Category: Skills and Learning

The third diagonal column in Pascal’s Triangle (r = 2 in the usual way of labeling and numbering) consists of the triangular numbers (1, 3, 6, 10, …) – numbers that can be arranged in 2-dimensional triangular patterns. The fourth column of Pascal’s triangle gives us triangular-based pyramidal numbers (1, 4, 10, 20, …), built by stacking the triangular numbers. The columns further out give “higher dimensional” triangular numbers that arise from stacking the triangular numbers from the previous dimension.

It is not by coincidence that the triangular and higher-dimensional triangular numbers appear in Pascal’s Triangle. If you think about layering of polygonal numbers in terms of equations, you get

In the above equation p^d_(k,n) is the nth k-polygonal number of dimension d. Triangular numbers are the 3-polygonal numbers of dimension 2, square numbers are the 4-polygonal numbers of dimension 2, “square based pyramidal numbers” would be denoted as p^3_(4,n).
from the sum above, you can obtain this equation:

Which looks very much like the Pascal Identity C(n,r) = C(n-1,r-1) + C(n-1,r), except for some translation of the variables. To be precise, if we consider the case where k=3 and use r = d and n‘ = n+d-1 we can translate the triangular numbers into the appropriate positions in Pascal’s Triangle.

Along with the definitions for the end columns, the Pascal Identity allows us to generate the whole triangle. This suggests the following strategy for calculating the higher k-Polygonal numbers: create a modified Pascal’s Triangle whose first column is equal to k-2 (instead of 1), and whose last column is equal to 1 (as usual). This modified Pascal’s Triangle is generated using these initial values and the usual Pascal Identity.

Here is an example with k=5, which sets the first column values equal to 3 (except for the top value, which we keep as 1) and yields the pentagonal numbers (column 3) and the higher pentagonal numbers.

The formula for these modified Pascal Triangles is given by this equation:

If we apply the change of variables mentioned above, we can obtain this general formula for the higher polygonal numbers in terms of combinations:

This formula illustrates how polygonal numbers are built out of triangular numbers. It says that the nth d-dimensional k-polygonal number is equal to the nth d-dimensional triangular number, plus (k-3) copies of the n-1 d-dimensional triangular number. This is a little easier to understand when you forget about the higher-dimensions and look at the regular 2-dimensional polygonal number.

For more such insights, log into www.international-maths-challenge.com.

*Credit for article given to dan.mackinnon*

Today I’d like to share an idea. It’s a very simple idea. It’s not fancy and it’s certainly not new. In fact, I’m sure many of you have thought about it already. But if you haven’t—and even if you have!—I hope you’ll take a few minutes to enjoy it with me. Here’s the idea:

So simple! But we can get a lot of mileage out of it.

To start, I’ll be a little more precise: every matrix corresponds to a weighted bipartite graph. By “graph” I mean a collection of vertices (dots) and edges; by “bipartite” I mean that the dots come in two different types/colors; by “weighted” I mean each edge is labeled with a number.

The graph above corresponds to a 3×23×2 matrix MM. You’ll notice I’ve drawn three greengreen dots—one for each row of MM—and two pinkpink dots—one for each column of MM. I’ve also drawn an edge between a green dot and a pink dot if the corresponding entry in MM is non-zero.

For example, there’s an edge between the second green dot and the first pink dot because M21=4M21=4, the entry in the second row, first column of MM, is not zero. Moreover, I’ve labeled that edge by that non-zero number. On the other hand, there is no edge between the first green dot and the second pink dot because M12M12, the entry in the first row, second column of the matrix, is zero.

Allow me to describe the general set-up a little more explicitly.

Any matrix MM is an array of n×mn×m numbers. That’s old news, of course. But such an array can also be viewed as a function M:X×Y→RM:X×Y→R where X={x1,…,xn}X={x1,…,xn} is a set of nn elements and Y={y1,…,ym}Y={y1,…,ym} is a set of mm elements. Indeed, if I want to describe the matrix MM to you, then I need to tell you what each of its ijijth entries are. In other words, for each pair of indices (i,j)(i,j), I need to give you a real number MijMij. But that’s precisely what a function does! A function M:X×Y→RM:X×Y→R associates for every pair (xi,yj)(xi,yj) (if you like, just drop the letters and think of this as (i,j)(i,j)) a real number M(xi,yj)M(xi,yj). So simply write MijMij for M(xi,yj)M(xi,yj).

