# Christmas conundrum #2

It’s time for the second Chalkdust Christmas conundrum. But first of all, we can proudly announce last week’s winners. There were 82 entries to last week’s competition, of which 67 were correct. The randomly selected winners are:

• Mike Fuller
• Stewart Robertson
• Catriona Shearer
• Steven Peplow

Congratulations! Chalkdust T-shirts are on their way! The solution to last week’s puzzle can be found at the bottom of this blog post.

Now on to today’s puzzle. Four lucky people who submit the correct answer to the puzzle will win a copy of The Indisputable Existence of Santa Claus by Hannah Fry and Thomas Oléron Evans. If you want to know how great this book is, you can read our review of it here. The deadline for entries is Friday 15 December at 6pm.

In the Christmas tree below, the rectangle, baubles, and the star at the top each contain a number. The square baubles contain square numbers; the triangle baubles contain triangle numbers; and the cube bauble contains a cube number.

The numbers in the rectangles (and the star) are equal to the sum of the numbers below them. For example, if the following numbers are filled in:

then you can deduce the following:

With the information given in the tree, you can work out the rest of the numbers.

Once you have solved the puzzle, enter the number in the star at the top in the form below for a chance to win. The deadline for entries is Friday 15 December at 6pm. The winners will be announced in next week’s conundrum post, when you will also have a chance to win a copy of The Element in the Room by Helen Arney and Steve Mould.

## The solution to conundrum #1

In last week’s conundrum, you were asked to find out who the icosahedral present was for. Stop reading now if you don’t want to know the answer yet.

Clue 5 told you that “the edges of Dominika’s present have an integer length, and her present has an integer volume”. The only Platonic solid that satisfies this is the cube, so that must be for Dominika.

Clues 1 and 2 tell you that Atheeta’s present has more faces than Emma’s, but fewer vertices. There are only two possible pairs of presents that satisfy this: the octahedron and the cube; or the icosahedron and the dodecahedron. As the cube is already taken by Dominika, the icosahedron must be Atheeta’s and the docedahedron must be Emma’s.

Finally, clues 3 and 5 tell us that the tetrahedron is Bernd’s and the octahedron is Colin’s.

So, the owner of the icosahedron was Atheeta.

# Christmas conundrum #1

It’s finally time for the first Chalkdust Christmas conundrum. Four lucky people who submit the correct answer to the puzzle will win Chalkdust T-shirts. The deadline for entries is Friday 8th December at 6pm.

It’s nearly Christmas and you have wrapped up five presents for your five best friends: Atheeta, Bernd, Colin, Dominika and Emma. You are especially happy this year as each present is the shape of a different Platonic solid.

But in your excitement, you have just forgotten which present is for which friend. You can only remember the following facts:

1. Atheeta’s present has more faces than Emma’s present.
2. Atheeta’s present has fewer vertices than Emma’s present.
3. The faces of Colin’s present are triangles.
4. Three faces meet at every vertex of Bernd’s present.
5. The edges of Dominika’s present have an integer length, and her present has an integer volume.

Who is the icosahedral present for?

Once you have solved the puzzle, enter your answer below for a chance to win. The deadline for entries is Friday 8th December at 6pm. The winners will be announced in next week’s conundrum post, when you will also have a chance to win a copy of The Indisputable Existence of Santa Claus by Hannah Fry and Thomas Oléron Evans.

This competition is now closed.

# Prize crossnumber, Issue 06

Our original prize crossnumber is featured on pages 52 and 53 of Issue 06.

Correction: The pdf was incorrect and 5D did not match the clues below. This has now been fixed.
Clarification: Added brackets to 29A and 34D to reduce ambiguity.

### Rules

• 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 8 January 2018. Only one entry per person will be accepted. Winners will be notified by email and announced on our blog by 22 January 2018.

# Prize crossnumber, Issue 05

Our original prize crossnumber is featured on pages 58 and 59 of Issue 05.

### Rules

• 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 22 July 2017. Only one entry per person will be accepted. Winners will be notified by email and announced on our blog by 30 July 2017.

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…

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.

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.

### 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…