# 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 $n^2$.

# 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 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…

# 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 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.

# 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…

# Top ten vote issue 04

What is the best part of a circle?

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# Problem solving 101

From the outside looking in, maths problem solving can seem like a kind of magic. Here is a typical image: a lone genius, peering at a vexing problem, rubs their chin, paces up and down; then a bolt of inspiration hits and the solution falls neatly into place.

And while it’s true that inspiration can strike the lucky few, for the rest of us this is no more than an illusion (and often a carefully cultivated illusion at that). In reality, problem solving is usually much more prosaic, nothing more than a careful application of well-known, and often quite elementary, techniques.

So what are these elementary techniques? In this article, I’ll look at some of the simplest and easiest to understand. Happily, they are also some of the most powerful and widely applicable. These techniques will be explained by way of example problems; I strongly encourage you to attempt the problems yourself before reading the solutions. Continue reading

# What’s hot and what’s not, Issue 04

Maths is a fickle world. Stay à la mode with our guide to the latest trends.

### HOT Mathematical crochet

Hyperbolic surfaces! Klein bottles! Impress fellow passengers with your DIY manifolds!
[Picture: Flickr user Pandaeskimo, CC BY-NC 3.0]

### NOT Paper Möbius strips

As Dirichlet says, Möbius strips should be band.

# Prime jewellery

I was recently given a copy of Crafting Conundrums: Puzzles and Patterns for the Bead Crochet Artist by Ellie Baker and Susan Goldstine. This was pretty exciting for me, as although I knew nothing about bead crochet (I’d never heard of it), I’m a mathematician who enjoys exploring mathematical ideas through craft. So naturally I rushed out and bought lots of beads and thread, and a very tiny crochet hook (1.5mm, if you’re really interested). Continue reading