Et voila. A matrix is a function.

For more such insights, log into www.international-maths-challenge.com.

*Credit for article given to Tai-Danae Bradley*

Brian Hayes recently provided some evidence that there are still many out there who are confounded by the Monty Hall problem.

The Monty Hall problem, perhaps the best-known counter-intuitive probability problem, gets a nice treatment in Jeffery Rosenthal’s Struck by Lightning: The Curious World of Probabilities, and is also explained well (perhaps better) in Mark Haddon’s novel The Curious Incident of the Dog in the Night-time. Professor Rosenthal has some further Monty Hall explanations here.

I just found an alternate version of the problem in Martin Gardner’s The Second Scientific American Book of Mathematical Puzzles and Diversions (published also by Penguin in a slightly different form as More Mathematical Puzzles and Diversions). In the chapter “Probability and Ambiguity” (chapter 19 in both versions of the book), Gardner describes the problem of the three prisoners. Here is a condensed description of the problem:

Three prisoners, A, B, and C are in separate cells and sentenced to death. The governor has selected one of them at random to be pardoned. Finding out that one is to be released, prisoner A begs the warden to let him know the identity of one of the others who is going to be executed. “If B is to be pardoned, give me C’s name. If C is to be pardoned, give me B’s name. And if I’m to be pardoned, flip a coin to decide whether to name B or C.”

The warden tells A that B is to be executed. Prisoner A is pleased because he believes that his probability of surviving has gone up from 1/3 to 1/2. Prisoner A secretly tells C the news, who is also happy to hear it, believing that his chance of survival has also risen to 1/2.

Are A and C correct? No. Prisoner A’s probability of surviving is still 1/3, but prisoner C’s probability of receiving the pardon is 2/3.

It is reasonably easy to see that the 3 prisoners problem is the same as the Monty Hall problem. Seeing the problem in this different formulation might help those who continue to struggle with it.

It is a nice activity to simulate both the 3 prisoners problem and the Monty Hall problem in Fathom – try it and confirm the surprising results that the “second prisoner” is pardoned 2/3 of the time, and that 2/3 of the time, the winning curtain is not the one you selected first.

There are many, many, ways to write these simulations. Here are the attributes and formulas for a Fathom implementation of the three prisoners:

The table below (click on it to see a larger version) shows a separate simulation for the Monty Hall problem. Here we are assuming three curtains “1”, “2”, and “3”, one of which has a prize behind it. You pick one, and then Monty reveals the contents behind one of the other curtains (the curtain with the prize behind it is not shown). In the game, you have the option of switching your choice for the curtain that has not been revealed.

After creating the attributes, you can “run the simulation” by adding data to the collection (Collection->New Cases…), the more the better.

Incidentally, Gardner’s use of A, B, and C reminds me of Stephen Leacock’s “A, B, and C: The Human Element in Mathematics.”

Addendum

A quick search shows that the connection between the Monty Hall problem and the Three Prisoners is well known (see the wikipedia entries on Monty Hall and the Three Prisoners), and that both are alternate formulations of an older problem, known as Bertrand’s box paradox.

For more such insights, log into www.international-maths-challenge.com.

*Credit for article given to dan.mackinnon*

You ordered a pizza for your party, but the restaurant forgot to slice it – these mathematical tricks can help you cut it evenly, says Katie Steckles.

Fairness is important – in life, and in pizza. If you want to cut a pizza into equal-sized pieces, the difficulty will depend on how many people you need to share it between. Luckily, mathematics has some tricks to keep things equal.

For example, if the number of people you are sharing a pizza between is a power of two – one, two, four, eight, 16 – cutting the pizza into as many slices is easy. For one piece, obviously no cuts are needed. For each larger power of two, a cut across through the centre of the pizza – cutting all of the existing pieces exactly in half – will result in pieces of equal size.

Some numbers will be much harder: prime numbers, by definition, can’t be divided easily. Luckily, geometry can help.

If you need to cut a pizza into five equal pieces, first grab a long, thin, rectangular strip of paper. Tie the paper in an ordinary overhand knot, like you would tie in a piece of string. Then, keeping the ends flat, pull gently to tighten the knot. The whole thing will flatten and come together – stop pulling when you can’t go any further without it wrinkling.

