considerably. – Mark K Cowan Feb 28 '18 at 11:01
• Don't they teach sum of an arithmetic progression in high school? – Gerry Myerson Feb 28 '18 at 11:23
• @GerryMyerson Well, it probably differs from country to country. Where I'm from there are pretty many different types of "high school" lines. To my knowledge only the "math"-line will teach it. But still many will forget about it if they don't continue with some kind of math related activity (e.g. math at university). But, well, you may be correct that this is too simple. On the other hand I think some of the other suggestions are too hard as non-mathematicians is likely to never be able to solve them. – 4386427 Feb 28 '18 at 11:35
• $T_{100}=1+2+3+...+99+100=\frac{1}{2}(100)(100+1)$ – VortexYT Feb 28 '18 at 17:38
• @4386427: Since the question specified struggle to solve, problems that non-mathematicians are unlikely to solve are acceptable. – PJTraill Feb 28 '18 at 22:22

This problem is described in Professor Stewart's Hoard of Mathematical Treasures as "a time-honoured way to make money in a pub, requiring three cups and one mug. (The mug is human, and should be moderately intoxicated for greater gullibility.)"

You have three cups. Two of them are inverted and one is upright. You can only flip two cups at a time. Your objective is to flip all the cups down.

This is impossible because of a parity argument (the number of inverted cups stays even). This same argument shows that the problem starting from $n$ upright cups, where each move inverts $m$ cups, is impossible iff $n$ is odd and $m$ is even.

• I think this is a good answer. This is usually taught in 1st year university courses (I learned it in linear algebra and later saw it again in group theory) – user370967 Feb 28 '18 at 10:05
• Great answer! Cheers. – Zestylemonzi Feb 28 '18 at 10:05
• I suspect sober non-mathematicians would often work out that it's impossible. – J.G. Feb 28 '18 at 10:21
• @J.G. That's the point made when Stewart recommends that the mug (i.e. mark, victim, etc.) be intoxicated. In the book, this is presented as part of a confidence trick, with the con-artist demonstrating first that it is possible to go from two-upright/one-inverted to all-inverted using the given operation. – Parcly Taxel Feb 28 '18 at 10:28
• "You can flip two cups at a time" or "You must flip two cups at a time"? – DOMiguel Feb 28 '18 at 13:24

There are more rational numbers than natural ones? And what about real numbers? (Cantor's diagonal argument)

Can you prove there are 2 people in the UK with the same number of hairs? (Pigeonhole principle)

• Good example. Generally anything involving "infinity" is quite inaccessible to common sense or ad-hoc reasoning. Without the rigorous tools math provides one is lost. – Peter - Reinstate Monica Feb 28 '18 at 14:31
• I don't think the first type of question is "fair," as it's not something that makes sense without mathematical context. And I'm not sure the second is well posed. I would guess the OP ideally wants problems where non-mathematicians could at least verify an answer presented to them. – Kimball Feb 28 '18 at 19:23
• Do bald people count? – Gerry Myerson Feb 28 '18 at 20:32
• The second one requires an approximate knowledge of the number of hairs on a human head, which some mathematicians may well not know, especially if they are more interested in absolutes than contingencies and a little detached from their surroundings. – PJTraill Feb 28 '18 at 22:24
• @PJTraill It also requires the person to know approximately how many people live in the UK. But I'd say it's reasonable to know that there are about 60 million Brits and that there can't be that many hairs on a human head. – loading... Feb 28 '18 at 22:36

## Example 1:

This question never fails to confuse non-mathematicians. It is pretty good at confusing even beginner math students, people who were "good" at math in high school.

You have a summer vacation home by a lake and through the back window you can only see exactly one-half of the lake. One summer, some algae starts growing on the surface of the lake where you can't see it from the window. The surface area of the algae doubles every day such that in thirty days it will cover the entire lake. How many days before you'll see the algae through your window?

