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2019-11-24
This year, the front page of mscroggs.co.uk will once again feature an Advent calendar, just like
last year, the year before, 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: while out on a test flight, Santa's sled was damaged and Santa, Rudolph and Blitzen fell to the ground over the Advent Isles.
You need to find Santa and his reindeer before Christmas is ruined for everyone.
You have gathered one inhabitant of the four largest Advent Isles—Rum, Land, Moon and County—and they are going to give you a series of clues about where Santa and his reindeer landed.
However, one or more of the islanders you have gathered may have been involved in damaging Santa's sled and causing it to crash: any islander involved in this will lie to you to attempt to stop
you from finding Santa and his reindeer.
Once you are ready to search for Santa, Rudolph and Blitzen, you can find the map by following this link.
Each of the clues will be about Santa's, Rudolph's or Blitzen's positions in Advent Standard Coordinates (ASC): ASC are given by six two-digit numbers with dots inbetween, for example
12.52.12.13.84.55.
For this example coordinate, the islanders will refer to
(the first) 12 as the first coordinate,
52 as the second coordinate,
(the second) 12 as the third coordinate,
13 as the fourth coordinate,
84 as the fifth coordinate, and
55 as the sixth coordinate.
To find a point's ASC coordinates, split a map of the islands into a 9×9 grid, then number the rows and columns 1 to 9: the first two digits of ASC give the vertical then horizontal position of a square in this grid.
The next two digits then give a smaller square when this square is then itself split into a 9×9 grid, and so on. An example is show below.
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 islanders tells you.
You must use these clues to find Santa and his two reindeer.
Ten randomly selected people who solve all the puzzles, find Santa and his reindeer, and fill in the entry form behind the door on the 25th will win prizes!
The winners will be randomly chosen from all those who submit the entry form before the end of 2019. 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 2019. 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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Thank you, I was worried. Thanks also for the puzzles, makes a change from my usual sudokus. I especially liked 3, 9, 11, 16, 21 and 23.
(anonymous)
@(anonymous): You don't get a confirmation email but if you hit submit you'll be entered. (I'll add confirmation emails before next year...)
Matthew
Do we get a confirmation email after submission of the entry form? I never received one, so I'm not sure if I am entered.
(anonymous)
@Seth: If you find them on the map then that's all you need to do. (And if you didn't find them yet then the entry form won't appear so you definably did everything you need to do.)
Matthew
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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.
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.
My next highlight is a geometry problem that appears to be about polygons, but actually it's about a secret circle.
12 December
There are 2600 different ways to pick three vertices of a regular 26-sided shape. Sometimes 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.
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).
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.
1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 | 16 | 17 | 18 | 19 | 20 | 21 | 22 | 23 | 24 |
Day |
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 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 Luey | Fred Metcalfe | Jo Ranger | Kip Urples | Meg 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:
1 | Sarah Brook |
2 | Mihai Zsisku |
3 | Bhavik Mehta |
4 | Peter Byrne |
5 | Martin Harris |
6 | Gert-Jan de Vries |
7 | Lyra |
8 | James O'Driscoll |
9 | Harry Poole |
10 | Albert 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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2018-11-25
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, Fred Metcalfe, Bob Luey, Kip Urples, 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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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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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 .
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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2017-11-28
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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Did you know Wally is called Waldo in the US, Walter in Germany, and Charlie in France?