# Christmas conundrum #3

It’s time for the next Chalkdust Christmas Conundrum! But first it’s time to announce the lucky winners of last week’s competition. There were 206 entries, of which 189 were correct. The four randomly selected winners are:

• Josie Neal
• Maureen Monroe
• Antonio S. Macias
• Tom Lock

You will all receive copies of The Indisputable Existence of Santa Claus by Hannah Fry and Thomas Oléron Evans. Hope you enjoy them! The answer to last week’s conundrum is at the bottom of this post.

The prizes up for grabs this week are four copies of The Element in the Room by Helen Arney and Steve Mould. To encourage you to try to win this, we will be publishing a review of it on Thursday.

This week’s conundrum is all about nonograms (AKA griddler, picross, tsunami, or “oh, that puzzle”). If you are unfamiliar with nonograms, we’ve prepared a little seasonal introduction to them. Otherwise, feel free to scroll down and dive straight into the third conundrum.

## An Introduction to Nonograms

A nonogram is a logic puzzle where the object is to fill in squares to create a picture using the numbers around the edge of the puzzle.

Consider the following puzzle.

Twinkle Twinkle…

The first column has a single one above it. This means there will be a single square filled in this column. Similarly for the first row. The second column has $1 \, 2$ written above the column. This means there will be one filled in square followed by a group of two filled in squares (working downwards).  The key point here is that there must be at least one blank square between the single square and the group—however it could be more.

Here is another puzzle to try.

Ding dong…

Now that you are familiar with nonograms, it’s time for this week’s conundrum:

## Chalkdust Christmas conundrum #3

The following nonogram puzzle doesn’t have a unique answer: there is more than one pattern of coloured squares that satisfied the clues. How many different solutions are there?

A non-unique nonogram.

Once you’ve worked out how many possible solutions there are, please enter this into the form below for a chance to win one of four copies of The Element in the Room. The deadline for entries is 12:00 (midday) on Friday 22 December. The winners will be announced on Friday afternoon, when we will be publishing the fourth conundrum and giving you a chance to win a copy of Bletchley Park Brainteasers by Sinclair McKay.

This competition is now closed.

## The solution to conundrum #2

In the second conundrum, we asked you to find the number in the star at the top of three. Stop reading now if you don’t want to know the answer yet.

The two numbers between 14 are a cube number and a triangle number: these must be 8 and 6. Next you can see that 23 is the sum of 8, two times a triangle number and a square number: the triangle and square numbers must be 3 and 9.

Next, call the square number at the bottom left $a$, the square number at the top left $b$, and the triangle number at the top right $c$. Adding upwards, we find that $a+b+45=106$ and $a+141+c=198$; and so $a+b=61$ and $a+c=57$. The only two square numbers that add to 61 are 25 and 36. Therefore $c$ must be 21 or 32, but must be 21 as 32 is not a triangle number. And so $a$ is 36 and $b$ is 25.

Putting all these numbers into the tree gives the top number as 433. This is fitting because 433 is in fact a star number:

433 dots arranged to make a star.

# 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.

This competition is now closed.

## 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.