# The mathematical con artist

You’re walking down the street one fine morning, and you see that someone has set up a little table and gathered a small crowd. It’s that sneaky trickster Kyle again. His inspired variants on the shell game have brought him something of a following amongst the local mathematicians. Today, though, he’s got something else entirely. Stepping closer, you see that he’s written out the alternating harmonic series…
$$1 – \frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6} \dots$$ You hear his patter. “This, ladies and gentlemen, is a perfectly ordinary convergent series. Now, I will pay one hundred pounds — no tricks — to anyone who can tell me to what, exactly, this series converges. Don’t think I’m trying to mess you around with a divergent series. If any one of you fine fellows can show that it sums to infinity, they will receive the one hundred pounds just the same as if they showed that it converged. Just five pounds a guess. Who’s first?”

Is Kyle finally going straight, or has he got something up his sleeve? Continue reading

# Maths and music, together in harmony

There’s a deep and well-established link between music and mathematics. For example, both deal with frequencies, waves, and ratios – even if they use different terminology. Yet, when these subjects are taught at school, the connection is usually completely severed. However, reuniting the two can only increase the enjoyment students get from learning mathematics, while also enriching and deepening all aspects of their music-making, from casual singing to composing their own pieces. Here, we take a look at some of the connections to motivate the teaching of the two subjects together, in perfect harmony. Continue reading

# Review of the year 2017

It’s been another big year for Chalkdust, and it started strongly with Adam’s review of the mathematical socks he got for Christmas. The socks had some really nice maths on them—including a proof of Pythagoras’ theorem—and some nonsense, such as $\int(x)=\tan x$.

Shortly after this, TD told us about the maths in the film Mean Girls. Later in the year, this blog post was transformed into TD’s five minute MathsJam talk. But that’s enough about MathsJam for now…

In February, Simon Allen showed us an interesting property of twin prime numbers: the digital root (or repeated digital sum) of the product of every pair of prime numbers is 8. If you’re wondering why this is true, then talk a look at the proof in Simon’s post. Continue reading

# Solution to Christmas conundrum #4

As 2017 comes to an end, it’s time to announce the winners of the fourth and final Chalkdust Christmas Conundrum!

There were 57 entries, 56 of which were correct. The five randomly selected winners are:

• Elena Petrunina
• Andy Sloane
• Molly Barker
• Israel Cerro
• Rata Ingram

You will all soon be proud owners of signed copies of Bletchley Park Brainteasers by Sinclair McKay.

## The solution to conundrum #4

In the fourth conundrum, we asked you discover our secret message. Stop reading now if you don’t want to know the answer yet.

### Part 1: prime factors

These factors were:

1. 2 (An even prime number)
2. 3 (The smallest odd prime number)
3. 3 (A number whose factors add to 4)
4. 13 (A number that Blur, Black Sabbath, Havoc, and many others have named albums after)
5. ? (The key that was used to encode the message in part 4)
6. 17 (The sum of four consecutive prime numbers)
7. 193 (The smallest prime number that is 182 more than another prime number)

The ? must be either 13 or 17 (as the factors are in order). The product of these numbers is therefore either 9980802 or 13051818. It must be 13051818 as we were told it is a 8 digit number.

### Part 2: a Vigenère cipher

The key was Fry, giving the answer Twenty-five million, one hundred and ninety thousand, three hundred and fifteen (25190315).

### Part 3: 12 days of Christmas

1. 12 (After all 12 days, how many partridges has my true love given me?)
2. 28 (How many items does my true love give me on the seventh day?)
3. 12 (After 12 days, how many drummers drumming has my true love given me?)
4. 364 (After 12 days, how many items has my true love given me?)
5. 40 (After 12 days, how many gold rings has my true love given me?)
6. 9 (On which day does the total number of items that my true love has given me first exceed 150?)
7. 120 (After 8 days, how many items has my true love given me?)
8. 55 (How many items does my true love give me on the tenth day?)

This gives the number 18160915

### Part 4: a Caesar shift

Using 17 from the first part to decode this gives fourteen million, one hundred and thirty thousand, onehundred and nineteen (14130119)

### The final message

Concatenating the four answers gives 13051818251903151816091514130119, revealing the message “Merry Scorpion-mas“.

# Christmas conundrum #4

It’s time for the next Chalkdust Christmas Conundrum! But first it’s time to announce the lucky winners of the third conundrum competition. There were 26 entries, of which 18 were correct. The four randomly selected winners are:

• Shelby Noonan
• Jodie Fromage
• Suzie Brown
• Josh Farrant

You will all soon be proud owners of signed copies of The Element in the Room by Helen Arney and Steve Mould. The answer to conundrum #3 is at the bottom of this post.

This week’s prize, perfect for more Christmas puzzling!

The prizes up for grabs for the final conundrum are five copies of Bletchley Park Brainteasers by Sinclair McKay. Inspired by the prize, this conundrum builds up to a hidden message, and two of its parts involve breaking ciphers. Continue reading

# Review of The Element in the Room

As a fan of the Festival of the Spoken Nerd comedy troupe, I was keen to read this book written by two thirds of the nerds.  The stage show focusses on maths and physics, but the book is far wider in its scientific scope. Perhaps unsurprisingly given its title, the book includes a fair amount of chemistry, alongside biology, along with physics and a modest amount of maths.  All of the material is presented to be fun and easily accessible to a non-specialist, and in many cases there are experiments described that one could try at home. Continue reading

# Christmas conundrum #3

It’s time for the next Chalkdust Christmas Conundrum! But first it’s time to announce the lucky winners of last week’s competition. There were 206 entries, of which 189 were correct. The four randomly selected winners are:

• Josie Neal
• Maureen Monroe
• Antonio S. Macias
• Tom Lock

You will all receive copies of The Indisputable Existence of Santa Claus by Hannah Fry and Thomas Oléron Evans. Hope you enjoy them! The answer to last week’s conundrum is at the bottom of this post.

