mscroggs.co.uk
mscroggs.co.uk

subscribe

Blog

 2024-11-21 
The mscroggs.co.uk Advent Calendar is back for its tenth year! 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: there's been a major malfunction in multiple machines in Santa's toy factory, and not enough presents have been made. Santa has a backup warehouse full of wrapped presents that can be used in the case of severe emergency, but the warehouse is locked. You need to help Santa work out the code to unlock the warehouse so that he can deliver the presents before Christmas is ruined for everyone.
The information needed to work out the code to the warehouse is known by Santa and his three most trusted elves: Santa is remembering a three-digit number, and each elf is remembering a one-digit and a three-digit number. If Santa and the elves all agree that the emergency warehouse should be opened, they can work out the code for the door as follows:
But this year, there is a complication: the three elves are on a diplomatic mission to Mars to visit Martian Santa and cannot be contacted, so you need to piece together their numbers from the clues they have left behind.
Behind each day (except Christmas Day), there is a puzzle with a three-digit answer. Each of these answers forms part of a clue about Santa's and the elves' numbers. You must use these clues to work out the code for the warehouse. You can use this page to try opening the door. If you enter an incorrect code three times, the door mechanism locks until the following day.
Ten randomly selected people who solve all the puzzles, open the warehouse, and fill in the entry form behind the door on the 25th will win prizes!
The prizes will include an mscroggs.co.uk Advent 2024 T-shirt. If you'd like one of the T-shirts from a previous Advent, they are available to order at merch.mscroggs.co.uk.
The winners will be randomly chosen from all those who submit the entry form before the end of 2024. 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!
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.
To win a prize, you must submit your entry before the end of 2024. Only one entry will be accepted per person. If you have any questions, ask them in the comments below, on Bluesky, or on Mastodon. If you'd like to chat with other solvers, we'll be discussing the Advent Calendar in the #scroggs-advent-calendar channel in the Finite Group Discord: you can join the Discord by following the link in this post on Patreon (you'll need to become a free member on Patreon to unlock the post).
So once December is here, get solving! Good luck and have a very merry Christmas!
×9      ×2                  ×2
(Click on one of these icons to react to this blog post)

You might also enjoy...

Comments

Comments in green were written by me. Comments in blue were not written by me.
@Ben: Thanks, I've added the "before"
Matthew
   ×2              Reply
I am so appreciative that you continue to make these puzzles every year. I think I started doing them in 2017 and always enjoy them. Thank you!
Jessica
×2                 Reply
Thanks Scroggs - first time I've done this and very much enjoyed the days and also the meta-puzzling. Brilliant!!. If you run in future years I have one request for a tiny tweak - I find the numbers on the advent calendar for the days very small for my ageing eye-sight - any chance of a bigger font? And last suggestion - when it gets to the end, provide a link to your "buy me a cup of tea" page as this is more than worth a few £s :-). Thanks again :-)
Justin
×1                 Reply
@Seth Cohen: Thanks Seth, I solved it by looking at the 5×5 picture and thinking harder. Thanks to Matthew for another enjoyable set of puzzles. I look forward to reading the proper solution.
Reza
                 Reply
Had a great time doing this puzzles again this year! 23 was particularly fun :) Thanks for taking the time to make this!
Bill V.
                 Reply
I enjoyed working the advent puzzles. Thank you for providing such fun entertainment and math challenges! Attempted most without any programming help but some begged for a programming solution. Refreshed some former Python skills to happily solve a few puzzles. Looking forward to math solutions. Merry Christmas and Happy Holidays!
Tony
×1                 Reply
 Add a Comment 


I will only use your email address to reply to your comment (if a reply is needed).

Allowed HTML tags: <br> <a> <small> <b> <i> <s> <sup> <sub> <u> <spoiler> <ul> <ol> <li> <logo>
To prove you are not a spam bot, please type "ratio" in the box below (case sensitive):
 2024-02-20 
Back in November, I wrote about making 2n-page zines. Thanks to some conversations I had at Big MathsJam in later November, I've been able to work out how many 128-page zines there are: 315434.

The insight

At Big MathsJam, Colin Beveridge pointed out something he'd noticed about the possible zines: when drawing the line connecting the pages in order, there were some line segments that were always included. For example, here are all of the possible 64-page zines:
Every single one of these includes these line segments:
Colin conjectured that for a zine of any size, a pattern like this of alternative horizontal segments must always be included. He was close to justifying this, and since MathsJam I've been able to fill in the full justificication.

