This post was part of the Chalkdust 2016 Advent Calendar.

Welcome to the fourteenth day of the 2016 Chalkdust Advent Calendar.  Today, we bring you some more fascinatingT&Cs apply facts, randomly generated by Santa’s elves.  Remember to send us your favourite scientific curiosities either on Facebook, Twitter or email and we’ll feature the best in a blog next year.

Osiris was ripped into 14 pieces by his brother Set.

### The number 14

The numbers fourteen and forty sound very similar. They often get confused when speaking on the telephone. This is a common affliction caused by all teens and I’m sure the reader can think of many more.

### Pesky bumblebees defying nature

Bumblebees. Obviously capable of flying.

According to an engineer (obviously) a long time ago (probably also obviously), bumblebees can’t fly. This is based on the assumption that the air remains attached as it goes over a bee’s wings and that the same aerodynamics takes place as occurs for flow over aeroplane wings. This leads to the result that a bumblebee would not be able to generate enough lift to stay in the air. Unfortunately, however, they can. This was all quite mysterious until someone had the bright idea to stick some bees in a wind tunnel and see what is going on. It turns out that bumblebees don’t flap their wings just up and down, but forwards and backwards too; and a vortex evolves on the upper side of the wing, allowing the bee to generate more lift. The moral of the story is to never listen to engineers. You can find a cool video of bumblebees in a wind tunnel here.

### The computer of the Ancient Greeks

Our forefathers’ computer.

The Antikythera mechanism is the first (analogue) computer, built by the ancient Greeks prior to 100BC. Either that, or it was left behind by alien visitors who decided that our civilisation 2,000 years ago wasn’t worth bothering about. Incorporating much of the mathematical and astronomical knowledge of the Greeks, interlocking gearwheels turned a minimum of seven dials that told celestial time (whatever that means). Presumably we then got bored of making scientific magic and went back to waging war on each other, which meant that we didn’t come up with anything that mind-blowingly intricate until at least the invention of the pocket watch.

### The perpetually surprising trapezium rule

The trapezium rule is often used for numerical integration and involves summing up the areas of many many trapeziums. The more trapeziums you use, the more accurate your answer will be. Bizarrely, despite it being just a simple sum, it turns out that in some cases adding more trapeziums allows you to converge exponentially to the right answer. My brain has just exploded.

I’ll be back on the 21st when I’ve picked up all the pieces.

[Pictures: Osiris, public domain; Bumblebees, CC0 license; Antikythera mechanism, CCA-SA 3.0, Marsyas]

# Which mathematician are you?

This post was part of the Chalkdust 2016 Advent Calendar.

On the thirteenth day of December, Chalkdust gave to me… a super awesome personality quiz. For those of you who have always wondered which mathematician they were in their previous life (and for those who haven’t, but should) Chalkdust has the answer. Just answer a few carefully selected questions and find out for yourself!

[playbuzz-item url=”//www.playbuzz.com/nikoletak10/which-mathematician-are-you”]

[Pictures:

Cover Picture Test – adapted from Flickr.com – Great Mathematicians by  Andrew O’BrienCC-BY 2.0;

1.adapted from public domain – Cauchy;

4.adapted from Flickr.com – Emmy Noether by Open Logic, Public Domain Mark 1.0;

5.adapted from Flickr.com – Vintage Ad #2,067: The Apple that Rocked the World by Jamie, CC-BY 2.0;

other pictures by Chalkdust]

This post was part of the Chalkdust 2016 Advent Calendar.

Welcome to the twelfth day of the 2016 Chalkdust Advent Calendar. Today, we have another puzzle for you to enjoy, plus the answer to the puzzles from 06 December.

Today’s puzzle is taken from Daniel Griller‘s talk at the MathsJam conference earlier this year.

### Odd factors

Pick a number. Call it $n$. Write down all the numbers from $n+1$ to $2n$ (inclusive). Under each of these, write its largest odd factor. What is the sum of these odd factors?

Now for the solutions to the puzzles from 06 December.