The prizes up for grabs this week are four copies of The Element in the Room by Helen Arney and Steve Mould. To encourage you to try to win this, we will be publishing a review of it on Thursday.

This week’s conundrum is all about nonograms (AKA griddler, picross, tsunami, or “oh, that puzzle”). If you are unfamiliar with nonograms, we’ve prepared a little seasonal introduction to them. Otherwise, feel free to scroll down and dive straight into the third conundrum.

## An Introduction to Nonograms

A nonogram is a logic puzzle where the object is to fill in squares to create a picture using the numbers around the edge of the puzzle.

Consider the following puzzle.

Twinkle Twinkle…

The first column has a single one above it. This means there will be a single square filled in this column. Similarly for the first row. The second column has $1 \, 2$ written above the column. This means there will be one filled in square followed by a group of two filled in squares (working downwards).  The key point here is that there must be at least one blank square between the single square and the group—however it could be more.

Here is another puzzle to try.

Ding dong…

Now that you are familiar with nonograms, it’s time for this week’s conundrum:

## Chalkdust Christmas conundrum #3

The following nonogram puzzle doesn’t have a unique answer: there is more than one pattern of coloured squares that satisfied the clues. How many different solutions are there?

A non-unique nonogram.

Once you’ve worked out how many possible solutions there are, please enter this into the form below for a chance to win one of four copies of The Element in the Room. The deadline for entries is 12:00 (midday) on Friday 22 December. The winners will be announced on Friday afternoon, when we will be publishing the fourth conundrum and giving you a chance to win a copy of Bletchley Park Brainteasers by Sinclair McKay.

This competition is now closed.

## The solution to conundrum #2

In the second conundrum, we asked you to find the number in the star at the top of three. Stop reading now if you don’t want to know the answer yet.

The two numbers between 14 are a cube number and a triangle number: these must be 8 and 6. Next you can see that 23 is the sum of 8, two times a triangle number and a square number: the triangle and square numbers must be 3 and 9.

Next, call the square number at the bottom left $a$, the square number at the top left $b$, and the triangle number at the top right $c$. Adding upwards, we find that $a+b+45=106$ and $a+141+c=198$; and so $a+b=61$ and $a+c=57$. The only two square numbers that add to 61 are 25 and 36. Therefore $c$ must be 21 or 32, but must be 21 as 32 is not a triangle number. And so $a$ is 36 and $b$ is 25.

Putting all these numbers into the tree gives the top number as 433. This is fitting because 433 is in fact a star number:

433 dots arranged to make a star.

# The science behind Santa Claus

Christmas is coming! My favourite season of the year has begun, which means that is time to start celebrating with friends, buying presents for my loved ones, sending Christmas cards, etc. But most importantly, it is the time of year when our beloved Father Christmas, Santa Claus, along with his elves, have to start planning a long and exhausting journey that begins the night of 24 December in order to deliver presents to all the good children in every household around the world.

Santa getting ready to deliver presents. Image: public domain.

Here at Chalkdust we are already in Christmas mood (see our series of Christmas conundrums), but this time we wanted to show you a bit about the science behind Santa’s journey, in case you ever wondered. In addition, if you think you spend a lot of money during the Christmas period, we’ll show you how much money Santa has to spend on presents each year (according to our quick calculations).

We are not pretending to understand how Santa works his magic, but we thought it might be fun to apply some maths to the problem. So let’s begin.

# Christmas conundrum #2

It’s time for the second Chalkdust Christmas conundrum. But first of all, we can proudly announce last week’s winners. There were 82 entries to last week’s competition, of which 67 were correct. The randomly selected winners are:

• Mike Fuller
• Stewart Robertson
• Catriona Shearer
• Steven Peplow

Congratulations! Chalkdust T-shirts are on their way! The solution to last week’s puzzle can be found at the bottom of this blog post.

Now on to today’s puzzle. Four lucky people who submit the correct answer to the puzzle will win a copy of The Indisputable Existence of Santa Claus by Hannah Fry and Thomas Oléron Evans. If you want to know how great this book is, you can read our review of it here. The deadline for entries is Friday 15 December at 6pm.

In the Christmas tree below, the rectangle, baubles, and the star at the top each contain a number. The square baubles contain square numbers; the triangle baubles contain triangle numbers; and the cube bauble contains a cube number.

The numbers in the rectangles (and the star) are equal to the sum of the numbers below them. For example, if the following numbers are filled in:

then you can deduce the following:

With the information given in the tree, you can work out the rest of the numbers.

Once you have solved the puzzle, enter the number in the star at the top in the form below for a chance to win. The deadline for entries is Friday 15 December at 6pm. The winners will be announced in next week’s conundrum post, when you will also have a chance to win a copy of The Element in the Room by Helen Arney and Steve Mould.

This competition is now closed.

## The solution to conundrum #1

In last week’s conundrum, you were asked to find out who the icosahedral present was for. Stop reading now if you don’t want to know the answer yet.

Clue 5 told you that “the edges of Dominika’s present have an integer length, and her present has an integer volume”. The only Platonic solid that satisfies this is the cube, so that must be for Dominika.

Clues 1 and 2 tell you that Atheeta’s present has more faces than Emma’s, but fewer vertices. There are only two possible pairs of presents that satisfy this: the octahedron and the cube; or the icosahedron and the dodecahedron. As the cube is already taken by Dominika, the icosahedron must be Atheeta’s and the docedahedron must be Emma’s.

Finally, clues 3 and 5 tell us that the tetrahedron is Bernd’s and the octahedron is Colin’s.

So, the owner of the icosahedron was Atheeta.