The unstated assumption is that I am asking for the worst case scenario so the algae starts at the point furthest away from the visible part of the lake. The solution is:

You will not the see the algae until day 29.

People are shocked and amazed because people love linear relationships and just assume linear relationships by default. In this case, I tell them it is doubling everyday so it is a geometric relationship but everyone still says something like 15 days. Or sometimes they realize it is a "trick" question and the answer can't be as obvious as 15 so it is 14 or 16 or something. Anyone who is decently trained to listen carefully and think critically should get this after about five seconds.

I always use this example whenever I am talking about exponential growth/decay, especially when I am explaining global warming, population growth, depletion of resources, polar ice caps melting, etc. The moral of the story is, sometimes we will simply not even see the problem until it is too late.

## Example 2:

Another category which always confuses the general public is percentages. I have managed to confuse small business-owners, cashiers, and salesmen who deal with buying, selling, discounts, taxes, sales, markups, growth rates, profits, and losses every single day. The biggest shock for me was someone who had been in business for more than a decade. He was confused and couldn't give me a correct answer despite repeated attempts, with a calculator.

You walk into a store and purchase an item. The total at the counter, including sales tax, was \$100. The sales tax rate is 10%. What was the tag price of the item on the shelf? The cultural assumption is that the displayed tag price of the item doesn't include the sales tax so sales tax is added at the counter. I know that different parts of the world have different norms but this is how it is done in the USA. Inevitably the first, and the intuitive, answer is always \$90. Then I point out that \$90 plus 10% gives you \$99 so that cannot be the correct answer. After the shock and the amazement, they think I am bamboozling them with the language. I point out that it is a straightforward, simple, and short question. It is not a trick question. They always ask me to repeat and they listen super carefully the second time. Then the (cop-out) answer is that "oh, it is just a bit more than \$90". I say yes but can you tell me exactly what it is. Then they grab a calculator and start incrementally increasing \$90 and adding ten percent to it until they get \$100. Some are quick enough to use a bisection-method-variant to quickly hone in on the correct amount in a couple of tries. Anyone who remembers beginner algebra should quickly setup the equation$x+0.1x=100$and get the original tag price to be \$ 90$\frac{10}{11} \approx \$90.91 $. This results in the total amount being \$100 to two decimal places.