The justificiation

First, consider the left-most column of pages. They must be connected like this:
If they were connected in any other way, there would be two vertical connections in a row, which would create a page that is impossible to open (as every other connection must be a horizontal that ends up in the spine). Additionally, the horizontal lines in this diagram must all be in the spine (as otherwise we again get pages that cannot be opened).
Next, consider a horizontal line that's in the spine (shown in red below), and we can look at all the possible ways to draw the line through the highlighted page, paying particular attention to the dashed blue line:
The six possible ways in which the line could travel through the highlighted page are:
The three options in the top row do not give a valid zine: the leftmost diagram has two vertical connections in a row (leading to pages that do not open). The other two diagrams in the top row have the horizontal line that we know is in the spine, followed by a horizontal line not in the spine, then a vertial line: this vertical line should be in the spine, but as it is vertical it cannot be (without making a page that doesn't open).
In each of the diagrams in the bottom row, the connection shown in dashed blue is included and must be in the spine: in the leftmost diagram, the horizontal line that we know is in the spine is followed by a horizontal not in the spine, then the horizinal in the dashed blue position that must therefore be in the spine. The othe other two diagrams in the bottom row, the dashed blue position is connected to a vertical line: this means that the dashed blue connection must be in the spine (as otherwise the vertical would cause a page that doesn't open).
Overall, we've now shown that the leftmost column of lines must always be included and must all be in the spine; and for each horizontal line in the spine, the line to the right of it after a single gap must also be included and in the spine. From this, it follows that all the horizontal lines in Colin's pattern must always be included.

Calculating the number of 128-page zines

Now that I knew that all these horizonal lines are always included, I was able to update the code I was using to find all the possible zines to use this. After a few hours, it had found all 315434 possibilites. I was very happy to get this total, as it was the same as the number that Luna (another attendee of Big MathsJam) had calculated but wasn't certain was correct.
The sequence of the number of 2n-page zines, including the newly calculated number, is now published on the OEIS. I think calculating number of 256-page zines is still beyond my code though...
      ×8      ×5      ×5      ×8
(Click on one of these icons to react to this blog post)

You might also enjoy...

Comments

Comments in green were written by me. Comments in blue were not written by me.
 Add a Comment 


I will only use your email address to reply to your comment (if a reply is needed).

Allowed HTML tags: <br> <a> <small> <b> <i> <s> <sup> <sub> <u> <spoiler> <ul> <ol> <li> <logo>
To prove you are not a spam bot, please type "h" then "e" then "x" then "a" then "g" then "o" then "n" in the box below (case sensitive):
 2024-01-07 
Welcome to 2024 everyone! Now that the Advent calendar has disappeared, it's time to reveal the answers and announce the winners. But first, some good news: with your help, the machine was fixed in time for Santa to deliver presents and Christmas was saved!
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 and a couple of other interesting bits and pieces.

Highlights

My first highlight is the puzzle from 4 December. I like this puzzle, because at first it looks really difficult, and the size of the factorial involved is impossibly large, but the way of solving it that I used essentially just ignores the factorial leading to a much easier question.

4 December

If \(n\) is 1, 2, 4, or 6 then \((n!-3)/(n-3)\) is an integer. The largest of these numbers is 6.
What is the largest possible value of \(n\) for which \((n!-123)/(n-123)\) is an integer?

Show answer


My next pair of highlights are the puzzles from 6 and 7 December. I always enjoy a surprise appearance of the Fibonacci sequence, and a double enjoyed a double appearance in two contexts that at first look completely different.

6 December

There are 5 ways to tile a 4×2 rectangle with 2×1 pieces:
How many ways are there to tile a 12×2 rectangle with 2×1 pieces?

Show answer

7 December

There are 8 sets (including the empty set) that contain numbers from 1 to 4 that don't include any consecutive integers:
\(\{\}\), \(\{1\}\), \(\{2\}\), \(\{3\}\), \(\{4\}\), \(\{1,3\}\), \(\{1,4\}\), \(\{2, 4\}\)
How many sets (including the empty set) are there that contain numbers from 1 to 14 that don't include any consecutive integers?

Show answer & extension


My next highlight is the puzzle from 13 December. I love a good crossnumber, and had a lot of fun making this small one up. (If you enjoyed this one, you should check out the crossnumbers I write for Chalkdust.)

13 December

Today's number is given in this crossnumber. No number in the completed grid starts with 0.

Show answer


My final highlight is the puzzle from 22 December. I enjoy that you can use one of the circle theorems to solve this, despite there being no circles directly involved in the question.

