Solve this and you could be the lucky winner of a signed copy of The Indisputable Existence of Santa Claus

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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.

Click here to download a printable PDF of this week’s puzzle 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.

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.

Matthew Scroggs Matthew Scroggs is a postdoctoral researcher in the Department of Engineering at the University of Cambridge working on finite and boundary element methods. His website, mscroggs.co.uk, is full of maths.

@mscroggs mscroggs.co.uk + More articles by Matthew