The problem here is of course, that people always assume 10% of the final amount, which is \$100. We don't want that. The sales tax is calculated on the tag price and then added to get the final price. The final price should be divided by 1.1 instead of multiplied by 0.9. The reciprocal of 1.1 is not 0.9. Adding ten percent of a quantity and then subtracting ten percent of the sum will not give you the original quantity. You will subtract a tiny bit more than you should. Sure, for a mathematician this is not mathematics. It is just arithmetic. But for the general public this is indeed "math". ## Example 3: Another one, which I always give as extra credit, confuses even advanced undergrads. Give me two numbers which add up to ten and multiply out to a hundred. The unspoken rule here is that don't assume numbers to be the integers only, positive or otherwise. This is a super short, easy to understand question. Everyone starts with thinking something like ten times ten is a hundred but ten plus ten is twenty. Okay, not quite, but the answer should be close enough to ten and ten. Maybe something like eleven and nine. Then they start increasing one, decreasing the other. Some people start systematically writing tables. But of course, there is no solution in the integers. A mathematician, after trying a few integers as above, should setup the system, \begin{array}{cr} x+y&=&10\\ xy&=&100 \end{array} obtaining the equation$x^2-10x+100=0$which gives you \begin{array}{cl} x&=&5+i\sqrt{75}\\ y&=&5-i\sqrt{75}. \end{array} These indeed add up to ten and multiply out to one hundred but they just happen to be complex numbers. The problem here is the "number" assumption. Upon hearing the word "number" everyone assumes that "number" means a positive integer. A mathematician should always think about the assumptions that the question or the answer statement assumes. Okay so maybe this is a "trick" question. A bit disingenuous to ask the general public but math undergrads should always get this. It is my favorite extra credit problem to give. • When someone asks me to pick a number between one and ten, I usually pick pi. Never fails to annoy. – Gerry Myerson Feb 28 '18 at 20:27 • Good thing that algae is stubbornly avoiding the half you can see... It could've started just on the other side of the boundary, and then the first doubling could spread over the boundary and make it visible. – user253751 Feb 28 '18 at 23:36 • What's i, and what does the square root of 75 times i plus five equal? – Mazura Mar 1 '18 at 0:39 • @Maz,$i$is a square root of minus one (and$5+i\sqrt{75}$equals$5+i\sqrt{75}$– it has been simplified as much as it can be). Do a websearch for "complex numbers" or "imaginary numbers", and all will be revealed. – Gerry Myerson Mar 1 '18 at 1:49 • In 29 days the algae covers half the lake. If the half it covers is the half you don't see, then you still won't see the algae on day 29. You'll only see it on day 30. – Joel Reyes Noche Mar 4 '18 at 13:09 All existing answers are about math problems. Studying mathematics, however, improves your ability to solve all kinds of logic puzzles. You'll be surprised how many people who didn't study math at university will get the answer to this puzzle wrong: You are given cards with only a letter on one side and only a digit on the other. Your task is to flip as few cards as possible when checking whether it is true that if there is an odd number on one side, there is a vowel on the other. The card faces you see are: R 8 E 3 Which cards do you need to flip? If you studied math at all, however, you get the answer right instantly. • As I haven't studied mathematics, I'm not sure about my solution. I'd pick 3 and R as it wasn't stated that every vowel has to have an odd number on the other side. Therefore I just have to check if the other side of R is even to confirm the rule. – Adyrem Feb 28 '18 at 13:09 • @Adyrem That's correct. But this isn't actually about misunderstanding the question. There are many people who understand the question perfectly well and only choose two cards but choose the one showing an odd number and the one showing a vowel. They don't understand implications and want to make sure the one showing a vowel actually has an odd number on the other side, even though this is of no relevance. – UTF-8 Feb 28 '18 at 15:02 • Zero. My task is to flip as few cards as possible. when you want me to check for that other thing, let me know. :) – Mazura Feb 28 '18 at 20:31 • @Mazura Then why is your answer zero? Shouldn't it be the empty set? ;-) – UTF-8 Feb 28 '18 at 22:21 • This is an interesting question. Indeed, many people flounder with it. On the other hand, you can ask an isomorphic question that the same people tend to find easy: suppose the law says you can only drink alcohol if you're aged at least 18. We see 4 people drinking, and we are told: Pete is drinking lemonade, Doug is 17, Emma is 20, Alice is drinking beer. To check everything's legal, which people do we need to know more about? – James Martin Feb 28 '18 at 23:28 The 14-15 puzzle was introduced by Sam Lloyd* in the early 20th Century. It was a 4x4 sliding-square puzzle, like the picture puzzles you find in museums. The squares were numbered: $$\begin{array}{cccc} 1&2&3&4\\ 5&6&7&8\\ 9&10&11&12\\ 13&15&14& \end{array}$$ (The last square is empty so that you can slide the