This article first appeared in issue 03 of Chalkdust. I highly recommend reading the rest of the magazine (and trying to solve the crossnumber I wrote for the issue).
It all began in December 1956, when an article about hexaflexagons was published in Scientific American. A hexaflexagon is a hexagonal paper toy which can be folded and then opened out to reveal hidden faces. If you have never made a hexaflexagon, then you should stop reading and make one right now. Once you've done so, you will understand why the article led to a craze in New York; you will probably even create your own mini-craze because you will just need to show it to everyone you know.
The author of the article was, of course, Martin Gardner.
A Christmas flexagon.
Martin Gardner was born in 1914 and grew up in Tulsa, Oklahoma. He earned a bachelor's degree in philosophy from the University of Chicago and after four years serving in the US Navy during the Second World War, he returned to Chicago and began writing. After a few years working on children's magazines and the occasional article for adults, Gardner was introduced to John Tukey, one of the students who had been involved in the creation of hexaflexagons.
Soon after the impact of the hexaflexagons article became clear, Gardner was asked if he had enough material to maintain a monthly column. This column, Mathematical Games, was written by Gardner every month from January 1956 for 26 years until December 1981. Throughout its run, the column introduced the world to a great number of mathematical ideas, including Penrose tiling, the Game of Life, public key encryption, the art of MC Escher, polyominoes and a matchbox machine learning robot called MENACE.


Gardner regularly received topics for the column directly from their inventors. His collaborators included Roger Penrose, Raymond Smullyan, Douglas Hofstadter, John Conway and many, many others. His closeness to researchers allowed him to write about ideas that the general public were previously unaware of and share newly researched ideas with the world.
In 1970, for example, John Conway invented the Game of Life, often simply referred to as Life. A few weeks later, Conway showed the game to Gardner, allowing him to write the first ever article about the now-popular game.
In Life, cells on a square lattice are either alive (black) or dead (white). The status of the cells in the next generation of the game is given by the following three rules:
For example, here is a starting configuration and its next two generations:
The first three generations of a game of Life.
The collection of blocks on the right of this game is called a glider, as it will glide to the right and upwards as the generations advance. If we start Life with a single glider, then the glider will glide across the board forever, always covering five squares: this starting position will not lead to the sad ending where everything is dead. It is not obvious, however, whether there is a starting configuration that will lead the number of occupied squares to increase without bound.
Gosper's glider gun.
Originally, Conway and Gardner thought that this was impossible, but after the article was published, a reader and mathematician called Bill Gosper discovered the glider gun: a starting arrangement in Life that fires a glider every 30 generations. As each of these gliders will go on to live forever, this starting configuration results in the number of live cells perpetually increasing!
This discovery allowed Conway to prove that any Turing machine can be built within Life: starting arrangements exist that can calculate the digits of pi, solve equations, or do any other calculation a computer is capable of (although very slowly)!

Encrypting with RSA

To encode the message \(809\), we will use the public key:
$$s=19\quad\text{and}\quad r=1769$$
The encoded message is the remainder when the message to the power of \(s\) is divided by \(r$:

Decrypting with RSA

To decode the message, we need the two prime factors of \(r\) (\(29\) and \(61\)). We multiply one less than each of these together:
\begin{align*} a&=(29-1)\times(61-1)\\[-2pt] &=1680. \end{align*}
We now need to find a number \(t\) such that \(st\equiv1\mod a\). Or in other words:
$$19t\equiv1\mod 1680$$
One solution of this equation is \(t=619\) (calculated via the extended Euclidean algorithm).
Then we calculate the remainder when the encoded message to the power of \(t\) is divided by \(r\):


Another concept that made it into Mathematical Games shortly after its discovery was public key cryptography. In mid-1977, mathematicians Ron Rivest, Adi Shamir and Leonard Adleman invented the method of encryption now known as RSA (the initials of their surnames). Here, messages are encoded using two publicly shared numbers, or keys. These numbers and the method used to encrypt messages can be publicly shared as knowing this information does not reveal how to decrypt the message. Rather, decryption of the message requires knowing the prime factors of one of the keys. If this key is the product of two very large prime numbers, then this is a very difficult task.

Something to think about

Gardner had no education in maths beyond high school, and at times had difficulty understanding the material he was writing about. He believed, however, that this was a strength and not a weakness: his struggle to understand led him to write in a way that other non-mathematicians could follow. This goes a long way to explaining the popularity of his column.
After Gardner finished working on the column, it was continued by Douglas Hofstadter and then AK Dewney before being passed down to Ian Stewart.
Gardner died in May 2010, leaving behind hundreds of books and articles. There could be no better way to end than with something for you to go away and think about. These of course all come from Martin Gardner's Mathematical Games:

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This post explains how to make a trihexaflexagon with and images you like on the three faces.

Making the template

To make a flexagon template with your images on, visit On this page, you will be able to choose three images (png, jp(e)g or gif) which will appear on the faces of your flexagon. Once you have created the template, save and print the image it gives you.
The template may fail to load if your images are too large; so if your template doesn't appear, resize your images and try again.