### Digital sums

Source: mscroggs.co.uk Advent calendar, day 6
When you add up the digits of a number, the result is called the digital sum.

How many different digital sums do the numbers from 1 to 1091 have?

As this puzzle is part of a larger advent calendar (with prizes!), I’m not going to give you the answer here!

### Wipeout

Source: nrich Secondary Advent calendar, day 10
You are given the numbers 1,2,3,4,5,6 and are allowed to erase one. If you erase 5, the mean of the remaining numbers will be 3.2. Is it possible to erase a number so that the mean of the remaining number is an integer?

If you are given the numbers 1,2,3,4,…,$N$, can you erase one number so that the mean of the remaining numbers is an integer?

For the first part, erasing 6 will leave numbers that sum to 15, with a mean of 3.

For the second part, if $N$ is even, erasing $N$ from the list 1,2,3,4,…,$N$ will leave numbers that sum to $\tfrac12N(N-1)$. $N$ is even, so $\tfrac12N$ is an integer; therefore $N-1$ is a factor of the sum, so the numbers have an integer mean.

If $N$ is odd, removing the middle number from the list leaves an integer mean. I’ll let you work out why this is and will return in a few days with the answer to today’s puzzle…

# Christmas cracker joke I

This post was part of the Chalkdust 2016 Advent Calendar.

‘Twas the function’s Christmas party at the Leibnitz bar. The various functions were happily drinking and discussing. At some point, Sine and Cosine noticed Exponential sitting alone in the corner. They approached him and asked him to join them, upon which he replied: “I tried integrating, but just ended up with myself”.

# Binary magic card trick

This post was part of the Chalkdust 2016 Advent Calendar.

Ah, the Christmas holidays. A time to be spent hiding from the cold, wearing pyjamas, eating too much food and solving integro-differential equations. (At least that’s how I think everyone spends Christmas, right?) Come Christmas Day, after you’ve finally cracked Chalkdust Issue 4’s Crossnumber and stuffed your face with turkey, why don’t you stun your (slightly drunk) family and friends with a very simple math-based magic trick?

### The Trick

This trick follows a fairly standard magic setup: you are going to successfully guess (telepathically, of course) a number decided in secret by the unsuspecting watcher. Hand your target a set of cards with numbers on (an example set of cards can be seen below):

Tell them to think of a number in between the smallest and largest number present on the cards (in our case, between 1 and 63). Then ask they hand you every single card with that number on it. Add up the first number on each of the cards given to you, et voilà, you’ve ‘magically’ obtained their number.

### Why does it work?

Afterwards, you’ll be asked how it works. Spend a good bit of time insisting you’re in touch with the Other Side, were recently struck by lightning and can now hear other peoples’ thoughts, or are simply VERY good at guessing. But since, as mathematicians, we need to try and convince people maths is interesting and, more importantly, not a form of witchcraft, consider explaining to them how it works (I’ve found talking over the television during the Christmas edition of Eastenders particularly effective).

Those of you with a keen eye will have clocked it already. Certainly, a big hint is that the first number on each card is a power of 2. The cards have been designed such that each combination of cards uniquely represents a number in binary.

### Binary

For those of you who are not sure what binary is, it is the base two numeral system. You will certainly be familiar with the base ten numeral system since that’s how we normally represent numbers. A good explanation on base systems can be found here.

As an example, we can represent forty-seven in binary and base-ten as

\begin{eqnarray}
\text{forty seven} &=& 4 \times 10^1 + 7 \times 10^0 \\
&=& 47 \,\,\,\, (\text{base ten}) \\
&=& 1 \times 2^5 + 0 \times 2^4 + 1 \times 2^3 + 1 \times 2^2 + 1 \times 2^1 + 1 \times 2^0 \\
&=& 101111 \,\,\,\, (\text{base two})
\end{eqnarray}

So, let us re-order and label the cards as follows

Card $N$ is characterised as containing all the numbers which contain a “$1 \times$” in front of the term $2^N$. Taking our example 47 (base ten), we see that 47 appears on cards [0,1,2,3,5]. That is the say

\begin{eqnarray}
47 \,\,\, \text{(base 10)} &=& 1 \times 2^5 + 0 \times 2^4 + 1 \times 2^3 + 1 \times 2^2 + 1 \times 2^1 + 1 \times 2^0 \\
&=& 101111 (\text{base two})
\end{eqnarray}

Ensuring that card $N$ has $2^N$ as the first number on the card (for your ease of reference), simply adding all the first digits on the cards handed back to you will give the correct answer.

