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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...
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, ...)?
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).
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."
White shirt says: "Yesterday's elf lied: Carol wants 4, 9 or 6."
10
Orange hat says: "249 is my favourite number."
Orange hat says: "249 is my favourite number."
5
Red shoes says: "Alex lives at 1, 9 or 6."
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."
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."
Red shoes says: "Carol wants a factor of 120. I am yesterday's elf."
4
Blue shoes says: "495 is my favourite number."
Blue shoes says: "495 is my favourite number."
15
Blue shoes says: "Carol lives at 9, 6 or 8."
Blue shoes says: "Carol lives at 9, 6 or 8."
22
Purple trousers says: "Carol wants a factor of 294."
Purple trousers says: "Carol wants a factor of 294."
11
White shirt says: "497 is my favourite number."
White shirt says: "497 is my favourite number."
6
Pink shirt says: "Ben does not live at the last digit of 106."
Pink shirt says: "Ben does not live at the last digit of 106."
9
Blue shoes says: "Ben lives at 5, 1 or 2."
Blue shoes says: "Ben lives at 5, 1 or 2."
20
Orange hat says: "Carol wants the first digit of 233."
Orange hat says: "Carol wants the first digit of 233."
1
Red shoes says: "Alex wants 1, 2 or 3."
Red shoes says: "Alex wants 1, 2 or 3."
24
Green hat says: "The product of the six final presents and homes is 960."
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."
Grey trousers says: "Alex wants the first digit of 194."
14
Pink shirt says: "One child lives at the first digit of 819."
Pink shirt says: "One child lives at the first digit of 819."
3
White shirt says: "Alex lives at 2, 1 or 6."
White shirt says: "Alex lives at 2, 1 or 6."
18
Green hat says: "Ben wants 1, 5 or 4."
Green hat says: "Ben wants 1, 5 or 4."
7
Green hat says: "Ben lives at 3, 4 or 3."
Green hat says: "Ben lives at 3, 4 or 3."
12
Grey trousers says: "Alex lives at 3, 1 or 5."
Grey trousers says: "Alex lives at 3, 1 or 5."
19
Purple trousers says: "Carol lives at 2, 6 or 8."
Purple trousers says: "Carol lives at 2, 6 or 8."
8
Red shoes says: "The digits of 529 are the toys the children want."
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."
Green hat says: "One child lives at the first digit of 755."
2
Red shoes says: "Alex wants 1, 4 or 2."
Red shoes says: "Alex wants 1, 4 or 2."
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:
# | Location | Present |
1 | Maidstone, Kent | a partridge |
2 | Burcot, Worcestershire | turtle doves |
3 | Three Holes, Norfolk | French hens |
4 | Balfour, Orkney | calling birds |
5 | Fivehead, Somerset | gold rings |
6 | Sixpenny Handley, Dorset | geese |
7 | Sevenhampton, Glos | swans |
8 | Leighton Buzzard, Beds | maids |
9 | Nine Elms, Wiltshire | ladies |
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:
1 | M Oostrom |
2 | Rosie Paterson |
3 | Jonathan Winfield |
4 | Lewis Dyer |
5 | Merrilyn |
6 | Sam Hartburn |
7 | Hannah Charman |
8 | David |
9 | Thomas Smith |
10 | Jessica 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, Ed, 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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2017-11-28
This year, the front page of mscroggs.co.uk will once again feature an advent calendar, just like last year 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: 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.
Santa has called on you to help him work out the details he has forgotten. 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 elves tells you. You must work out which combination of clothes each elf wears, which one lies on each day, then put all the clues together to work out which presents need delivering to Alex, Ben and Carol, and where to deliver them.
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! A selection of the prizes are shown below, and will be added to throughout December.
The ten winners will also will one of these winners' medals:
As you solve the puzzles, your answers will be stored. This year, there is a new feature allowing you to synchronise your answers between multiple computers: simply 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 2017. 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 2017. 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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Me again
Just for info (clarification?): I read question on 22nd as
22 is two times an odd number. Today's number is the mean of all the answers, on days (including today), that are two times an odd number."
Note my added commas. I was averaging the answers, not the dates. Certainly ambiguous as far as I am concerned.
