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 2019-01-01 
It's 2019, and the Advent calendar has disappeared, so it's time to reveal the answers and annouce the winners. But first, some good news: with your help, Santa was able to work out who had stolen the presents and save Christmas:
Now that the competition is over, the questions and all the answers can be found here. Before announcing the winners, I'm going to go through some of my favourite puzzles from the calendar, reveal the solution and a couple of notes and Easter eggs.

Highlights

My first highlight is the first puzzle in the calendar. This is one of my favourites as it has a pleasingly neat solution involving a surprise appearance of a very famous sequence.

1 December

There are 5 ways to write 4 as the sum of 1s and 2s:
  • 1+1+1+1
  • 2+1+1
  • 1+2+1
  • 1+1+2
  • 2+2
Today's number is the number of ways you can write 12 as the sum of 1s and 2s.

Show answer


My next highlight is a puzzle that I was particularly proud of cooking up: again, this puzzle at first glance seems like it'll take a lot of brute force to solve, but has a surprisingly neat solution.

10 December

The equation \(x^2+1512x+414720=0\) has two integer solutions.
Today's number is the number of (positive or negative) integers \(b\) such that \(x^2+bx+414720=0\) has two integer solutions.

Show answer


My next highlight is a geometry problem that appears to be about polygons, but actually it's about a secret circle.

12 December

These three vertices form a right angled triangle.
There are 2600 different ways to pick three vertices of a regular 26-sided shape. Sometime the three vertices you pick form a right angled triangle.
Today's number is the number of different ways to pick three vertices of a regular 26-sided shape so that the three vertices make a right angled triangle.

 

Show answer


My final highlight is a puzzle about the expansion of a fraction in different bases.

22 December

In base 2, 1/24 is 0.0000101010101010101010101010...
In base 3, 1/24 is 0.0010101010101010101010101010...
In base 4, 1/24 is 0.0022222222222222222222222222...
In base 5, 1/24 is 0.0101010101010101010101010101...
In base 6, 1/24 is 0.013.
Therefore base 6 is the lowest base in which 1/24 has a finite number of digits.
Today's number is the smallest base in which 1/10890 has a finite number of digits.
Note: 1/24 always represents 1 divided by twenty-four (ie the 24 is written in decimal).

Show answer

Notes and Easter eggs

I had a lot of fun this year coming up with the names for the possible theives. In order to sensibly colour code each suspect's clues, there is a name of a colour hidden within each name: Fred Metcalfe, Jo Ranger, Bob Luey, Meg Reeny, and Kip Urples. Fred Metcalfe's colour is contained entirely within his forename, so you may be wondering where his surname came from. His surname is shared with Paul Metcalfe—the real name of a captain whose codename was a certain shade of red.
On 20 December, Elijah Kuhn emailed me to point out that it was possible to solve the final puzzle a few days early: although he could not yet work out the full details of everyone's timetable, he had enough information to correctly work out who the culprit was and between which times the theft had taken place.
Once you've entered 24 answers, the calendar checks these and tells you how many are correct. This year, I logged the answers that were sent for checking and have looked at these to see which puzzles were the most and least commonly incorrect. The bar chart below shows the total number of incorrect attempts at each question.
You can see that the most difficult puzzles were those on 13, 24, and 10 December; and the easiest puzzles were on 5, 23, 11, and 15 December.
I also snuck a small Easter egg into the door arrangement: the doors were arranged to make a magic square, with each row and column, plus the two diagonals, adding to 55.