squares around.) Challenge. Swap the 14 and the 15, so that all the squares end up in the correct order. The solution is quite elegant, if you know enough maths: Solution. This is impossible. Swapping two squares corresponds to an "odd" permutation. However, every permutation which takes the blank square back to itself is an "even" permutation (why?). A related, and more modern, puzzle would be: can you rotate a single square in a Rubik's cube? The answer is no, for the same reason as above. *A specific case of a more general puzzle, which he had nothing to do with but claimed credit for. See wiki for more details... • Thanks for your answer! Unfortunately I remembered solving this when I was an undergraduate so couldn't test out if it worked on myself. – Zestylemonzi Feb 28 '18 at 9:59 • “can you rotate a single square in a Rubik's cube?“ clearly you've never taken one apart – DonQuiKong Feb 28 '18 at 13:26 • "can you rotate a single square in a Rubik's cube?" - mathematician: no. Non-mathematician: yes, because you can pull off and re-apply the stickers. – Burnsba Feb 28 '18 at 15:34 • @Burnsba At which the mathematician responds: "That square just looks rotated, but fundamentally it is in the same position. Relabelling does not change anything fundamental." – user1729 Feb 28 '18 at 15:54 • When Rubik's cubes were popular, I got to be quite expert at disassembling them and putting the parts together correctly to "solve" it. – Yoshi Bro Feb 28 '18 at 19:44 Use six pencils to form four equilateral triangles each with sides a pencil length. If this is a bar bet, use toothpicks rather than pencils. • Can I use beer bottles? – PatrickT Feb 28 '18 at 17:25 • You may need some tape – Bonnaduck Feb 28 '18 at 20:24 • @Bonnaduck Not for proof of concept, which would win the bet. – Ethan Bolker Feb 28 '18 at 22:03 The Monty Hall problem probably fits the bill. • But even some mathematicians had a lot of trouble with that one. Look up what happened when Marilyn Vos Savant published the right answer and got told off by several mathematicians. – Gerry Myerson Feb 28 '18 at 11:50 • @GerryMyerson, true, but that's largely because the assumptions behind the problem were not explained very clearly, in particular that Monty knew where the car was and would always open a door to a goat. – PatrickT Feb 28 '18 at 12:37 • When I was presented with this problem the first time, the rules for which doors the host could open and under which conditions was not given; this is a very important piece of information that is (as far as my experience goes) is left out. If it is left out, your assumption on what the correct answer is will not be a valid one and you probably can't tell people apart from this question. – Clearer Feb 28 '18 at 13:16 • @PatrickT. This is a convenient prevarication because there was an intense discussion and clarification about the rules and still most of the people, including statistical (!) mathematicians and very smart people insisted on the wrong solution, including Paul Erdős and Cecil Adams from Straight Dope. It really should not be portrayed as a problem mathematicians solve instinctively. – Thorsten S. Feb 28 '18 at 16:57 • Does it have to be solvable instinctively? I mean, if you write out all the probabilities on a piece of paper it becomes a basic exercise in probability. – Orpheus Feb 28 '18 at 17:21 Here's one I like. You have an electronic scale, and ten piles of ten coins each. Exactly one pile out of the ten is made of fake coins; the other nine piles are made of regular coins. A regular coin weighs 10 grams; a fake one weighs only 9 grams. You are allowed to take any number of coins and weigh them once. How do you find out which pile is made of fake coins? • It took me 5 seconds. You number the piles, take one coin from the first pile, 2 from the second, 3 from the 3rd and so on. The grams short from 550 grams will tell you which pile has the fake coins – Krauss Feb 28 '18 at 10:19 • @Krauss congrats, you're a mathematician ;) – Glougloubarbaki Feb 28 '18 at 10:21 • To be fair, I think a lot of "code monkeys" could solve this problem quickly too. I have heard of some phone interview questions like this one. This is true for many of the answers here. Whether OP counts them as "mathematicians", I don't know. – bames Feb 28 '18 at 10:44 • @bames I agree, unfortunately... – Glougloubarbaki Feb 28 '18 at 20:37 • ... but you can only use the scale once. Otherwise I take one coin from every pile and "weigh them once". Why would I weigh a coin twice? ;p – Mazura Mar 1 '18 at 0:25 I have found that what really confuses people who have not studied mathematics is questions that involve maturity of chances. A good example of this is the Monte Carlo Fallacy. It basically states that "Red has come up 10 times in a row on this roulette, the next time it is bound to be black". A mathematician of course knows why this doesn't work, but as the casinos in Las Vegas show a large enough portion of people don't know this, or choose to ignore this. • It was 26 times, not 10! And it was black not red! – PatrickT Feb 28 '18 at 17:27 • @PatrickT Who said anything about 3628800? – Dason Feb 28 '18 at 19:18 I have two boxes: box A and box B. I place \$1 bills into box A. For each \$1 bill placed int A, I place a \$10 into B. Suppose I do this infinitely many times. Which box contains more value?