The flat shape you are looking at should now be vaguely familiar, if you ignore the two ends of paper sticking out. Fold these ends into the middle, or cut them off, and you will have a shape with five straight edges, created purely by the shape of the knot. Yes, that is right – you have made a perfect regular pentagon, with five equal-length sides and five equal angles at the corners.

It is possible to prove this mathematically by showing that all the folds you make in the paper strip are at 72 degrees to the parallel edges of the strip. But for simplicity, because the paper is the same width everywhere, and weaves in and out five times in the right way, these will be five equal edges. And more importantly, the pentagon’s corners are equally spread around a circle – making it the perfect guide for pizza slicing.

Place your pentagon in the centre of the pizza, then cut along lines radiating out from the centre of the pentagon and through each corner. And presto: you have a pentagonal pizza party for five. This paper-strip method can be used whenever you are in a pentagon-based emergency.

You can use the same technique to produce a shape with any odd number of sides by creating a more complex knot with the strip passing through the middle more times, although the strip of paper needs to be increasingly thin and it takes a lot more patience to pull the ends through and carefully flatten out the shape.

Combined with our existing halving methods, you can now produce any number of slices you like. The same results can be extended to any other round food – thanks to maths, the world is your cheesecake.

For more such insights, log into www.international-maths-challenge.com.

*Credit for article given to Katie Steckles *

If you want your cup of tea to stay as hot as possible, should you put milk in immediately, or wait until you are ready to drink it? Katie Steckles does the sums.

Picture the scene: you are making a cup of tea for a friend who is on their way and won’t be arriving for a little while. But – disaster – you have already poured hot water onto a teabag! The question is, if you don’t want their tea to be too cold when they come to drink it, do you add the cold milk straight away or wait until your friend arrives?

Luckily, maths has the answer. When a hot object like a cup of tea is exposed to cooler air, it will cool down by losing heat. This is the kind of situation we can describe using a mathematical model – in this case, one that represents cooling. The rate at which heat is lost depends on many factors, but since most have only a small effect, for simplicity we can base our model on the difference in temperature between the cup of tea and the cool air around it.

A bigger difference between these temperatures results in a much faster rate of cooling. So, as the tea and the surrounding air approach the same temperature, the heat transfer between them, and therefore cooling of the tea, slows down. This means that the crucial factor in this situation is the starting condition. In other words, the initial temperature of the tea relative to the temperature of the room will determine exactly how the cooling plays out.

When you put cold milk into the hot tea, it will also cause a drop in temperature. Your instinct might be to hold off putting milk into the tea, because that will cool it down and you want it to stay as hot as possible until your friend comes to drink it. But does this fit with the model?

Let’s say your tea starts off at around 80°C (176°F): if you put milk in straight away, the tea will drop to around 60°C (140°F), which is closer in temperature to the surrounding air. This means the rate of cooling will be much slower for the milky tea when compared with a cup of non-milky tea, which would have continued to lose heat at a faster rate. In either situation, the graph (pictured above) will show exponential decay, but adding milk at different times will lead to differences in the steepness of the curve.

Once your friend arrives, if you didn’t put milk in initially, their tea may well have cooled to about 55°C (131°F) – and now adding milk will cause another temperature drop, to around 45°C (113°F). By contrast, the tea that had milk put in straight away will have cooled much more slowly and will generally be hotter than if the milk had been added at a later stage.

Mathematicians use their knowledge of the rate at which objects cool to study the heat from stars, planets and even the human body, and there are further applications of this in chemistry, geology and architecture. But the same mathematical principles apply to them as to a cup of tea cooling on your table. Listening to the model will mean your friend’s tea stays as hot as possible.

For more such insights, log into www.international-maths-challenge.com.

*Credit for article given to Katie Steckles*

Starting with p (p a positive integer) equally distributed dots (vertices) around a circle. Connecting each dot to the next as you move around the circle will give you a regular p-gon – a p-sided polygon with p vertices. If, however, instead of connecting each dot to the one next to it you skip over a fixed number of dots, then you might end up with a star-like pattern, like the ones shown above. In this process, imagine that you start with a particular vertex and move in a counter-clockwise direction. If there are any dots left-over when you get back to the dot you started from, just throw away the unconnected dots.