Making the flexagon

First, cut out your printed tempate. For this example, I used plain blue, green and purple images.
Then fold and glue your template in half lengthways.
Next, fold diagonally across the blue diamond, being careful to line the fold up with the purple diamond. This will bring two parts of the purple picture together.
Do the same again with the blue diamond which has just been folded into view.
Fold the green triangle under the purple.
And finally tuck the white triangle under the purple triangle it is covering. This will bring the two white triangles into contact. Glue these white triangles together and you have made a flexagon.

Flexing the flexagon

Before flexing the flexagon, fold it in half through each pair of corners. This will get it ready to flex in the right places.
Now fold your flexagon into the following position.
Then open it out from the centre to reveal a different face.
This video shows how to flex a flexagon in more detail.

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Last weekend, I attended Electromagnetic Field, a camp for hackers, geeks, makers and the interested. On the Sunday, I gave a talk on four mathematical ideas/tasks which I have encountered over the past few years: Flexagons, Folding Tube Maps, Braiding and Sine Curves. I'd love to see photos, hear stories, etc from anyone who tries these activities: either comment on here or tweet @mscroggs.


It's probably best to start by showing you what a flexagon is...
What you saw there is called a trihexaflexagon. Tri- because it has three faces; -hexa- because it is a hexagon; and -flexagon because it can be flexed to reveal the other faces.
The story goes that, in 1939, Arthur H. Stone, who was an Englishman studying mathematics at Princeton, was trimming the edges off his American paper to fit in his English folder. He was fiddling with the offcuts and found that if he folded the paper under itself in a loop, he could make a hexagon; and when this hexagon was folded up as we saw, it would open out to reveal a different face.
The way it flexes can be shown on a diagram: In the circles, the colour on either side of the flexagon is shown and the lines show flexes which can be made.
When Stone showed his flexagon to other students at Harvard, they were equally amazed by it, and they formed what they called 'The Flexagon Committee'. Members of the committee included Richard Feynman, who was then still a graduate student. The committee could meet regularly and soon discovered other flexagons, the first of which was the hexahexaflexagon: Again shaped like a hexagon, but this time with six faces.
A hexahexaflexagon is created by taking a longer strip of paper and rolling it around itself like this. The shorter strip at the end is then folded and glued in the same way the trihexaflexagon was. Once made, the hexahexaflexagon can be flexed. From some positions, the flexagon can be flexed in different ways to reveal different faces. Due to this, finding some of the faces can be quite difficult.
The committee went on to find other flexagons which could be made, again made by first folding into a shorter strip, then folding up like the trihexaflexagon.
The committee later found that hexaflexagons with any number of faces could be made by starting with a certain shaped strip, rolling it up then folding it like a trihexaflexagon.

Resources & further reading

An excellent article by Martin Gardner on flexagons can be found in this book.
Trihexaflexagon templates (click to enlarge then print):
Our second story starts with me sitting on the tube reading Alex's Adventures in Numberland by Alex Bellos on the tube. In his book, Alex describes how to fold a tetrahedron, or triangle-based pyramid, from two business cards. With no business cards to hand, I picked up two tube maps and followed the steps: first, I folded it corner to corner; then I folded the overlaps over.I made another one of these, but the second a mirror image of the first, slotted them together and I had my tetrahedron.
Then I made a tube map cube by making six squares like so and slotting them together.
While making these shapes, I discovered an advantage of tube maps over business cards: Due to the pages, folded tube maps have slots to tuck the tabs into, so the solids are pretty sturdy.
Making these shapes got me wondering: what other Platonic solids could I make?
In 2D, we have regular shapes: shapes with all the sides of the same length and all angles equal. Platonic solids are sort of the 3D equivalent of this: they are 3D shapes where every face is the same regular shape and at each vertex the same number of faces meet.
For example, our tetrahedron is a Platonic solid because every side is a regular triangle, and three triangles meet at every vertex. Our cube is a Platonic solid because every side is a square (which is a regular shape) and three squares meet at every vertex.
In order to fold all the Platonic solids, we must first find out how many there are.
To do this, we're going to start with a triangle, as it is the 2D shape with the smallest number of sides, and make Platonic solids.
If we try to put two triangles at each vertex, then they'll squash flat; so that's no good. We've seen that three triangles at each vertex makes a tetrahedron. If we put four triangles at each vertex then we get an octahedron.
Five triangles at each vertex gives us an icosahedron.
Each angles in an equilateral triangle is 60°. So if we put six triangles at each vertex the angles add up to 360°, a full turn. This means that the triangles will lie flat, giving us a nice pattern for a kitchen floor, but not a solid. Any more than 6 triangles will add up to more than 360 and also not give a solid. So we have found all the Platonic solids whose faces are triangles.
Next, four sided faces. Three squares at each vertex gives us a cube. Four squares at each vertex will add up to 4 times 90°... 360° again, so another kitchen floor and as before we have all the Platonic solids whose faces are squares.
Now moving up again to five sided faces. Three pentagons at each vertex will gives us a dodecahedron, which looks like this.
This is the best I could do.
(After the talk, I was shown a few better ways to fold pentagons. Watch this space for my attempts...) Now if we try four pentagons around a vertex: the internal angle in a pentagon is 108°. 4 times 108° is 432°. This is more than a full turn, so we don't get a solid.
Moving up again, if we take three hexagons we get another tessellation. Shapes with more than six sides will all have larger angles than this so three make more than a full turn. Therefore, we have found and folded all the Platonic solids.
In 2012, I posted this on my blog and got the following comment:
I'm pretty sure this was a joke, but one hour, 48 tube maps and a lot of glue later:

Resources & further reading

Alex's Adventures in Numberland by Alex Bellos introduced business card folding and takes it further, finishing with a business card Menger sponge.