• You've destroyed the value of the dollar so they're both worthless. – Dason Feb 28 '18 at 19:16
• You can't "do [something] infinitely many times". You can do something N times, but even as N approaches infinity, box B will contain ten times the dollar amount of box A. So my answer is half-"box B" and half-"Syntax Error". – Sam Feb 28 '18 at 20:27
• To be fair, they did also create two boxes with an infinite capacity, which would surely be a Nobel Prize in Physics-winning event. @Dason – Nij Mar 1 '18 at 5:41
• @Nij if they have the ability to dictate the size of of the bill they can just cut the size in half every time and the whole process will only take finite space. – Dason Mar 1 '18 at 13:24
• Well if you won't circulate ever then both boxes really are worthless. I was right all along! :D – Dason Mar 1 '18 at 19:20

There is also Hamiltonian Path problem from the Graph Theory. Although I could not find a name for this question, I can explain it in a detailed way:

Hamilton, the prisoner, is in the $5 \times 8$ dungeon. Can the prisoner start at $A$ and end at $B$, visiting each of the $40$ cells exactly once by moving upward, downward, leftward and rightward only? Moreover, it also can be asked that in which conditions, this can be done? By conditions I mean, for example we have $m \times n$ dungeon and for which $(m,n)$ pairs, prisoner can go from upperleft corner to lowerleft corner by visiting each of $mn$ cells exactly once?

This was famous when I was in primary school and it was shown us as a "mind game". However, a good mathematician can solve this with a simple argument and can reach a general result easily using the same argument:

For the specific case of $5 \times 8$, it is not possible because in total, there are $39$ cells that prisoner should visit (because he visits $A$ in the beginning). Now assume that $u$ is the total upward moves that prisoner does, $d$ is the downward, $r$ is the rightward and $l$ is the leftward moves, similarly. Then we need to have $u+d+r+l = 39$ (since he should visit each cell exactly once). But also notice that when he finishes his travel, net displacement of him is $4$ cells downward and $0$ cell rightward-leftward direction. Therefore, we have $d-u = 4$ and $r = l$. Now it is easy to see that since $d-u$ is an even number, $d+u$ is also even and since $r = l$, $r+l$ is also an even number. But $u+d+r+l = 39$ is an odd number so he can't have a way to travel each cell exactly once, starting from $A$ and ending at $B$.

• I think a mathematician can do better than that. Make it a checkerboard, note there's as many white squares as black, note that every move changes color, note that A and B are the same color, and you're done. – Gerry Myerson Feb 28 '18 at 11:52
• Yes, that argument is another way to solve that (I have also seen some professors that have done it in your way). Although the argument that I've written seems long, a mathematician can do it even without using pen and paper I believe. – ArsenBerk Feb 28 '18 at 11:56
• Can you reduce the grid to 2x3 (or by any even number of rows and columns) and call it equivalent? – JollyJoker Feb 28 '18 at 14:37

Anything with statistics and probability throws off the general public.

For example, the question:

Does men have more sex partners than women, on average?

Most mathematicians should figure this out, but non-mathematicians might go with the incorrect gut feeling.

It is generally believed that men are more promiscuous than women, so most people would say yes. However, (assuming no same-sex relations), it is clear that every new sex partner for a guy, also implies a new partner for his female partner, thus keeping the averages exactly the same.