Only fixed it by 'cheating'. Trying best guessses of averages until I got the correct one.
Just for info (clarification?): I read question on 22nd as
22 is two times an odd number. Today's number is the mean of all the answers, on days (including today), that are two times an odd number."
Note my added commas. I was averaging the answers, not the dates. Certainly ambiguous as far as I am concerned.
Only fixed it by 'cheating'. Trying best guessses of averages until I got the correct one.
neal (@zbvif)
Wow. Just discovered I meisread 15th Dec puzzle.
I can tell you that the number of combinations of n As and Bs which contain at at least one uninterrupted sequence of 3 As is 2^n - F3(n+3) where F3 is the fibonaccia variant adding 3 numbers (1,1,2,4,7,13,24 etc.).
Only took me about 8 hours (with some small help form OEIS for the 2 As problem)
I can tell you that the number of combinations of n As and Bs which contain at at least one uninterrupted sequence of 3 As is 2^n - F3(n+3) where F3 is the fibonaccia variant adding 3 numbers (1,1,2,4,7,13,24 etc.).
Only took me about 8 hours (with some small help form OEIS for the 2 As problem)
Neal (@zbvif)
With todays puzzle does the pancake have any thickness i.e can we slice the pancake into 2 circular pancakes each with half the thickness or are we to assume its 2D
Alex
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2016-12-28
More than ten correct solutions to this year's Advent calendar puzzle competition were submitted on Christmas Day, so the competition is now over.
(Although you can still submit your answers to get me to check them.) Thank-you to everyone who took part in the puzzle, I've had a lot of
fun watching your progress and talking to you on Twitter, Reddit, etc. You can find all the puzzles and answers (from 1 January) here.
The (very) approximate locations of all the entries I have received so far are shown on this map:
This year's winners have been randomly selected from the 29 correct entries on Christmas Day. They are:
1 | Jack Jiang |
2 | Steve Paget |
3 | Joe Gage |
4 | Tony Mann |
5 | Stephen Cappella |
6 | Cheng Wai Koo |
7 | Demi Xin |
8 | Lyra |
9 | David Fox |
10 | Bob Dinnage |
Your prizes will be on their way in early January.
Now that the competition has ended, I can give away a secret. Last year, Neal
suggested that it would be fun if a binary picture was hidden in the answers. So this year I hid one. If you write all the answers in binary, with
each answer below the previous and colour in the 1s black, you will see this:
I also had a lot of fun this year making up the names, locations, weapons and motives for the final murder mystery puzzle. In case you missed them these were:
# | Murder suspect | Motive |
1 | Dr. Uno (uno = Spanish 1) | Obeying nameless entity |
2 | Mr. Zwei (zwei = German 2) | To worry others |
3 | Ms. Trois (trois = French 3) | To help really evil elephant |
4 | Mrs. Quattro (quattro = Italian 4) | For old unknown reasons |
5 | Prof. Pum (pum = Welsh 5) | For individual violent end |
6 | Miss. Zes (zes = Dutch 6) | Stopping idiotic xenophobia |
7 | Lord Seacht (seacht = Irish 7) | Suspect espied victim eating newlyweds |
8 | Lady Oito (oito = Portuguese 8) | Epic insanity got him today |
9 | Rev. Novem (novem = Latin 9) | Nobody in newsroom expected |
# | Location | Weapon |
1 | Throne room | Wrench (1 vowel) |
2 | Network room | Rope (2 vowels) |
3 | Beneath reeds | Revolver (3 vowels) |
4 | Edge of our garden | Lead pipe (4 vowels) |
5 | Fives court | Neighbour's sword (5 vowels) |
6 | On the sixth floor | Super banana bomb (6 vowels) |
7 | Sparse venue | Antique candlestick (7 vowels) |
8 | Weightlifting room | A foul tasting poison (8 vowels) |
9 | Mathematics mezzanine | Run over with an old Ford Focus (9 vowels) |
Finally, well done to Scott,
Matthew Schulz,
Michael Gustin,
Daniel Branscombe,
Kei Nishimura-Gasparian,
Henry Hung,
Mark Fisher,
Jon Palin,
Thomas Tu,
Félix Breton,
Matt Hutton,
Miguel,
Fred Verheul,
Martine Vijn Nome,
Brennan Dolson,
Louis de Mendonca,
Roni,
Dylan Hendrickson,
Martin Harris,
Virgile Andreani,
Valentin Valciu,
and Adia Batic for submitting the correct answer but being too unlucky to win prizes this year. Thank you all for taking part and I'll see you
next December for the next competition.