22 December

There are 4 ways to pick three vertices of a regular quadrilateral so that they form a right-angled triangle:
In another regular polygon with \(n\) sides, there are 14620 ways to pick three vertices so that they form a right-angled triangle. What is \(n\)?

Show answer

Hardest and easiest puzzles

Once you've entered 24 answers, the calendar checks these and tells you how many are correct. 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.
It looks like the hardest puzzles were on 23 and 12 December; and the easiest puzzles were on 1, 3, 5, and 11 December.

Fixing the machine

To finish the Advent calendar, you were tasked with fixing the machine. The answers to all the puzzles were required to be certain of which combination of parts were needed to fix the machine, but it was possible to reduce the number of options to a small number and get lucky when trying these options. This graph shows how many people fixed the machine on each day:

The winners

And finally (and maybe most importantly), on to the winners: 180 people managed to fix the machine. That's slightly fewer than last year:
From the correct answers, the following 10 winners were selected:
Congratulations! Your prizes will be on their way shortly.
The prizes this year include 2023 Advent calendar T-shirts. If you didn't win one, but would like one of these, I've made them available to buy at merch.mscroggs.co.uk alongside the T-shirts from previous years.
Additionally, well done to 100118220919, Aaron, Adam NH, Aidan Dodgson, AirWrek, Alan Buck, Alejandro Villarreal, Alek2ander, Alex, Alex Hartz, Allan Taylor, Andrew Roy, Andrew Thomson, Andrew Turner, Andy Ennaco, Ashley Jarvis, Austin Antoniou, Becky Russell, Ben, Ben Boxall, Ben Reiniger, Ben Tozer, Ben Weiss, Bill Russ, Bill Varcho, Blake, Bogdan, Brian Wellington, Carl Westerlund, Carmen, Carnes Family, Cathy Hooper, Chris Eagle, Chris Hellings, Colin Brockley, Connors of York, Corbin Groothuis, Dan Colestock, Dan May, Dan Rubery, Dan Swenson, Dan Whitman, Daphne, David and Ivy Walbert, David Ault, David Berardo, David Fox, David Kendel, David Mitchell, Deborah Tayler, Diane, Donald Anderson, Duncan S, Dylan Madisetti, Ean, Elise Raphael, Emelie, Emily Troyer, Emma, Eric, Eric Kolbusz, Ewan, Frank Kasell, Fred Verheul, Gabriella Pinter, Gareth McCaughan, Gary M, Gary M. Gerken, George Witty, Gert-Jan, Grant Mullins, Gregory Wheeler, Guillermo Heras Prieto, Heerpal Sahota, Helen, Herschel, Iris Lasthofer, Ivan Molotkov, Jack, Jack H, Jacob Y, James Chapman, Jan Z, Jay N, Jean-Sébastien Turcotte, Jen Sparks, Jenny Forsythe, Jessica Marsh, Jim Ashworth, Joe Gage, Johan, Jon Palin, Jonathan Chaffer, Jonathan Thiele, Jorge del Castillo Tierz, K Brooks, Kai, Karen Climis, Kevin Docherty, Kevin Fray, Kirsty Fish, Kristen Koenigs, lacop, Lazar Ilic, Lewis Dyer, Lisa Stambaugh, Lise Andreasen, Lizzie McLean, Louis, Magnus Eklund, Marco van der Park, Mark Fisher, Mark Stambaugh, Martijn O., Martin Harris, Martin Holtham, Mary Cave, Matthew Schulz, Max, Merrilyn, Mihai Zsisku, Mike Hands, Miles Lunger, Mr J Winfield, Nadine Chaurand, Naomi Bowler, Nathan Whiteoak, Nick C, Nick Keith, Niji Ranger, Pamela Docherty, Pierce R, Qaysed, Rashi, Ray Arndorfer, rea, Reuben Cheung, Riccardo Lani, Richard O, Rob Reynolds, Robby Brady, Roger Lipsett, Roni, Rosie Paterson, RunOnFoot, Ruth Franklin, Ryan Wise, Sage Robinson, Sam Dreilinger, Sarah, Scott, Sean Henderson, Seth Cohen, Shivanshi, Shreevatsa, Stephen Cappella, Steve Blay, TAS, Tehnuka, The Johnston Family, Tina, Tony Mann, Trent Marsh, tripleboleo, Valentin VĂLCIU, Vinny R, William Huang, Yasha, and Yuliya Nesterova, who all also completed the Advent calendar but were too unlucky to win prizes this time or chose to not enter the prize draw.
See you all next December, when the Advent calendar will return.
×9      ×10      ×5      ×5      ×5
(Click on one of these icons to react to this blog post)

You might also enjoy...