• There are usually more women who are sexually active than men in any given society. Given that, men in general have more sexual partners than women. – Clearer Feb 28 '18 at 13:21
• For those struggling with that solution as I was - the number of men and women is assumed to be equal. – Adam Barnes Feb 28 '18 at 13:45
• @Adam: Something which is generally false. It's almost equal, but once you start bringing "real world stuff" into math, "almost" is no longer something you can ignore... – Asaf Karagila Feb 28 '18 at 14:26
• I doubt most mathematicians would just assume the numbers are equal and homosexuality doesn't exist. They tend to be more precise than non-mathematicians. – JollyJoker Feb 28 '18 at 14:31
• Also, are there gender differences on what's considered a sex partner? Or can you do something that makes A a partner of B but not vice versa? – JollyJoker Feb 28 '18 at 14:40

The pigeonhole principle is probably a rich source for generating questions which meet your criterion.

Suppose that you're in a 2 meter by 2 meter room and you throw five pennies in the air. After they land is it possible that each penny is further than 1.5 meters from its nearest neighbor?

Taken from Puzzling, all credits to NL628: link

Suppose you have 100 lbs of cucumbers and these cucumbers consist of 99% water. You decide to leave the cucumbers in the sun for a while until they consist of 98% water. You bring the cucumbers back in, and you think, "Now the cucumbers should weigh a little less than they were before, right?" But, you try as hard as you can, and you still can't figure out how much they weight. How much do the cucumbers weigh?

Solution discovered by Votbear:

I'm gonna say:

50 lbs

Explanation:

(Assuming the 99% water is by weight)

- The % of X in the cucumbers is calculated as Weight of X / Weight of the cucumbers.

- In the start there's 99% water and 1% solids in 100 lbs of cucumbers.

- Only the Weight of water will change in the process. The Weight of solids won't change after evaporation.

- Going from 99% water to 98% water means the % of solids doubled from 1% to 2%.

- Recall that % of solids = Weight of solids / Weight of cucumbers.

- Since the % of solids was doubled, and Weight of solids didn't change, that could only mean the Weight of cucumbers is halved.

As such, the remaining total weight is 100 / 2 = 50 lbs.

A really simple question is "What's larger: 32% of 25, or 25% of 32?"

A general strategy for finding examples would be to look at "real world" applications of mathematical fields. Combinatorics is an area with a lot of such problems. So, for instance, "How many ways can the letters of 'puppies' be arranged in different sequence of letters?" should qualify.

For real analysis, an example would be "If someone walks up a hill starting at 1:00 PM and ending at 2:00 PM, and walks down from 2:00 PM to 3:00, is there any time at which they were at the same point, separated be exactly one hour?"

For group theory, there are a lot of examples based on the orbits of group actions. For instance, "You have N objects in a circle. Is there a sequence of moves that reverses the order of the objects, if each move consists of picking up one object, sliding the next two back one space each, and then putting the object you picked up back in the space you opened up?" Or "Suppose you have a set of non-zero numbers. You can replace any number with the difference between it and another number. Is there a way to set all the numbers to zero?"

For statistics: "Do men or women have the largest average deviation from the mean height? That is, if for each man you were to give them a positive or negative number representing how much taller or shorter they are than the average man, and do the same for women, which group would have the larger average such number?"

Rotate a sphere eastward by ninety degrees. Then rotate it forward by another ninety degrees. The net result is a rotation around some axis. The question is: by how many degrees? (Note: this question can be answered without using paper and pencil. Also, what we concern here is the net change. A rotation by 361 degrees is regarded the same as a rotation by 1 degree, or by 359 degrees if you consider the opposite clock direction.)

I don't know how a non-mathematician would approach it. Based on my experience, a non-mathematician (or strictly speaking, a person without mathematical training at undergraduate level) may even have trouble understanding why the composition of two rotations is still a rotation.

• I think this becomes a lot easier with a cube instead of a sphere. – Paŭlo Ebermann Mar 1 '18 at 0:04

The one that is understandable for both has to do with the well-known "fly between two trains" problem. Especially the variant given in this post.

Two trains 150  miles apart are traveling toward each other along the same track. The first train goes 60 miles per hour; the second train rushes along at 90 miles per hour. A fly is hovering just above the nose of the first train. It buzzes from the first train to the second train, turns around immediately, flies back to the first train, and turns around again. It goes on flying back and forth between the two trains until they collide. If the fly's speed is  120 miles per hour, how many times does it touch the trains?

