*This post was part of the Chalkdust 2016 Advent Calendar.*

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.

Continue reading

A magazine for the mathematically curious

*This post was part of the Chalkdust 2016 Advent Calendar.*

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.

Continue reading

*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.

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.

*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!

*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…

*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.

*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.

*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:

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.

*This post was part of the Chalkdust 2016 Advent Calendar.*

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.

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…

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

These puzzles appeared in **Issue 04** of the magazine.

Our original prize crossnumber is featured on pages 48 and 49 of Issue 04.

- Download Crossnumber #4 as a PDF, or read on!
~~Submit your answer~~This competition has now ended.

- Although many of the clues have multiple answers, there is only one solution to the completed crossnumber. As usual, no numbers begin with 0. Use of Python, OEIS, Wikipedia, etc. is advised for some of the clues.
- One randomly selected correct answer will win a
**£100**Maths Gear goody bag. Three randomly selected runners up will win a Chalkdust t-shirt. The prizes have been provided by Maths Gear, a website that sells nerdy things worldwide, with free UK shipping. Find out more at mathsgear.co.uk - To enter, submit the
**sum of the across clues**via this form by**7 January 2017**. Only one entry per person will be accepted. Winners will be notified by email and announced on our blog by 21 January 2017.

When learning A-level maths, much time is devoted to learning how to differentiate and integrate. For this week’s blog post, I have collected some puzzles based on these skills. They should be fun to solve, present a few surprises and maybe even provide a teacher or two with an extra challenge for capable students.

The answers to these puzzles will appear here from Sunday at 8am.

This month I’ve put together a mini crossnumber (click here to download it). It should be possible to complete using only a pen and paper, so it would be perfect to take along to this week’s Maths Jam!

There’s just over a month left to submit answers to the full size crossnumber from issue 3. If you haven’t solved it yet, get cracking!

Welcome to the 133rd Carnival of Mathematics, the monthly round up of maths blogs organised by The Aperiodical. Next month’s Carnival will be hosted by Comfortably Numbered. You can submit items for next month here.