Comments

Comments in green were written by me. Comments in blue were not written by me.
In your solution for the 12th, I think there's still a little work to do: to check that the answer is the smallest integer that works. For that, because 241 is prime, you only have a handful of values to check.
Ben Reiniger
×4   ×4   ×4   ×4   ×4     Reply
(you've left the "drones" in at the beginning of the Winners section)
Ben Reiniger
×2   ×2   ×3   ×1   ×2     Reply
On the 6th and 7th, there's also a direct bijection: in the tiling, horizontal tiles must occur in aligned pairs (else they split left/right into odd number of 1x1 blocks). Encode a tiling with the set of horizontal locations of the left ends of the horizontal-tile-pairs.
Ben Reiniger
×2   ×2   ×2   ×2   ×2     Reply
 Add a Comment 


I will only use your email address to reply to your comment (if a reply is needed).

Allowed HTML tags: <br> <a> <small> <b> <i> <s> <sup> <sub> <u> <spoiler> <ul> <ol> <li> <logo>
To prove you are not a spam bot, please type "a" then "x" then "e" then "s" in the box below (case sensitive):
 2023-12-08 
In November, I spent some time (with help from TD) designing this year's Chalkdust puzzle Christmas card.
The card looks boring at first glance, but contains 10 puzzles. By colouring in the answers to the puzzles on the front of the card in the colours given (each answer appears four time), you will reveal a Christmas themed picture.
If you're in the UK and want some copies of the card to send to your maths-loving friends, you can order them at mscroggs.co.uk/cards.
If you want to try the card yourself, you can download this printable A4 pdf. Alternatively, you can find the puzzles below and type the answers in the boxes. The answers will automatically be found and coloured in...
13 36 8 13 32 34 18 18 81 81 32 7 11 1 20 40 75 12 94 36 2 2 11 20 7 1 34 11 10 18 64 88 94 60 94 64 94 88 88 60 88 64 2 64 43 2 43 40 49 49 12 60 75 10 49 32 81 18 49 20 34 36 32 40 75 12 43 40 43 12 60 75 36 34 4 11 20 7 10 10 8 7 13 13 4 8 8 1 4 4 1 81
×21      ×15      ×7      ×8      ×13
(Click on one of these icons to react to this blog post)

You might also enjoy...

Comments

Comments in green were written by me. Comments in blue were not written by me.
Incorrect answers are treated is correct.

Looking at the JavaScript code, I found that any value that is a key in the array "regions" is treated as correct for all puzzles.
Lars Nordenström
×4   ×4   ×4   ×4   ×4     Reply
My visual abilities fail me - managed to solve the puzzles but cannot see what the picture shows
Gantonian
×4   ×4   ×4   ×4   ×4     Reply
@nochum: It can't, so the answer to that one probably isn't 88.
Matthew
×4   ×5   ×5   ×5   ×5     Reply
how can a dodecagon with an area of 88 fit inside anything with an area of 62.83~?
nochum
×4   ×4   ×4   ×4   ×2     Reply
 Add a Comment 


I will only use your email address to reply to your comment (if a reply is needed).

Allowed HTML tags: <br> <a> <small> <b> <i> <s> <sup> <sub> <u> <spoiler> <ul> <ol> <li> <logo>
To prove you are not a spam bot, please type "emirp" backwards in the box below (case sensitive):
 2023-11-22 
This year, the front page of mscroggs.co.uk will once again feature an Advent calendar, just like in each of the last eight years. 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: a machine in Santa's toy factory has malfunctioned, and is unable to finish building all the presents that Santa needs. You need to help Santa work out how to fix the broken machine so that he can build the presents and deliver them before Christmas is ruined for everyone.
Inside the broken machine, there were five toy production units (TPUs) installed at sockets labelled A to E. During the malfunction, these TPUs were so heavily damaged that Santa is unable to identify which TPU they were when trying to fix the machine. The company that supplies TPUs builds 10 different units, numbered from 0 to 9. You need to work out which of the 10 TPUs needs to be installed in each of the machine's sockets, so that Santa can fix the machine. It may be that two or more of the TPUs are the same.
Behind each day (except Christmas Day), there is a puzzle with a three-digit answer. Each of these answers forms part of a clue about the machine's TPUs. You must use these clues to work out which TPU to install in each socket. You can use this page to plug in five TPUs and test the machine. It takes a significant amount of Santa's time to test the machine, so you can only run a very small number of tests each day.
Ten randomly selected people who solve all the puzzles, fix the machine, and fill in the entry form behind the door on the 25th will win prizes!
The prizes will include an mscroggs.co.uk Advent 2023 T-shirt. If you'd like one of the T-shirts from a previous Advent, they are available to order at merch.mscroggs.co.uk.
The winners will be randomly chosen from all those who submit the entry form before the end of 2023. 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!
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.
To win a prize, you must submit your entry before the end of 2023. Only one entry will be accepted per person. If you have any questions, ask them in the comments below, on Twitter, or on Mastodon.
So once December is here, get solving! Good luck and have a very merry Christmas!
×9                  ×3      ×4
(Click on one of these icons to react to this blog post)

