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2018-01-02

## Christmas (2017) is over

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, ...)?
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

2016'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."

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, 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.

### Similar posts

 Christmas card 2017 Christmas (2017) is coming! Christmas (2016) is over Christmas card 2016

### 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 .
Lewis
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2017-12-18

## Christmas card 2017

Just like last year, I spent some time in November this year designing a puzzle Christmas card for Chalkdust.
The card looks boring at first glance, but contains 10 puzzles. Converting the answers to base 3, writing them in the boxes on the front, then colouring the 1s black and 2s orange will reveal a Christmassy picture.
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 converted to base 3 and coloured...
 # Answer (base 10) Answer (base 3) 1 0 0 0 0 0 0 0 2 0 0 0 0 0 0 0 3 0 0 0 0 0 0 0 4 0 0 0 0 0 0 0 5 0 0 0 0 0 0 0 6 0 0 0 0 0 0 0 7 0 0 0 0 0 0 0 8 0 0 0 0 0 0 0 9 0 0 0 0 0 0 0 10 0 0 0 0 0 0 0
1. In a book with 116 pages, what do the page numbers of the middle two pages add up to?
2. What is the largest number that cannot be written in the form $$14n+29m$$, where $$n$$ and $$m$$ are non-negative integers?
3. How many factors does the number $$2^6\times3^{12}\times5^2$$ have?
4. How many squares (of any size) are there in a $$15\times14$$ grid of squares?
5. You take a number and make a second number by removing the units digit. The sum of these two numbers is 1103. What was your first number?
6. What is the only three-digit number that is equal to a square number multiplied by the reverse of the same square number? (The reverse cannot start with 0.)
7. What is the largest three-digit number that is equal to a number multiplied by the reverse of the same number? (The reverse cannot start with 0.)
8. What is the mean of the answers to questions 6, 7 and 8?
9. How many numbers are there between 0 and 100,000 that do not contain the digits 0, 1, 2, 3, 4, 5, or 6?
10. What is the lowest common multiple of 52 and 1066?

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 Christmas card 2016 Christmas (2017) is over Christmas (2017) is coming! Christmas (2016) is over

### Comments

Comments in green were written by me. Comments in blue were not written by me.
2018-01-01
C - look up something called Frobenius numbers. This problem's equivalent to finding the Frobenius number for 14 and 29.
Lewis
2017-12-28
I can solve #2 with code, but is there a tidy maths way to solve it directly?
C
2017-12-23
My efforts were flightless.
NHH
2017-12-20
What a fun diversion! I have to admit I can't remember how to solve #2 - I finally had to write some code. (So primitive.)
Heather
2017-12-19
Thank you for the clarification!

I don't want to spoil the answer for anyone, but my Christmassy picture looks like a well dressed bird.
Claudio
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2017-11-28

## Christmas (2017) is coming!

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!

### Similar posts

 Christmas (2017) is over Christmas card 2017 Christmas (2016) is over Christmas card 2016

### Comments

Comments in green were written by me. Comments in blue were not written by me.
2017-12-27
Thanks, I've added a clarification to 22
Matthew
2017-12-24
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.
neal (@zbvif)
2017-12-24
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)
Neal (@zbvif)
2017-12-18
Assume the pancake is 2D
Matthew
2017-12-18
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

## Christmas (2016) is over

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.

### Similar posts

 Christmas (2017) is over Christmas card 2017 Christmas (2017) is coming! Christmas card 2016

### Comments

Comments in green were written by me. Comments in blue were not written by me.
2017-02-03
Thanks for the prizes. Fascinating books!
Steve Paget
2017-01-19
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!
SC
2016-12-28
Thanks, Matthew! The puzzles were really fun, and piecing the clues was very interesting too!
Jack
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2016-12-20

## Christmas card 2016

Last week, I posted about the Christmas card I designed on the Chalkdust blog.
The card looks boring at first glance, but contains 12 puzzles. Converting the answers to base 3, writing them in the boxes on the front, then colouring the 1s green and 2s red will reveal a Christmassy picture.
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 converted to base 3 and coloured...
 # Answer (base 10) Answer (base 3) 1 0 0 0 0 0 0 0 0 0 2 0 0 0 0 0 0 0 0 0 3 0 0 0 0 0 0 0 0 0 4 0 0 0 0 0 0 0 0 0 5 0 0 0 0 0 0 0 0 0 6 0 0 0 0 0 0 0 0 0 7 0 0 0 0 0 0 0 0 0 8 0 0 0 0 0 0 0 0 0 9 0 0 0 0 0 0 0 0 0 10 0 0 0 0 0 0 0 0 0 11 0 0 0 0 0 0 0 0 0 12 0 0 0 0 0 0 0 0 0
1. The square number larger than 1 whose square root is equal to the sum of its digits.
2. The smallest square number whose factors add up to a different square number.
3. The largest number that cannot be written in the form $$23n+17m$$, where $$n$$ and $$m$$ are positive integers (or 0).
4. Write down a three-digit number whose digits are decreasing. Write down the reverse of this number and find the difference. Add this difference to its reverse. What is the result?
5. The number of numbers between 0 and 10,000,000 that do not contain the digits 0, 1, 2, 3, 4, 5 or 6.
6. The lowest common multiple of 57 and 249.
7. The sum of all the odd numbers between 0 and 66.
8. One less than four times the 40th triangle number.
9. The number of factors of the number $$2^{756}$$×$$3^{12}$$.
10. In a book with 13,204 pages, what do the page numbers of the middle two pages add up to?
11. The number of off-diagonal elements in a 27×27 matrix.
12. The largest number, $$k$$, such that $$27k/(27+k)$$ is an integer.

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 Christmas card 2017 Christmas (2017) is over Christmas (2017) is coming! Christmas (2016) is over

### Comments

Comments in green were written by me. Comments in blue were not written by me.
2016-12-20
Thank you for the prompt response! It makes sense now and perhaps I should have read a little closer!
Dan Whitman
2016-12-20
Find the difference between the original number and the reverse of the original. Call this difference $$a$$. Next add $$a$$ to the reverse of $$a$$...
Matthew
2016-12-20
In number 4 what are we to take the difference between? Do you mean the difference between the original number and its reverse? If so when you add the difference back to the reverse you simply get the original number, which is ambiguous. I am not sure what you are asking us to do here.
Dan Whitman
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© Matthew Scroggs 2018