We chose 63 as it is of the form $2^6 – 1$. It is best advised to choose a number of the form $2^M -1$ as your maximum number as it ensures each card has $M/2$ digits on it, raising the least suspicion.

# Snowflake, the symbol of winter: different sizes, infinite shapes

This post was part of the Chalkdust 2016 Advent Calendar.

Winter is coming. Or, to be more precise, the winter season begins on 21 December. Some people hate it, while others (like me) love it. Every time we hear the word winter, we think about that cold time of the year when we wear our scarves, jackets and coats, and we gather with our loved ones and eat (lots) of delicious food. But if you were asked to describe the word ‘winter’ with a simple symbol, what would that symbol be?  Most of us would probably think of a white, beautiful snowflake. And we are not the only ones thinking of that: if you Google the word ‘winter’ and go to the image section, you will find lots of pictures of them.

First pictures of snoflakes taken in 1931 by Wilson Bentley

Recent pictures of snoflakes (made in a laboratory) taken by Prof. Kenneth G. Libbrecht

The snowflake is the most iconic symbol representing cold weather, and is also a traditional image used during the Christmas period. It is well known that most snowflakes have six-sides (hexagonal pattern) and many branches around them; but apart from that snowflakes come in a large variety of shapes and sizes, leading to the common phrase “no two snowflakes are alike”. But how is a snowflake formed? In this blog, we will describe the process of snowflake formation, and explain why they exist in a variety of shapes and sizes. To do that, we just need to understand some basic concepts: humidity and supersaturation.

Construct your own snowflakes using just paper and scissors. Follow
the instructions here.

This post was part of the Chalkdust 2016 Advent Calendar.

Welcome to the sixth day of the 2016 Chalkdust Advent Calendar. Today, we have another two puzzles for you to enjoy, plus the answer to the puzzle from the 02 December.

If you’ve spent time browsing the internet recently, you will have noticed that we’re not the only site running an advent calendar. Today’s puzzles are taken from two of our rival calendars, run by Matthew Scroggs (who?) and nrich.

First a puzzle from my own advent calendar.

### Digital sums

Source: mscroggs.co.uk Advent calendar, day 6
When you add up the digits of a number, the result is called the digital sum.

How many different digital sums do the numbers from 1 to 1091 have?

Second, a puzzle from the excellent nrich Advent calendar.

### Wipeout

Source: nrich Secondary Advent calendar, day 10
You are given the numbers 1,2,3,4,5,6 and are allowed to erase one. If you erase 5, the mean of the remaining numbers will be 3.2. Is it possible to erase a number so that the mean of the remaining number is an integer?

If you are given the numbers 1,2,3,4,…,N, can you erase one number so that the mean of the remaining numbers is an integer?

Back on 02 December, I posted a longer version of the following puzzle:

### Decorations

You love big equilateral triangles but hate small equilateral triangles. Can you arrange ten red and blue baubles in a triangle so that no three baubles of the same colour form the vertices of an equilateral triangle?

This is not possible. To see this, first pick a colour for the central bauble. I’ve picked red.

Now we try to colour the rest without making a triangle. One of the three baubles on the following triangles must be red (otherwise there is a blue triangle). Pick one of them to make red. If a different one is red, rotate the triangle to make this one red.

The baubles must be coloured as follows. In each step, the colour is chosen to avoid a triangle.

Now, the bauble shown in green below cannot by either colour, as in each case it makes a triangle.

Hence, it is impossible to find a triangle without a smaller triangle.