You might also enjoy...

Comments

Comments in green were written by me. Comments in blue were not written by me.
Thank you Matthew. 23rd was my favourite puzzle as the cuisenaire rods helped me and I worked with my son to get a final answer. Happy New Year.
Jenny
   ×1   ×2   ×1   ×1     Reply
loving the new and harder types of puzzles this year :)
V
   ×1   ×1   ×3   ×3     Reply
@Seth Cohen: Even with those hints I just can't seem to get this one!
Steve
×6   ×6   ×6   ×6   ×6     Reply
I really like 22, and will be using it with my top set Year 10s when I do circle theorems next term :)
Artie Smith
×7   ×6   ×6   ×7   ×6     Reply
I love doing your puzzles, your advent ones as well as the Chalkdust Crossnumbers - thank you!
Merrilyn
×9   ×8   ×8   ×8   ×8     Reply
 Add a Comment 


I will only use your email address to reply to your comment (if a reply is needed).

Allowed HTML tags: <br> <a> <small> <b> <i> <s> <sup> <sub> <u> <spoiler> <ul> <ol> <li> <logo>
To prove you are not a spam bot, please type "v" then "e" then "c" then "t" then "o" then "r" in the box below (case sensitive):

Archive

Show me a random blog post
 2025 

Mar 2025

How to write a crossnumber

Jan 2025

Christmas (2024) is over
Friendly squares
 2024 
▼ show ▼
 2023 
▼ show ▼
 2022 
▼ show ▼
 2021 
▼ show ▼
 2020 
▼ show ▼
 2019 
▼ show ▼
 2018 
▼ show ▼
 2017 
▼ show ▼
 2016 
▼ show ▼
 2015 
▼ show ▼
 2014 
▼ show ▼
 2013 
▼ show ▼
 2012 
▼ show ▼

Tags

pac-man graphs matt parker fence posts harriss spiral kings european cup pizza cutting boundary element methods bodmas london fractals trigonometry dragon curves javascript speed numbers national lottery folding tube maps realhats cambridge london underground golden spiral matrix of minors probability wool sport map projections countdown runge's phenomenon books bempp computational complexity weather station radio 4 numerical analysis world cup exponential growth edinburgh royal institution correlation the aperiodical matrix multiplication misleading statistics turtles football coins plastic ratio games matrices youtube programming gaussian elimination electromagnetic field sound binary martin gardner noughts and crosses logic triangles inverse matrices python convergence mathslogicbot stirling numbers asteroids tmip pascal's triangle hats preconditioning newcastle hyperbolic surfaces wave scattering people maths interpolation frobel reuleaux polygons folding paper hexapawn regular expressions ternary pi squares flexagons logs standard deviation royal baby dataset a gamut of games bubble bobble mathsteroids arithmetic pythagoras datasaurus dozen php weak imposition mathsjam geometry zines finite group cross stitch machine learning sorting logo data light live stream chess graph theory christmas gerry anderson big internet math-off sobolev spaces crossnumber gather town phd approximation accuracy manchester game of life data visualisation talking maths in public finite element method 24 hour maths final fantasy mean go craft nine men's morris geogebra menace dates matrix of cofactors estimation christmas card polynomials dinosaurs tennis video games databet puzzles ucl crochet quadrilaterals friendly squares raspberry pi news hannah fry propositional calculus errors statistics recursion error bars game show probability manchester science festival pi approximation day chalkdust magazine braiding stickers advent calendar curvature draughts signorini conditions crossnumbers golden ratio rhombicuboctahedron chebyshev guest posts simultaneous equations captain scarlet inline code determinants fonts latex oeis platonic solids bots reddit crosswords palindromes anscombe's quartet rugby

Archive

Show me a random blog post
▼ show ▼
© Matthew Scroggs 2012–2025