The solution

The solutions to all the individual puzzles can be found here. Using the clues, you can work out that everyone's seven activities formed the following timetable.
Bob LueyFred MetcalfeJo RangerKip UrplesMeg Reeny
0:00–1:21
Billiards
0:00–2:52
Maths puzzles
0:00–2:33
Maths puzzles
0:00–1:21
Billiards
0:00–1:10
Ice skating
1:10–2:33
Skiing
1:21–2:52
Ice skating
1:21–2:52
Stealing presents
2:33–4:45
Billiards
2:33–4:45
Billiards
2:52–3:30
Lunch
2:52–3:30
Lunch
2:52–3:30
Lunch
3:30–4:45
Climbing
3:30–4:45
Climbing
3:30–4:45
Climbing
4:45–5:42
Curling
4:45–5:42
Curling
4:45–5:42
Curling
4:45–5:42
Curling
4:45–5:42
Lunch
5:42–7:30
Maths puzzles
5:42–7:30
Ice skating
5:42–7:30
Chess
5:42–7:30
Chess
5:42–7:30
Maths puzzles
7:30–10:00
Skiing
7:30–9:45
Chess
7:30–8:45
Skiing
7:30–10:00
Maths puzzles
7:30–9:45
Chess
8:45–9:45
Lunch
9:45–10:00
Table tennis
9:45–10:00
Table tennis
9:45–10:00
Table tennis
Following your investigation, Santa found all the presents hidden under Kip Urples's bed, fired Kip and sucessfully delivered all the presents on Christmas Eve.

The winners

And finally (and maybe most importantly), on to the winners: 73 people submitted answers to the final logic puzzle. Their (very) approximate locations are shown on this map:
From the correct answers, the following 10 winners were selected:
1Sarah Brook
2Mihai Zsisku
3Bhavik Mehta
4Peter Byrne
5Martin Harris
6Gert-Jan de Vries
7Lyra
8James O'Driscoll
9Harry Poole
10Albert Wood
Congratulations! Your prizes will be on their way shortly. Additionally, well done to Alan Buck, Alex Ayres, Alex Bolton, Alex Lam, Alexander Ignatenkov, Alexandra Seceleanu, Andrew Turner, Ashwin Agarwal, Becky Russell, Ben Reiniger, Brennan Dolson, Carl Westerlund, Cheng Wai Koo, Christopher Embrey, Corbin Groothuis, Dan Whitman, David, David Ault, David Kendel, Dennis Oltmanns, Elijah Kuhn, Eric, Eric Kolbusz, Evan Louis Robinson, Felix Breton, Fred Verheul, Gregory Loges, Hannah, Jean-Noël Monette, Jessica Marsh, Joe Gage, Jon Palin, Jonathan Winfield, Kai Lam, Louis de Mendonca, M Oostrom, Martine Vijn Nome, Matt Hutton, Matthew S, Matthew Wales, Michael DeLyser, MikeKim, Naomi Bowler, Pranshu Gaba, Rachel Bentley, Raymond Arndorfer, Rick Simineo, Roni, Rosie Paterson, Sam Hartburn, Scott, Sheby, Shivanshi, Stephen Cappella, Steve Paget, Thomas Smith, Tony Mann, Valentin Vălciu, Yasha Ayyari, Zack Wolske, and Zoe Griffiths, who all also submitted the correct answer but were too unlucky to win prizes this time.
See you all next December, when the Advent calendar will return.

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 2018-12-08 
Just like last year and the year before, TD and I spent some time in November this year designing a Chalkdust puzzle Christmas card.
The card looks boring at first glance, but contains 10 puzzles. By splitting the answers into pairs of digits, then drawing lines between the dots on the cover for each pair of digits (eg if an answer is 201304, draw a line from dot 20 to dot 13 and another line from dot 13 to dot 4), you will reveal a Christmas themed picture. Colouring the region of the card labelled R red or orange will make this picture even nicer.
If you want to try the card yourself, you can download this pdf. Alternatively, you can find the puzzles below and type the answers in the boxes. The answers will be automatically be split into pairs of digits, lines will be drawn between the pairs, and the red region will be coloured...
If you enjoy these puzzles, then you'll almost certainly enjoy this year's puzzle Advent calendar.

