# Blog

## Archive

Show me a random blog post**2018**

### Jun 2018

World Cup stickers 2018, pt. 2### May 2018

A bad Puzzle for Today### Apr 2018

Building MENACEs for other games### Mar 2018

A 20,000-to-1 baby?World Cup stickers 2018

### Jan 2018

*Origins of World War I*

Christmas (2017) is over

**2017**

**2016**

**2015**

**2014**

**2013**

**2012**

## Tags

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## A bad Puzzle for Today

Every morning just before 7am, one of the Today Programme's presenters reads out a puzzle. Yesterday, it was this puzzle:

In a given month, the probability of a certain daily paper either running a story about inappropriate behaviour at a party conference or running one about somebody's pet being retrieved from a domestic appliance is exactly half the probability of the same paper containing a photo of a Tory MP jogging. The probability of no such photo appearing is the same as that of there being a story about inappropriate behaviour at a party conference. The probability of the paper running a story about somebody's pet being retrieved from a domestic appliance is a quarter that of its containing a photo of a Tory MP jogging. What are the probabilities the paper will (a) run the conference story, (b) run the pet story, (c) contain the jogging photo?

I'm not the only one to notice that some of Radio 4's daily puzzles are not great.
I think this puzzle is a great example of a terrible puzzle. You can already see the first problem with it: it's long and words and very hard to follow on the radio.
But maybe this isn't so important, as you can
read it here after it's been read out.

Once you've done this, you can re-write the puzzle as follows:
there are three news stories (\(A\), \(B\) and \(C\)) that the newspaper might publish in a month. We are given the following information:

$$\mathbb{P}(A\text{ or }B)=\tfrac12\mathbb{P}(C)$$
$$1-\mathbb{P}(C)=\mathbb{P}(A)$$
$$\mathbb{P}(B)=\tfrac14\mathbb{P}(C)$$
To solve this puzzle, we need use the formula \(\mathbb{P}(A\text{ or }B)=\mathbb{P}(A)+\mathbb{P}(B)-\mathbb{P}(A\text{ and }B)\).
These Venn diagrams justify this formula:

Using the information we were given in the question, we get:

\begin{align}
\tfrac12\mathbb{P}(C)&=\mathbb{P}(A\text{ or }B)\\
&=1-\mathbb{P}(C)+\tfrac14\mathbb{P}(C)-\mathbb{P}(A\text{ and }B)\\
\mathbb{P}(C)&=\tfrac45(1-\mathbb{P}(A\text{ and }B)).
\end{align}
At this point we have reached the second problem with this puzzle: there's no answer unless we make some extra assumptions, and the question doesn't make it clear what we can assume.
But let's give the puzzle the benefit of the doubt and try some assumptions.

### Assumption 1: The events are mutually exclusive

If we assume that the events \(A\) and \(B\) are mutually exclusive—or, in other words, only one of these two articles can be published,
perhaps due to a lack of space—then we can use the fact that

$$\mathbb{P}(A\text{ and }B)=0.$$
This means that
\(\mathbb{P}(C)=\tfrac45\),
\(\mathbb{P}(A)=\tfrac15\), and
\(\mathbb{P}(B)=\tfrac15\). There's a problem with this answer, though: the three probabilities add up to more than 1.

This wouldn't be a problem, except we assumed that only one of the articles \(A\) and \(B\) could be published.
The probabilities adding up to more than 1 means that either \(A\) and \(C\) are not mutually exclusive or \(A\) and \(B\) are not mutually exclusive,
so \(C\) could be published alongside \(A\) or \(B\). There seems to be nothing special about the three news stories to mean that only some combinations
could be published together, so at this point I figured that this assumption was wrong and moved on.

Today, however, the answer was posted, and
this answer was given (without an working out). So we have a third problem with this puzzle: the answer that was given is wrong, or at the very best
is based on questionable assumptions.