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I got my prize in the mail today. I really liked the stories from Gustave Verbeek; I thought that was pretty clever. I really appreciate you being willing to send the prizes internationally.
Thanks for setting this all up; I had a lot of fun solving the puzzles every day (and solving half them again when my cookie for the site somehow got deleted). I'll be sure to participate next time too!
Thanks for setting this all up; I had a lot of fun solving the puzzles every day (and solving half them again when my cookie for the site somehow got deleted). I'll be sure to participate next time too!
SC
Thanks, Matthew! The puzzles were really fun, and piecing the clues was very interesting too!
Jack
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2016-11-27
This year, the front page of mscroggs.co.uk will feature an advent calendar, just like last 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 murder mystery logic puzzle in which you have to work out the murderer, motive, location and weapon used: the answer to each of these murder facts is a digit from 1 to 9 (eg. The murderer could be 6, the motive 9, etc.).
As you solve the puzzles, your answers will be stored in a cookie. 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 on Christmas Day. Runners up will then be chosen from those who submit the correct answer on Christmas Day, then those who submit the correct answer on Boxing Day, then the next day, and so on. As the winners will be chosen randomly, there is no need to get up at 5am on Christmas Day this year!
The winner will win this array of prizes:
I will be adding to the pile of prizes throughout December. Runners up will get a subset of the prizes. The winner and runners up will also win an mscroggs.co.uk 2016 winners medal:
To win a prize, you must submit your entry before the end of 2016. Only one entry will be accepted per person. Once ten correct entries have been submitted, I will add a note here and below the calendar. 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!
Edit: added picture of this year's medals.
Edit: more than ten correct entries have been submitted, list of prize winners can be found here.
You can still submit your answers but the only prize left is glory.
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@Another Matthew: Ten correct submissions have been made. Just updating the pages to reflect this...
Matthew
Really enjoyed the extra bit at the end this year! Looking forward to 2017's calendar.
Louis
@Lyra: I'll email you if you are one of the winners to get the rest of your address!
Matthew
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2016-01-01
The deadline for entering the 2015 Advent calendar competition has now passed. You can see all the puzzles and their answers here.
Thank-you to everyone who took part in the competition! In total, 23 people submitted answers to the puzzles. Every one of these had correct answers to at least 18 of the puzzles. The winners are as follows:
# | Name | Details |
1 | Scott | All correct at 5:00:18 GMT on Christmas Day |
2 | Louis de Mendonca | All correct at 5:00:32 GMT on Christmas Day |
3 | Alex Bolton | All correct at 5:00:34 GMT on Christmas Day |
4 | Martin Harris | All correct at 6:15 GMT on Christmas Day |
5 | Linus Hamilton | All correct at 14:12 GMT on Christmas Day |
6 | Zephi | All correct at 20:40 GMT on Christmas Day |
7 | Daniel Chiverton | All but one (5 December) correct at 5:00:24 GMT on Christmas Day |
8 | Jon Palin | All but one (12 December) correct at 5:00:34 GMT on Christmas Day |
9 | Kathryn Coffin | All but one (5 December) correct at 6:28 GMT on Christmas Day |
10 | Félix Breton | All but one (15 December) correct at 9:05 GMT on Christmas Day |
I will be in touch will all the entrants in the next few days and I will post pictures of prizes here once they are on their way!
I have already started working on puzzles for next year's (in fact this year's) calendar, so make sure you're back here in December...
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@stephan: It looks like the code I wrote to check the solutions were unique contained errors. This might explain why first question was very difficult to solve with logic alone.
Matthew
Are you sure http://www.mscroggs.co.uk/puzzles/126 have unique solutions. See e.g. http://cryptarithms.awardspace.us/solv... which gives lots of solutions
stephan
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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 .