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Comments in green were written by me. Comments in blue were not written by me.
 2018-12-17 
@Carmel: It's not meant to check your answers. It only shows up red if the number you enter cannot be split into valid pairs (eg the number has an odd number of digits or one of the pairs of digits is greater than 20).
Reply
Matthew
 2018-12-16 
The script for checking the answers doesn't work properly
Reply
Carmel
 2018-12-14 
Thank you Shawn!
Reply
SueM
 2018-12-14 
So satisfying!
Reply
Heather
 2018-12-10 
Great puzzle problems! Hint on #9: try starting with an analogous problem using smaller numbers (e.g. 3a + 10b). This helped me to see what I had to do more generally.
Reply
Noah
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 2018-11-25 
This year, the front page of mscroggs.co.uk will once again feature an advent calendar, just like last year, the year before and the year before. Behind each door, there will be a puzzle with a three digit solution. The solution to each day's puzzle forms part of a logic puzzle:
It's nearly Christmas and something terrible has happened: one of Santa's five helpers—Jo Ranger, Kip Urples, Fred Metcalfe, Bob Luey, and Meg Reeny—has stolen all the presents during the North Pole's annual Sevenstival. You need to find the culprit before Christmas is ruined for everyone.
Every year in late November, Santa is called away from the North Pole for a ten hour meeting in which a judgemental group of elders decide who has been good and who has been naughty. While Santa is away, it is traditional for his helpers celebrate Sevenstival. Sevenstival gets its name from the requirement that every helper must take part in exactly seven activities during the celebration; this year's available activities were billiards, curling, having lunch, solving maths puzzles, table tennis, skiing, chess, climbing and ice skating.
Each activity must be completed in one solid block: it is forbidden to spend some time doing an activity, take a break to do something else then return to the first activity. This year's Sevenstival took place between 0:00 and 10:00 (North Pole standard time).
During this year's Sevenstival, one of Santa's helpers spent the time for one of their seven activities stealing all the presents from Santa's workshop. Santa's helpers have 24 pieces of information to give to you, but the culprit is going to lie about everything in an attempt to confuse you, so be careful who you trust.
Behind each day (except Christmas Day), there is a puzzle with a three-digit answer. Each of these answers forms part of a fact that one of the helpers tells you. You must work out who the culprit is and between which times the theft took place.
Ten randomly selected people who solve all the puzzles and submit their answers to the logic puzzle using the form behind the door on the 25th will win prizes!
The winners will also receive one of these medals:
As you solve the puzzles, your answers will be stored. To share your stored answers between multiple devices, enter your email address below the calendar and you will be emailed a magic link to visit on your other devices.
Behind the door on Christmas Day, there will be a form allowing you to submit your answers. The winner will be randomly chosen from all those who submit the correct answer before the end of 2018. Each day's puzzle (and the entry form on Christmas Day) will be available from 5:00am GMT. But as the winners will be selected randomly, there's no need to get up at 5am on Christmas Day to enter!
To win a prize, you must submit your entry before the end of 2018. Only one entry will be accepted per person. If you have any questions, ask them in the comments below or on Twitter.
So once December is here, get solving! Good luck and have a very merry Christmas!

