Once again, we are delighted to be able to announce the winners of the Chalkdust prize crossnumber #4! But before we do so, here is the solution:

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# Author Archives: Matthew Scroggs

# Is MEDUSA the new BODMAS?

Recently, Colin “IceCol” Beveridge blogged about something that’s been irking him for a while: those annoying social media posts that tell you to work out a sum, such as $3-3\times6+2$, and state that only $n$% of people will get it right (where $n$ is quite small). Or as he calls it “fake maths”.

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# 23 December

Welcome to the twenty third day of the 2016 Chalkdust Advent Calendar. Today, we have another puzzle for you to enjoy, plus the answer to the puzzles from 12 December.

Here’s today’s puzzle.

### Finding Santa

It’s Christmas Eve Eve and Santa is so nervous that he decides to hide in the top floor of a nearby hotel. Santa starts in a random room and every hour moves to one of the rooms next door to his current room. There are 17 rooms on the top floor of the hotel.

The hotel manager is a very busy lady and only has time to allow you to look into one room every hour. It’s 2pm on Christmas Eve Eve, so you have 34 hours to find Santa or Christmas is ruined. Can you find him in time?

Now for the solution to the puzzle from 12 December.

### 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?

Incredibly, the result will always be $n^2$.

To see why, imagine writing every number, $n+1\leq k\leq 2n$, in the form $$k=2^ab$$ where $b$ is an odd number and also the $k$’s largest odd factor. The next largest number whose largest odd factor is $b$ will be $2^{a+1}b=2k$. But this will be larger than $2n$, so outside the range. Therefore each number in the range has a different largest odd factor.

Each of the largest odd factors must be one of $1, 3, 5, …, 2n-1$, as they cannot be larger than $2n$. But there are $n$ odd numbers here and $n$ numbers in the range, so each number $1, 3, 5, …, 2n-1$ is the highest odd factor of one of the numbers (as the highest odd factors are all different).

Therefore, the sum of the odd factors is the sum of the first $n$ odd numbers, which is $n^2$.

# The Chalkdust Christmas card

Recently, some of you may have received a Chalkdust Christmas card. If not, it’s not because we hate you, it’s just that we couldn’t find your address… Unless we hate you, in which case it *is* because we hate you.

The card initially looks very boring: it is just a grid of squares with “Merry Christmas” written below it. Definitely NOT HOT… But there’s more. There’s a puzzle inside that leads you to add some colour to the squares to reveal a Christmassy picture.

Without giving any more away, here is the puzzle. If you’d like to give it to someone as a Christmas card (or just want to actually be able to colour it in), you can print and fold this lovely pdf.

### Christmas Card 2016

#### Instructions

- Solve the puzzles below.
- Convert the answers to base 3.
- Write the answers in the boxes on the front cover.
- Colour squares containing a 1 green. Colour squares containing a 2 red. Leave squares containing a 0 unshaded.

#### Puzzles

- The square number larger than 1 whose square root is equal to the sum of its digits.
- The smallest square number whose factors add up to a different square number.
- The largest number that cannot be written in the form $23n+17m$, where $n$ and $m$ are positive integers (or 0).
- Write down a three-digit number whose digits are decreasing. Write down the reverse of this number and find the difference. Add this difference to its reverse. What is the result?
- The number of numbers between 0 and 10,000,000 that do not contain the digits 0, 1, 2, 3, 4, 5 or 6.
- The lowest common multiple of 57 and 249.
- The sum of all the odd numbers between 0 and 66.
- One less than four times the 40th triangle number.
- The number of factors of the number 2
^{756}×3^{12}. - In a book with 13,204 pages, what do the page numbers of the middle two pages add up to?
- The number of off-diagonal elements in a 27×27 matrix.
- The largest number, $k$, such that $27k/(27+k)$ is an integer.

# 12 December

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 10^{91} 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…

# 06 December

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 10^{91} 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?

I’ll be back with answers and more puzzles later in Advent.

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.

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

# 02 December

Welcome to the second day of the 2016 Chalkdust Advent Calendar. Today, we have a lovely Christmas-themed puzzle for you to enjoy. Like so many good puzzles, this one is inspired by a puzzle that I found in a Martin Gardner book.

### Decorations

It’s the start of Advent, so you decide to decorate your flat with some homemade decorations. You have a large number of red and blue baubles that you bought in last year’s January sales.

You like equilateral triangles, so you decide to stick ten baubles together into a large equilateral as shown below.

You are not, however, happy with this arrangement of colours as you hate smaller equilateral triangles, and three of the red baubles lie on the vertices of an equilateral triangle (as shown below).

Is it possible to arrange ten baubles into a triangle so that no three baubles of the same colour form the vertices of an equilateral triangle?

I’ll be back later in Advent with the answer to this puzzle and more puzzles…

# Visualising mathematics with 3D printing

There are too many maths books out there that repeat the same old boring stories about Pythagoras drowning people for inventing irrational numbers, Gauss adding up numbers quickly, or Dirichlet eating too many grapes^{[citation needed]}. Luckily, *Visualising Mathematics with 3D Printing* (Amazon UK, USA) is not one of these: its author, Henry Segerman, has achieved the seemingly impossible and written something refreshingly original and different.

In this book, Segerman takes you on a tour of modern geometry, exploring symmetry, hyperbolic surfaces, four-dimensional shapes, knots and much more. These concepts are hard to visualise. While pictures in a book can help, there is only so much that can be shown in a 2D picture. This is where this book comes into its own: it focuses on using 3D objects to visualise these concepts, with instructions for 3D-printing the objects, as well as interactive simulations, available on the book’s website. For many objects, such as the tiled genus 3 surface shown below, the ability to explore these objects from multiple angles is key to understanding them.

One of my favourite concepts in the book (and one of Segerman’s favourites; perhaps this is why it comes across as the most interesting) is the stereographic projection. By placing a light source at the top point of a sphere, this projection maps the 2D surface of a 3D ball onto the 2D plane. Incredibly, this projection preserves angles: a right angle on the sphere is mapped to a right angle on the plane, as seen in the object below.

A 3D shape can be represented by lines drawn on a sphere; these lines can then be stereographically projected onto the 2D plane. The result of this projection for a dodecahedron is shown on the left.

The same process can be followed to project a 4D solid onto the 3D “plane”. Below, you can see a model of half of the 120-cell, a regular 4D shape made by connecting dodecahedra. Without a 3D model of this or the simulation below, it is impossible to get an idea of what this projection looks like.

There is so much more in this book that I don’t have space here to show you. My best advice is to go out and buy yourself a copy of the book. Be warned though, that you may find yourself also buying a very expensive 3D printer.

# Puzzle Solutions, Issue 04

# Maths at EMF camp

Header picture: Simon Arlott, CC-BY-ND 2.0

Recently, I went to a festival. More specifically, I went to Electromagnetic Field (EMF), where I spent three days surrounded by hackers, geeks, scientists, engineers and mathematicians, who had all brought along their coolest toys to show off. In this week’s blog post, I’m going to share my (mostly mathematical) highlights with you.