A few months ago, my mother showed my a way to make braids using a cardboard octagon with a slot cut on each side and a hole in the middle.
To make a braid, seven strands of wool are tied together, fed through the hole, then one tucked into each slot.
Now, we jump over two strands, pick the third strand and move it to the vacant slot. So first, we jump over the orange and green and move the red strand.
Then we jump the light blue and yellow and move the dark blue.
And so on..
After a while, the braid looks like this:
Once I'd made a few braids, I began to wonder which other numbers of threads could be used to make braids like this. To investigate this I found it useful to represent braids by drawing connections to show where a thread is moved. This shows the first move:
Then the second move:
And so on until you get:
After the octagon, I tried braiding on a hexagon, moving the second thread each time. Here's what happened:
I only moved the yellow and green threads and nothing interesting happened. When I drew this out as before, it demonstrated what had gone wrong: three slots are missed so three threads are never moved.
So we need to find out when slots are missed and when all the slots are hit. To do this, let's call the number of slots \(a\), and let \(b\) be the number thread we pick each time. For example, in the first braid that worked \(a\) was 8 and \(b\) was 3.
First we'll label the slots. Label the slot which starts empty 0, then number anti-clockwise. This numbering puts all the multiples of \(a\) at the bottom slot.
Now let's look at which slots we visit. We start at0, then visit \(b\), then \(2b\), then \(3b\) and so on. We visit all the multiples of \(b\).
Therefore we will reach the bottom slot again and finish our loop when we reach a common multiple of \(a\) and \(b\). The first time this happens will be at the lowest common multiple, or:
On our way to this slot, we visited one slot for every \(a\) we passed, so the number of slots we have visited is
and we will visit every slot if
or, equivalently if
This is true when, \(a\) and \(b\) have no common factors, or in other words are coprime; which can be written
So we've found that if \(a\) is the number of slots and \(b\) is the jump then the braid will not work unless \(a\) and \(b\) are coprime.
For example, if \(a\) is 6 and \(b\) is 2 then 2 is a common factor so the braid fails. And, if \(a\) is 8 and \(b\) is 3 then there are no common factors and the braid works. And, if \(a\) is 12 and \(b\) is 5 then there are no common factors and the braid works.
But, if \(a\) is 5 and \(b\) is 2 then there are no common factors but the braid fails.
The rule I've explained is still correct, and explains why some braids fail. But if \(a\) and \(b\) are coprime, we need more rules to decide whether or not the braid works.
And that's as far as I've got, so I'm going to finish braiding with two open questions: Why does the 5 and 2 braid fail? And for which numbers \(a\) and \(b\) does the braid work?

Sine curves

For the last part of the talk, I did a practical demonstration of how to draw a sine curve using five people.
I told the first person to stand on the spot and the second person to stand one step away, hold a length of string and walk.
The third person was instructed to stay in line with the second person, while staying on a vertical line.
The fourth person was told to walk in a straight line at a constant speed.
And the fifth person had to stay in line with both the third and fourth people. This led them to trace a sine curve.
To explain why this is a sine curve, consider the following triangle:
As our first two people are one step apart, the hypotenuse of this triangle is 1. And so the opposite (vertical) side is equal to the sine of the angle.
I like to finish with a challenge, and this task leads nicely into two challenge questions:
1. How could you draw a cosine curve with five people?
2. How could you draw a tan(gent) curve with five people?

Resources & further reading

People Maths: Hidden Depths is full of this kind of dynamic task involving moving people.

Similar posts

Tube map Platonic solids, pt. 3
Tube map Platonic solids, pt. 2
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Braiding, pt. 1


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Here is a collection of Christmas relates mathematical activities.


I first encountered flexagons sometime around October 2012. Soon after, we used this template to make them at school with year 11 classes who had just taken GCSE papers as a fun but mathematical activity. The students loved them. This lead me to adapt the template for Christmas:
And here is an uncoloured version of the template on that site if you'd like to colour it yourself and a blank one if you'd like to make your own patterns:
The excitement of flexagons does not end there. There are templates around for six faced flexagons and while writing this piece, I found this page with templates for a great number of flexagons. In addition, there is a fantastic article by Martin Gardner and a two part video by Vi Hart.

Fröbel stars

I discovered the Fröbel star while searching for a picture to be the Wikipedia Maths Portal picture of the month for December 2013. I quickly found these very good instructions for making the star, although it proved very fiddly to make with paper I had cut myself. I bought some 5mm quilling paper which made their construction much easier. With a piece of thread through the middle, Fröbel starts make brilliant tree decorations.

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