The Schläfli notation for polygons is very useful for describing regular connected star polygons, and provides an example of how sometimes calculations with notation match exactly with calculations done with diagrams. In this notation, regular polygons like triangle, square, pentagon, etc. are written as {3}, {4}, and {5} respectively. A regular p-gon is written as {p}. If when drawing your p-gon you connect to the second next vertex instead of the first, then you would write this as {p/2}. If you connect to the q’th next vertex, then you would write this polygon as {p/q}. Note that if you are just connecting to the next vertex to make a regular p-gon, this notation gives you {p/1} = {p}, as you would expect.

If you start playing with this process you will notice that {p/q} gives you the same polygon as {p/(p–q)} (as long as you ignore the orientation of the polygon). You may also notice that if q is larger than p, you end up repeating the same patterns, in particular {p/q} = {p/(q mod p)}. Also you will notice that if p and q have common factors, you end up having skipped vertices. In our process we are throwing these away to ensure that our polygons are connected, but you can extend the process and keep them (see note below).

The process described is straightforward to implement in a program. The images shown here were generated in Tinkerplots. To implement it in Tinkerplots, you need two sliders – p and q, and the following attributes:

n = caseIndex()
theta = 2*n*pi(1-q/p)
x = cos(theta)
y = sin(theta)

If you create a plot with y vertical and x horizontal, choose “show connecting lines” and add a filter n<=p+1, you can add a large number of cases to the collection (~200, say) and be able to slide p and q to create a wide variety of connected star polygons. The only restriction is that p must be less than the number of cases you have created. There is nothing special about using Tinkerplots here – any programming environment with reasonable graphics should do a reasonable job (Logo would be fine. :)).

The polygons below are the regular connected polygons based on 12 vertices. Because 12 is divisible by 2, 3, 4, and 6 we end up with regular polygons triangle {3}, square {4}, hexagon {6} and only one star polygon {12/5}. The “degenerate” polygon {2} is known as a “digon.” Here, drawing the diagram first and then seeing what polygon comes out will give you the same result as dividing p/q first and then drawing the corresponding polygon. In this sense, the notation and diagrams nicely reflect each other.

Contrast this with the family of star polygons that are generated when a prime number of vertices are used. The images below are the family of regular connected polygons generated on 13 vertices.

Note – by throwing away the unconnected dots in our process we are ignoring star polygons that are made of overlapping disjoint star or regular polygons, for example two overlapping triangles that make a star of David. These also work well with the Schläfli notation. To create these overlapping polygons, if you have any skipped vertices, you just begin your process again beginning with one of the vertices you skipped over. In the case of {6/2}, instead of getting one triangle {3} you will get two overlapping triangles, or 2{3}. To write a program that would draw these you would want to use something more sophisticated than Tinkerplots.

For more such insights, log into www.international-maths-challenge.com.

*Credit for article given to dan.mackinnon*

Mathematicians are celebrating a 1000-page proof of the geometric Langlands conjecture, a problem so complicated that even other mathematicians struggle to understand it. Despite that, it is hoped the proof can provide key insights across maths and physics.

The Langlands programme aims to link different areas of mathematics

Mathematicians have proved a key building block of the Langlands programme, sometimes referred to as a “grand unified theory” of maths due to the deep links it proposes between seemingly distant disciplines within the field.

While the proof is the culmination of decades of work by dozens of mathematicians and is being hailed as a dazzling achievement, it is also so obscure and complex that it is “impossible to explain the significance of the result to non-mathematicians”, says Vladimir Drinfeld at the University of Chicago. “To tell the truth, explaining this to mathematicians is also very hard, almost impossible.”

The programme has its origins in a 1967 letter from Robert Langlands to fellow mathematician Andre Weil that proposed the radical idea that two apparently distinct areas of mathematics, number theory and harmonic analysis, were in fact deeply linked. But Langlands couldn’t actually prove this, and was unsure whether he was right. “If you are willing to read it as pure speculation I would appreciate that,” wrote Langlands. “If not — I am sure you have a waste basket handy.”

This mysterious link promised answers to problems that mathematicians were struggling with, says Edward Frenkel at the University of California, Berkeley. “Langlands had an insight that difficult questions in number theory could be formulated as more tractable questions in harmonic analysis,” he says.

In other words, translating a problem from one area of maths to another, via Langlands’s proposed connections, could provide real breakthroughs. Such translation has a long history in maths – for example, Pythagoras’s theorem relating the three sides of a triangle can be proved using geometry, by looking at shapes, or with algebra, by manipulating equations.