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Comments

Comments in green were written by me. Comments in blue were not written by me.
 2018-12-24 
@Steve: Yes, the final door contains the entry form
Reply
Matthew
 2018-12-24 
Do we have to submit the answer on Christmas Day?
Reply
Steve
 2018-12-19 
@Elijah: yes
Reply
Matthew
 2018-12-19 
In day 19, is a "6-dimensional side" a 6d hypercube?
Reply
Elijah
 2018-12-08 
@Matthew: Oooh ...
Reply
Melli
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 2018-05-02 
Every morning just before 7am, one of the Today Programme's presenters reads out a puzzle. Yesterday, it was this puzzle:
In a given month, the probability of a certain daily paper either running a story about inappropriate behaviour at a party conference or running one about somebody's pet being retrieved from a domestic appliance is exactly half the probability of the same paper containing a photo of a Tory MP jogging. The probability of no such photo appearing is the same as that of there being a story about inappropriate behaviour at a party conference. The probability of the paper running a story about somebody's pet being retrieved from a domestic appliance is a quarter that of its containing a photo of a Tory MP jogging. What are the probabilities the paper will (a) run the conference story, (b) run the pet story, (c) contain the jogging photo?
I'm not the only one to notice that some of Radio 4's daily puzzles are not great. I think this puzzle is a great example of a terrible puzzle. You can already see the first problem with it: it's long and words and very hard to follow on the radio. But maybe this isn't so important, as you can read it here after it's been read out.
Once you've done this, you can re-write the puzzle as follows: there are three news stories (\(A\), \(B\) and \(C\)) that the newspaper might publish in a month. We are given the following information:
$$\mathbb{P}(A\text{ or }B)=\tfrac12\mathbb{P}(C)$$ $$1-\mathbb{P}(C)=\mathbb{P}(A)$$ $$\mathbb{P}(B)=\tfrac14\mathbb{P}(C)$$
To solve this puzzle, we need use the formula \(\mathbb{P}(A\text{ or }B)=\mathbb{P}(A)+\mathbb{P}(B)-\mathbb{P}(A\text{ and }B)\). These Venn diagrams justify this formula:
Venn diagrams showing that \(\mathbb{P}(A\text{ or }B)=\mathbb{P}(A)+\mathbb{P}(B)-\mathbb{P}(A\text{ and }B)\).
Using the information we were given in the question, we get:
\begin{align} \tfrac12\mathbb{P}(C)&=\mathbb{P}(A\text{ or }B)\\ &=1-\mathbb{P}(C)+\tfrac14\mathbb{P}(C)-\mathbb{P}(A\text{ and }B)\\ \mathbb{P}(C)&=\tfrac45(1-\mathbb{P}(A\text{ and }B)). \end{align}
At this point we have reached the second problem with this puzzle: there's no answer unless we make some extra assumptions, and the question doesn't make it clear what we can assume. But let's give the puzzle the benefit of the doubt and try some assumptions.

Assumption 1: The events are mutually exclusive

If we assume that the events \(A\) and \(B\) are mutually exclusive—or, in other words, only one of these two articles can be published, perhaps due to a lack of space—then we can use the fact that
$$\mathbb{P}(A\text{ and }B)=0.$$
This means that \(\mathbb{P}(C)=\tfrac45\), \(\mathbb{P}(A)=\tfrac15\), and \(\mathbb{P}(B)=\tfrac15\). There's a problem with this answer, though: the three probabilities add up to more than 1.
This wouldn't be a problem, except we assumed that only one of the articles \(A\) and \(B\) could be published. The probabilities adding up to more than 1 means that either \(A\) and \(C\) are not mutually exclusive or \(A\) and \(B\) are not mutually exclusive, so \(C\) could be published alongside \(A\) or \(B\). There seems to be nothing special about the three news stories to mean that only some combinations could be published together, so at this point I figured that this assumption was wrong and moved on.
Today, however, the answer was posted, and this answer was given (without an working out). So we have a third problem with this puzzle: the answer that was given is wrong, or at the very best is based on questionable assumptions.

Assumption 2: The events are independent

If we assume that the events are independent—so one article being published doesn't affect whether or not another is published—then we may use the fact that
$$\mathbb{P}(A\text{ and }B)=\mathbb{P}(A)\mathbb{P}(B).$$
If we let \(c=\mathbb{P}(C)\), then we get:
\begin{align} \tfrac12c&=\mathbb{P}(A)+\mathbb{P}(B)-\mathbb{P}(A\text{ and }B)\\ &=\mathbb{P}(A)+\mathbb{P}(B)-\mathbb{P}(A)\mathbb{P}(B)\\ &=1-c+\tfrac14c-\tfrac14(1-c)c\\ \tfrac14c^2-\tfrac32c+1&=0. \end{align}
You can use your favourite formula to solve this to find that \(c=3-\sqrt5\), and therefore \(\mathbb{P}(A)=\sqrt5-2\) and \(\mathbb{P}(B)=\tfrac34-\tfrac{\sqrt5}4\).
In this case, our assumption appears to be more reasonable—as over the course of a month the stories published by a paper probably don't have much of an effect on each other—but we have the fourth, and probably biggest problem with the puzzle: the question and answer are not interesting or surprising, and the method is a bit tedious.