I have had a long discussion about this with my friends and they still don't believe me. One reason might be that it is indirectly assumed the fly is just a point (it does not have a volume). Even after explanations about that they still say that the number must be finite without a proof.

• The fly does have a volume, but only until the trains meet. – PJTraill Feb 28 '18 at 22:40
• This question was quite a famous math competition question for 5th grade students (about 12-year-old) in china. – tsh Mar 1 '18 at 6:34

How about a little thinking outside the box?

"What's a complex number?" (Or any other definition that excludes most of humanity)

Most mathematicians will know (answer correctly instantly) and most non mathematicians will figure it out if motivated properly (look it up).

It does solve the question as given ;)

• -1. OP clearly asked for questions that "both mathematicians and non-mathematicians can understand." – user1892304 Mar 1 '18 at 13:59
• @user1892304 well, there's plenty of easy obscure definitions that a non mathematician can understand but doesnt know. – DonQuiKong Mar 1 '18 at 14:02
• Regardless of whether the majority of the non-mathematician population knows or even understands what complex numbers are (FWIW, I don't think so), your answer is not in the spirit of the question. The point was to come up with questions that aren't explicitly mathematical in nature that a mathematician would be much better equipped at answering than a non-mathematician. – user1892304 Mar 1 '18 at 14:07
• @user1892304 I know that I didn't answer the spirit but the wording of the question. I actually stated that in my answer ;) – DonQuiKong Mar 1 '18 at 14:35

Prove that there is always a place such that the point exactly the other side of the world has the same temperature and air pressure. This makes use of the intermediate value theorem. One of the most well known theorems to mathematicians.

Solution:

There is always a curve of antipodal points around the world where the temperature is the same by the intermediate value theorem. In that curve there are two antipodal points where the air pressure is the same. This is because it is continuous and there exists a point where it is colder and a place where it is warmer.

• This is generally given as an interpretation of the Borsuk-Ulam Theorem. I'm not completely convinced by your argument with the intermediate value theorem. – Arnaud D. Feb 28 '18 at 16:33
• In fact I don't see how you get antipodal points with the same pressure. – Arnaud D. Feb 28 '18 at 16:39
• The well-known hiker puzzle would make a better example use of intermediate value theorem. – user1551 Feb 28 '18 at 19:46
• This is all gibberish and I don't believe you. +1 – Mazura Mar 1 '18 at 1:54
• In fact, this proof is not correct. See mathoverflow.net/questions/251921/… – Arnaud D. Mar 1 '18 at 10:54

When I was in high school, I was given a math test that included the following problem:

A school has a line of 100 lockers labeled 001 through 100, all of which are closed. The first student goes by and opens every locker. The second student goes by and closes every other locker. The third student goes by and for every third locker, they open it if it was closed or close it if it was open. The fourth student goes by and for every fourth locker, they open it if it was closed or open it if it was open. This process continues, with the Nth student going to every Nth locker and opening it if it was closed and closing it if it was open.

After 100 students have gone by in this way, which lockers are open and which are closed?

.

A mathematician will quickly see that lockers 1, 4, 9, 16, 25, 36, 49, 64, 81, and 100 are open. They do this by observing that the Nth student will only toggle a door if there is a number A such that N*A is the number on the locker. This means there must also be a student A which also toggles the same locker, canceling out the first student's work. The exception to this are the numbers which are squares, where there will be a N*A where N==A, and thus there will be no other student to close the door.

Since I'm an engineer, not a mathematician, I suppose this means that the test doesn't prove you are a mathematician, but I also geek out over the Banach–Tarski paradox, so maybe I'm just weird.