As such, proving Langlands’s proposed connections has become the goal for multiple generations of researchers and led to countless discoveries, including the mathematical toolkit used by Andrew Wiles to prove the infamous Fermat’s last theorem. It has also inspired mathematicians to look elsewhere for analogous links that might help. “A lot of people would love to understand the original formulation of the Langlands programme, but it’s hard and we still don’t know how to do it,” says Frenkel.

One analogy that has yielded progress is reformulating Langlands’s idea into one written in the mathematics of geometry, called the geometric Langlands conjecture. However, even this reformulation has baffled mathematicians for decades and was itself considered fiendishly difficult to prove.

Now, Sam Raskin at Yale University and his colleagues claim to have proved the conjecture in a series of five papers that total more than 1000 pages. “It’s really a tremendous amount of work,” says Frenkel.

The conjecture concerns objects that are similar to those in one half of the original Langlands programme, harmonic analysis, which describes how complex structures can be mathematically broken down into their component parts, like picking individual instruments out of an orchestra. But instead of looking at these with harmonic analysis, it uses other mathematical ideas, such as sheaves and moduli stacks, that describe concepts relating to shapes like spheres and doughnuts.

While it wasn’t in the setting that Langlands originally envisioned, it is a sign that his original hunch was correct, says Raskin. “Something I find exciting about the work is it’s a kind of validation of the Langlands programme more broadly.”

“It’s the first time we have a really complete understanding of one corner of the Langlands programme, and that’s inspiring,” says David Ben-Zvi at the University of Texas, who wasn’t involved in the work. “That kind of gives you confidence that we understand what its main issues are. There are a lot of subtleties and bells and whistles and complications that appear, and this is the first place where they’ve all been kind of systematically resolved.”

Proving this conjecture will give confidence to other mathematicians hoping to make inroads on the original Langlands programme, says Ben-Zvi, but it might also attract the attention of theoretical physicists, he says. This is because in 2007, physicists Edward Witten and Anton Kapustin found that the geometric Langlands conjecture appeared to describe an apparent symmetry between certain physical forces or theories, called S-duality.

The most basic example of this in the real world is in electricity and magnetism, which are mirror images of one another and interchangeable in many scenarios, but S-duality was also used by Witten to famously unite five competing string theory models into a single theory called M-theory.

But before anything like that, there is much more work to be done, including helping other mathematicians to actually understand the proof. “Currently, there’s a very small group of people who can really understand all the details here. But that changes the game, that changes the whole expectation and changes what you think is possible,” says Ben-Zvi.

For more such insights, log into www.international-maths-challenge.com.

*Credit for article given to Alex Wilkins*

Looking at the last few digits that appear in the numbers that form the sequence b^0, b^1, b^2, b^3, … for b a positive integer, you’ll notice that the digits will always begin to repeat after a certain point. For example, looking at the last digit of the sequences for b = 2, 3, and 4 we have the sequences

b = 2: 1, 2, 4, 8, 6, 2, 4, 8, 6, …
b = 3: 1, 3, 9, 7, 1, 3, 9, 7, …
b = 4: 1, 4, 6, 4, 6, 4, 6, …

If we look at the sequence of last two digits of these sequence where b =2 we have

b = 2: 1, 2, 4, 8, 16, 32, 64, 28, 56, 12, 24, 48, 96, 92, 84, 68, 36, 72, 44, 88, 76, 52, 4, …

This sequence then repeats the loop that began at 4.

We can describe these sequences as T_b,d(n) = (b^n)mod 10^d. Recursively, T_b,d(n) = (T_b,d(n-1)*b)mod 10^d

These sequences are always eventually periodic. Although these sequences are simple to understand and calculate, there are several interesting ways of describing them.

For example, you can think of the elements of T_b,d as a commutative monoid, with multiplication defined as a*b = (a*b)mod 10^d. They form a monoid since 1 is always a member, and you can show that T_b,d is closed under the * operation. It turns out that for some values of b, and d, T_b,d is a group.

You can also think of this set as a finite state machine or graph, where each element is a node and the transition from one node to the next is defined by the operation *b mod 10^d. This provides a nice way of displaying the sequences. The pictures in this post were created by writing a short program to calculate the sequences, and then formatting the output to draw a di-graph in SAGE. The graph at the top of the post is for b=8, d=1, while the graph below is for b=2, d=2. The graph at the bottom of the page is for b=7, d=1.