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 2018-05-03 
@Stefan: It's possible that that was intended but it's not completely clear. If that was the intention, I stick to my point that it's odd that the other story can be published alongside these two.
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Matthew
 2018-05-03 
Doesn’t the word ‘either’ in the opening phrase make A and B explicity exclusive?
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Stefan
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 2018-01-02 
It's 2018, and the Advent calendar has disappeared, so it's time to reveal the answers and annouce the winners. But first, some good news: with your help, Santa was able to work out which present each child wanted, and get their presents to them just in time:
Now that the competition is over, the questions and all the answers can be found here. Before announcing the winners, I'm going to go through some of my favourite puzzles from the calendar.

4 December

Pick a three digit number whose digits are all different.
Sort the digits into ascending and descending order to form two new numbers. Find the difference between these numbers.
Repeat this process until the number stops changing. The final result is today's number.
This puzzle revealed the surprising fact that repeatedly sorting the digits of a three digit number into ascending and descending order then finding the difference will always give the same answer (as long as the digits of the starting number are all different). This process is known as the Kaprekar mapping.
If four digit starting numbers are chosen, then all starting numbers that do not have three equal digits will eventually lead to 6174. It's not as simple for five digit numbers, but I'll leave you to investigate this...

11 December

Two more than today's number is the reverse of two times today's number.
Ruben pointed something interesting out to me about this question: if you remove the constraint that the answer must be a three digit number, then you see that the numbers 47, 497, 4997, 49997, and in fact any number of the form 49...97 will have this property.

20 December

What is the largest number that cannot be written in the form \(10a+27b\), where \(a\) and \(b\) are nonnegative integers (ie \(a\) and \(b\) can be 0, 1, 2, 3, ...)?

Show answer & extension

If you didn't manage to solve this one, I recommend trying replacing the 10 and 27 with smaller numbers (eg 3 and 4) and solving the easier puzzle you get first, then trying to generalise the problem. You can find my write up of this solution here.
Pedro Freitas (@pj_freitas) sent me a different way to approach this problem (related to solving the same question with different numbers on this year's Christmas card). To see his method, click "Show Answer & Extension" in the puzzle box above.

24 December

Today's number is the smallest number with exactly 28 factors (including 1 and the number itself as factors).

Show answer

I really like the method I used to solve this one. To see it, click "Show Answer" above.
Solving all 24 puzzles lead to the following final logic puzzle:

Advent 2017 logic puzzle

2017's Advent calendar ended with a logic puzzle: It's nearly Christmas and something terrible has happened: Santa and his two elves have been cursed! The curse has led Santa to forget which present three children—Alex, Ben and Carol—want and where they live.
The elves can still remember everything about Alex, Ben and Carol, but the curse is causing them to lie. One of the elves will lie on even numbered days and tell the truth on odd numbered days; the other elf will lie on odd numbered days and tell the truth on even numbered days. As is common in elf culture, each elf wears the same coloured clothes every day.
Each child lives in a different place and wants a different present. (But a present may be equal to a home.) The homes and presents are each represented by a number from 1 to 9.
Here are the clues:
21
White shirt says: "Yesterday's elf lied: Carol wants 4, 9 or 6."
10
Orange hat says: "249 is my favourite number."
5
Red shoes says: "Alex lives at 1, 9 or 6."
16
Blue shoes says: "I'm the same elf as yesterday. Ben wants 5, 7 or 0."
23
Red shoes says: "Carol wants a factor of 120. I am yesterday's elf."
4
Blue shoes says: "495 is my favourite number."
15
Blue shoes says: "Carol lives at 9, 6 or 8."
22
Purple trousers says: "Carol wants a factor of 294."
11
White shirt says: "497 is my favourite number."
6
Pink shirt says: "Ben does not live at the last digit of 106."
9
Blue shoes says: "Ben lives at 5, 1 or 2."
20
Orange hat says: "Carol wants the first digit of 233."
1
Red shoes says: "Alex wants 1, 2 or 3."
24
Green hat says: "The product of the six final presents and homes is 960."
17
Grey trousers says: "Alex wants the first digit of 194."
14
Pink shirt says: "One child lives at the first digit of 819."
3
White shirt says: "Alex lives at 2, 1 or 6."
18
Green hat says: "Ben wants 1, 5 or 4."
7
Green hat says: "Ben lives at 3, 4 or 3."
12
Grey trousers says: "Alex lives at 3, 1 or 5."
19
Purple trousers says: "Carol lives at 2, 6 or 8."
8
Red shoes says: "The digits of 529 are the toys the children want."
13
Green hat says: "One child lives at the first digit of 755."
2
Red shoes says: "Alex wants 1, 4 or 2."