• You missed 49 and 100 in your solution. – Glen O Mar 1 '18 at 6:42
• @GlenO Fixed. Thank you! – Cort Ammon Mar 1 '18 at 7:31

I think a question like

Does the set of "all sets that do not contain itself" contain itself?

would more or less divide people that know set theory foundations from anyone else (based on how someone responds), though I feel this is more of a "trick question" approach.

• This is more of a deeply philosophical/axiomatic question. en.wikipedia.org/wiki/Russell%27s_paradox – PatrickT Feb 28 '18 at 17:32
• That obvious answer is "yes", if you allow such a set to be defined, because the set is itself. (There are many sets that do not contain themselves, but pick an arbitrary one and the answer is still "yes"). I believe you misquoted the question. – user253751 Feb 28 '18 at 23:41
• You’re probably thinking of Russell’s Paradox: “Does the set of all sets that do not contain themselves, contain itself?” (Mu.) – Davislor Mar 1 '18 at 0:25
• @immibis yes you are correct, thank you for pointing that out – Burnsba Mar 1 '18 at 13:16
• @Davislor yes you are correct, thank you for pointing that out – Burnsba Mar 1 '18 at 13:16

If you increase a value by $$10\%$$, then increase again by $$10\%$$, what is the final increase?

Most of the people would respond $$20\%$$.

But $$a\times1.1\times1.1 = a\times1.21$$. Hence the final increase is $$21\%$$.

• This is compound interest, which I think many people are familiar with. – elmer007 Mar 1 '18 at 13:54

How can we cut a hemisphere, by a plane parallel to its base, into two parts of equal curved surface areas?

We just need Archimedes' Hat-Box Theorem. The distance between the cutting plane and the base of the hemisphere is equal to half the radius of the hemisphere.

• I would wager that a large majority of mathematicians do not have Archimedes' Hat Box Theorem memorized and would therefore require much time and effort to arrive at the solution, instead of getting it instantly. – Paul Sinclair Feb 28 '18 at 18:04
• I think the Banach–Tarski paradox would be a followup question – Burnsba Feb 28 '18 at 21:32

When I was an undergrad, young and brash, I would ask in the first lecture of every math class, “Is this where we finally learn why 1 + 1 = 2?” Before I got to Peano arithmetic, one of my classmates finally said to me, “You do know that 1+1 is defined as 2?”

(At least, that’s how I remember him saying it. Interesting discussion in the comments about whether it’s better to define 2 as 1+1.)

• Surely 1 + 1 is usually defined as the successor of 1 + 0 as a special case of the general rule that m + n' = ( m + n )' ? – PJTraill Feb 28 '18 at 22:47
• Isn't it that 2 is defined as 1 + 1? – Paŭlo Ebermann Mar 1 '18 at 0:05
• @PaŭloEbermann I think he said it the other way around, but okay, I’ll change the punch line to be more accurate. – Davislor Mar 1 '18 at 0:11
• @Davislor you don't have to change it, both ways are valid ways of seeing it. (It depends on whether you start from a set with names for numbers, and then define the $+$ operation on it, or you start with a set without names, define $+$, and then define the names using $+$. Basically, you need to define all of $1$, $+$, $2$ and $=$ before discussing the truth of the equation, and there are several ways of doing that.) I prefer $2 := 1 + 1$ (in any ring with unit) to the other way, but that is my personal thing. – Paŭlo Ebermann Mar 1 '18 at 0:20

1) Brower fixed point in its "coffee cup" formulation: prove that when you put sugar in a coffe and then mix with a spoon, at every moment there is a gut of coffee that did not moved.

1.1) Also the hairy ball theorem could work

2) Prove that there exists a real number $x$ such that $x^3=2018x^2+8021x+3$

3) Prove that if you put the map of your city on the table then there is a (unique) point on the map which is exactly in its real postion (contractions theorem)

4) in how many ways one can arrange 5 objects on a line?

4.1) in how many ways one can choose 3 of 5 objects?

5) An annulus can be deformed to a cylinder?

6) Two polygons with the same perimeter do they have the same area?