For more such insights, log into www.international-maths-challenge.com.

*Credit for article given to dan.mackinnon*

Felker prefaces the quote by saying,

Giving up on math means you don’t believe that careful study can change the way you think.

He further notes that writing, like math, “is also not something that anyone is ‘good’ at without a lot of practice, but it would be completely unacceptable to think that your composition skills could not improve.”

Friends, this is so true! Being ‘good’ at math boils down to hard work and perseverance, not whether or not you have the ‘math gene.’ “But,” you might protest, “I’m so much slower than my classmates are!” or “My educational background isn’t as solid as other students’!” or “I got a late start in mathematics!”* That’s okay! A strong work ethic and a love and enthusiasm for learning math can shore up all deficiencies you might think you have. Now don’t get me wrong. I’m not claiming it’ll be a walk in the park. To be honest, some days it feels like a walk through an unfamiliar alley at nighttime during a thunderstorm with no umbrella. But, you see, that’s okay too. It may take some time and the road may be occasionally bumpy, but it can be done!

This brings me to another point that Felker makes: If you enjoy math but find it to be a struggle, do not be discouraged! The field of math is HUGE and its subfields come in many different flavors. So for instance, if you want to be a math major but find your calculus classes to be a challenge, do not give up! This is not an indication that you’ll do poorly in more advanced math courses. In fact, upper level math classes have a completely (I repeat, completely!) different flavor than calculus. Likewise, in graduate school you may struggle with one course, say algebraic topology, but find another, such as logic, to be a breeze. Case in point: I loathed real analysis as an undergraduate** and always thought it was pretty masochistic. But real analysis in graduate school was nothing like undergraduate real analysis (which was more like advanced calculus), and now – dare I say it? – I sort of enjoy the subject. (Gasp!)

All this to say that although Felker’s article is aimed at folks who may be afraid to take college-level math, I think it applies to math majors and graduate students too. I highly recommend you read it if you ever need a good ‘pick-me-up.’ And on those days when you feel like the math struggle is harder than usual, just remember:

Even the most accomplished mathematicians had to learn HOW to learn this stuff!

For more such insights, log into www.international-maths-challenge.com.

*Credit for article given to Tai-Danae Bradley*

A mathematical model of a particle that remembers its past so that it never travels the same path twice produces stunningly complex patterns.

A beautiful and surprisingly complex pattern produced by ‘mathematical billiards’

Albers et al. PRL 2024

In a mathematical version of billiards, particles that avoid retracing their paths get trapped in intricate and hard-to-predict patterns – which might eventually help us understand the complex movement patterns of living organisms.

When searching for food, animals including ants and slime moulds leave chemical trails in their environment, which helps them avoid accidentally retracing their steps. This behaviour is not uncommon in biology, but when Maziyar Jalaal at the University of Amsterdam in the Netherlands and his colleagues modelled it as a simple mathematical problem, they uncovered an unexpected amount of complexity and chaos.

They used the framework of mathematical billiards, where an infinitely small particle bounces between the edges of a polygonal “table” without friction. Additionally, they gave the particle “spatial memory” – if it reached a point where it had already been before, it would reflect off it as if there was a wall there.

The researchers derived equations describing the motion of the particle and then used them to simulate this motion on a computer. They ran over 200 million simulations to see the path the particle would take inside different polygons – like a triangle and a hexagon – over time. Jalaal says that though the model was simple, idealised and deterministic, what they found was extremely intricate.

Within each polygon, the team identified regions where the particle was likely to become trapped after bouncing around for a long time due to its “remembering” its past trajectories, but zooming in on those regions revealed yet more patterns of motion.

“So, the patterns that you see if you keep zooming in, there is no end to them. And they don’t repeat, they’re not like fractals,” says Jalaal.

Katherine Newhall at the University of North Carolina at Chapel Hill says the study is an “interesting mental exercise” but would have to include more detail to accurately represent organisms and objects that have spatial memory in the real world. For instance, she says that a realistic particle would eventually travel in an imperfectly straight line or experience friction, which could radically change or even eradicate the patterns that the researchers found.

For more such insights, log into www.international-maths-challenge.com.

*Credit for article given to Karmela Padavic-Callaghan*