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1. Cut out the two scales.

A magazine for the mathematically curious

1. Cut out the two scales.

This issue, Top Ten features the **top ten parts of a circle**! Then vote here on the top ten geometry instruments for Issue 06!

At 10, and dangerously close to having a “(sic)” added after it: the **center**.

Beating its American cover version to number 9 this week: Stuck in the **Centre** with You by Stealers Wheel.

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These solutions relate to puzzles found in **Issue 05** of the magazine.

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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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*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.

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*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…

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