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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.

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.

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.

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.)

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.

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.

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:

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.

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...

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!

This week, I've been at Talking Maths in Public (TMiP) in Newcastle. TMiP is a conference for anyone involved in—or interested in getting involved in—any sort of maths outreach, enrichment, or public engagement activity. It was really good, and I highly recommend coming to TMiP 2025.

The Saturday morning at TMiP was filled with a choice of activities, including a puzzle hunt written by me: the Tyne trial. At the start/end point of the Tyne trial, there was a locked box with a combination lock. In order to work out the combination for the lock, you needed to find some clues hidden around Newcastle and solve a few puzzles.

Every team taking part was given a copy of these instructions. Some people attended TMiP virtually, so I also made a version of the Tyne trial that included links to Google Street View and photos from which the necessary information could be obtained. You can have a go at this at mscroggs.co.uk/tyne-trial/remote. For anyone who wants to try the puzzles without searching through virtual Newcastle, the numbers that you needed to find are:

The solutions to the puzzles and the final puzzle are below. If you want to try the puzzles for yourself, do that now before reading on.

We are going to start with a number then repeat the following process: if the number you have is a palindrome, stop; otherwise add the number to itself backwards. For example, if we start with 219, then we do: $$219\xrightarrow{+912}1131\xrightarrow{+1311}2442.$$ If you start with the number \(10b+9\) (ie 59), what palindrome do you get?

There are 12 three-digit numbers whose digits are 1, 2, 3, 4, or 5 with exactly two digits that are ones. How many \(c\)-digit (ie 1838-digit) numbers are there whose digits are 1, 2, 3, 4, or 5 with exactly \(c-1\) digits (ie 1837) that are ones?

The largest value of \(n\) such that \((n!-2)/(n-2)\) is an integer is 4. What is the largest value of \(n\) such that \((n!-d)/(n-d)\) (ie \((n!-1931)/(n-1931)\)) is an integer?

We are going to start with a number then repeat the following process: if we've reached 0, stop; otherwise subtract the smallest prime factor of the current number. For example, if we start with 9, then we do: $$9\xrightarrow{-3}6\xrightarrow{-2}4\xrightarrow{-2}2\xrightarrow{-2}0.$$ It took 4 steps to get to 0. What is the smallest starting number such that this process will take \(e\) (ie 1619) steps?

I thought of a four digit number. I removed a digit to make a three digit number, then added my two numbers together. The result is \(200f+127\) (ie 9727). What was my original number?

If you roll two six-sided fair dice, the most likely total is 7. What is the most likely total if you rolled \(1470+g\) (ie 2470) dice?

The final puzzle involves using the answers to the five puzzles to find the four digit code that opens the box (and the physical locked box that was in the library on Saturday. To give hints to this code, each clue was given a "score".

The score of a number is the number of values of \(i\) such that the \(i\)th digit of the code is a factor of the \(i\)th digit of the number. For example, if the code was 1234, then the score of the number 3654 would be 3 (because 1 is a factor of 3; 2 is a factor of 6; and 4 is a factor of 4).

The seven clues to the final code are:

For the past couple of months, I've once again been spending an awful lot of my spare time working on Chalkdust. Today you can see the result of all this hard work: Chalkdust issue 17. I recommend checking out the entire magazine: you can read it online or order a physical copy.

My most popular contribution to the magazine is probably the crossnumber. I enjoyed writing this one; hope you enjoy solving it.

I also spent some time making this for the back page of the magazine. It's probably the most fun I've had making something stupid for Chalkdust for ages.