Show answer

Together the clues reveal what each elf was wearing:
Drawn by Alison Clarke
Drawn by Adam Townsend
and allow you to work out where each child lives and what they wanted. Thanks Adam and Alison for drawing the elves for me.
I had a lot of fun finding place names with numbers in them to use as answers in the final puzzle. For the presents, I used the items from The 12 Days of Christmas:
#LocationPresent
1Maidstone, Kenta partridge
2Burcot, Worcestershireturtle doves
3Three Holes, NorfolkFrench hens
4Balfour, Orkneycalling birds
5Fivehead, Somersetgold rings
6Sixpenny Handley, Dorsetgeese
7Sevenhampton, Glosswans
8Leighton Buzzard, Bedsmaids
9Nine Elms, Wiltshireladies
I also snuck a small Easter egg into the calendar: the doors were arranged in a knight's tour, as shown below.
And finally (and maybe most importantly), on to the winners: 84 people submitted answers to the final logic puzzle. Their (very) approximate locations are shown on this map:
From the correct answers, the following 10 winners were selected:
1M Oostrom
2Rosie Paterson
3Jonathan Winfield
4Lewis Dyer
5Merrilyn
6Sam Hartburn
7Hannah Charman
8David
9Thomas Smith
10Jessica Marsh
Congratulations! Your prizes will be on their way shortly. Additionally, well done to Alan Buck, Alessandra Zhang, Alex Burlton, Alex Hartz, Alex Lam, Alexander, Alexander Bolton, Alexandra Seceleanu, Arturo, Brennan Dolson, Carmen Günther, Connie, Dan Whitman, David Fox, David Kendel, Edison Yifeng He, Elijah Kuhn, Eva, Evan Louis Robinson, Felix Breton, Fred Verheul, Henry Hung, Joakim Cronvall, Joe Gage, Jon Palin, Kai Lam, Keith Sutherland, Kelsey, Kenson Li, Koo Zhengqun, Kristen Koenigs, Lance Nathan, Louis de Mendonca, Mark Stambaugh, Martin Harris, Martine Vijn Nome, Matt Hutton, Matthew Schulz, Max Nilsson, Michael DeLyser, Michael Smith, Michael Ye, Mihai Zsisku, Mike Walters, Mikko, Naomi Bowler, Pattanun Wattana, Pietro Alessandro Murru, Raj, Rick, Roni, Ross Milne, Ruben, Ryan Howerter, Samantha Duong, Sarah Brook, Shivanshi, Steve Paget, Steven Peplow, Steven Spence, Tony Mann, Valentin Vălciu, Virgile Andreani, and Yasha Asley, who all also submitted the correct answer but were too unlucky to win prizes this time.
See you all next December, when the Advent calendar will return.

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Comments

Comments in green were written by me. Comments in blue were not written by me.
 2018-01-02 
Thanks! The advent calendar was great fun to take part in - and winning something in the process is the cherry on top. The riddles themselves were interesting and varied, they fitted well together in the overall puzzle, and I learned some interesting new bits of maths in the process. And now, to try my hand at the other advent calendars...

I particularly liked the riddle on the 5th of December (with walking 13 units) - it was quite tricky at first, but then I solved it by seeing that you can't end up on a square an even distance away from the centre, so the possible areas are in "circles" from the center with odd side lengths . It was quite reminiscent of showing you can't cover a chessboard with dominoes when two opposite corners are removed .
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