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# Blog

2017-12-18
Just like last year, TD and 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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Comments in green were written by me. Comments in blue were not written by me.
2019-01-04
@Jose: There is a mistake in your answer: 243 (0100000) is the number of numbers between 10,000 and 100,000 that do not contain the digits 0, 1, 2, 3, 4, 5, or 6.

Matthew
2019-01-04
Thanks for the puzzle!
Is it possible that the question 9 is no correct?
I get a penguin with perfect simetrie except at answer 9 : 0100000 that breaks the simetry.
Is it correct or a mistake in my answer?
Thx

Jose
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

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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:
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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Comments in green were written by me. Comments in blue were not written by me.
2017-12-27
@Neal (@zbvif): 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
@Alex: 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
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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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
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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2016-12-20
@Matthew: Thank you for the prompt response! It makes sense now and perhaps I should have read a little closer!

Dan Whitman
2016-12-20
@Dan Whitman: 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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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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Comments in green were written by me. Comments in blue were not written by me.
2016-12-27
@Another Matthew: Ten correct submissions have been made. Just updating the pages to reflect this...

Matthew
2016-12-27
Have 10 correct submissions not been made yet?

Another Matthew
2016-12-25
Thank you, @Matthew: !

Lyra
2016-12-25
Really enjoyed the extra bit at the end this year! Looking forward to 2017's calendar.

Louis
2016-12-25
@Lyra: I'll email you if you are one of the winners to get the rest of your address!

Matthew