Maths is a fickle world. Stay à la mode with our guide to the latest trends.
HOT The new Chalkdust T-shirt
Available on this very website!
NOT The old Chalkdust T-shirt
Available in charity shops throughout the midlands.
Imagine, for a moment, that you have opportunity to build the house of your dreams. You are rich and powerful, you own a lot of land, and you are carefree. So carefree, in fact, that this is what you decide to call your house. You are Frederick the Great: an 18th century monarch, the creator of Sanssouci palace in Potsdam, Germany, and the first person on record to describe your dog as ‘man’s best friend’. Continue reading
It’s time for the next Chalkdust Christmas Conundrum! But first it’s time to announce the lucky winners of the third conundrum competition. There were 26 entries, of which 18 were correct. The four randomly selected winners are:
You will all soon be proud owners of signed copies of The Element in the Room by Helen Arney and Steve Mould. The answer to conundrum #3 is at the bottom of this post.
The prizes up for grabs for the final conundrum are five copies of Bletchley Park Brainteasers by Sinclair McKay. Inspired by the prize, this conundrum builds up to a hidden message, and two of its parts involve breaking ciphers. Continue reading
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:
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.
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.
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.
Now that you are familiar with nonograms, it’s time for this week’s conundrum:
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?
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.
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:
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:
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.
Click here to download a printable PDF of this week’s puzzle
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.
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.
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.
But in your excitement, you have just forgotten which present is for which friend. You can only remember the following facts:
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.
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.
Moonlighting agony uncle Professor Dirichlet answers your personal problems. Want the prof’s help? Send your problems to firstname.lastname@example.org.
I’ve just started my PhD at a well-known university, and I’m trying to make some friends. There are supposed to be 55 other students but nearly everyone in the PhD office refuses my offers of tea, sits in silence, and will barely talk to me unless I whisper them some very specific technical questions. I was hoping there would be some people in the group who enjoy everyday things: biscuits, beer, and just shooting the breeze. Is this really what academia is like?
— Pearl among swine, Withheld