### Assumption 2: The events are independent

If we assume that the events are independent—so one article being published doesn't affect whether or not another is published—then
we may use the fact that

$$\mathbb{P}(A\text{ and }B)=\mathbb{P}(A)\mathbb{P}(B).$$
If we let \(c=\mathbb{P}(C)\), then we get:

\begin{align}
\tfrac12c&=\mathbb{P}(A)+\mathbb{P}(B)-\mathbb{P}(A\text{ and }B)\\
&=\mathbb{P}(A)+\mathbb{P}(B)-\mathbb{P}(A)\mathbb{P}(B)\\
&=1-c+\tfrac14c-\tfrac14(1-c)c\\
\tfrac14c^2-\tfrac32c+1&=0.
\end{align}
You can use your favourite formula to solve this to find that \(c=3-\sqrt5\), and therefore
\(\mathbb{P}(A)=\sqrt5-2\) and
\(\mathbb{P}(B)=\tfrac34-\tfrac{\sqrt5}4\).

In this case, our assumption appears to be more reasonable—as over the course of a month the stories published by a paper probably don't have
much of an effect on each other—but we have the fourth, and probably biggest problem with the puzzle: the question and answer are not interesting or surprising, and
the method is a bit tedious.

### Similar posts

World Cup stickers 2018, pt. 2 | A 20,000-to-1 baby? | World Cup stickers 2018 | How much will I win on the new National Lottery? |

### Comments

Comments in green were written by me. Comments in blue were not written by me.

**2018-05-03**

Stefan

**Add a Comment**

**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...

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

White shirt says: "Yesterday's elf lied: Carol wants

**4**,**9**or**6**."10

Orange hat says: "

Orange hat says: "

**249**is my favourite number."5

Red shoes says: "Alex lives at

Red shoes says: "Alex lives at

**1**,**9**or**6**."16

Blue shoes says: "I'm the same elf as yesterday. Ben wants

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

Red shoes says: "Carol wants a factor of

**120**. I am yesterday's elf."4

Blue shoes says: "

Blue shoes says: "

**495**is my favourite number."15

Blue shoes says: "Carol lives at

Blue shoes says: "Carol lives at

**9**,**6**or**8**."22

Purple trousers says: "Carol wants a factor of

Purple trousers says: "Carol wants a factor of

**294**."11

White shirt says: "

White shirt says: "

**497**is my favourite number."6

Pink shirt says: "Ben does not live at the last digit of

Pink shirt says: "Ben does not live at the last digit of

**106**."9

Blue shoes says: "Ben lives at

Blue shoes says: "Ben lives at

**5**,**1**or**2**."20

Orange hat says: "Carol wants the first digit of

Orange hat says: "Carol wants the first digit of

**233**."1

Red shoes says: "Alex wants

Red shoes says: "Alex wants

**1**,**2**or**3**."24

Green hat says: "The product of the six final presents and homes is

Green hat says: "The product of the six final presents and homes is

**960**."17

Grey trousers says: "Alex wants the first digit of

Grey trousers says: "Alex wants the first digit of

**194**."14

Pink shirt says: "One child lives at the first digit of

Pink shirt says: "One child lives at the first digit of

**819**."3

White shirt says: "Alex lives at

White shirt says: "Alex lives at

**2**,**1**or**6**."18

Green hat says: "Ben wants

Green hat says: "Ben wants

**1**,**5**or**4**."7

Green hat says: "Ben lives at

Green hat says: "Ben lives at

**3**,**4**or**3**."12

Grey trousers says: "Alex lives at

Grey trousers says: "Alex lives at

**3**,**1**or**5**."19

Purple trousers says: "Carol lives at

Purple trousers says: "Carol lives at

**2**,**6**or**8**."8

Red shoes says: "The digits of

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

Green hat says: "One child lives at the first digit of

**755**."2

Red shoes says: "Alex wants

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**

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

**Add a Comment**

**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 |

- In a book with 116 pages, what do the page numbers of the middle two pages add up to?
- What is the largest number that cannot be written in the form \(14n+29m\), where \(n\) and \(m\) are non-negative integers?
- How many factors does the number \(2^6\times3^{12}\times5^2\) have?
- How many squares (of any size) are there in a \(15\times14\) grid of squares?
- 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?
- 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.)
- 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.)
- What is the mean of the answers to questions 6, 7 and 8?
- How many numbers are there between 0 and 100,000 that do not contain the digits 0, 1, 2, 3, 4, 5, or 6?
- What is the lowest common multiple of 52 and 1066?

### Similar posts

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**

Lewis

**2017-12-28**

C

**2017-12-23**

NHH

**2017-12-20**

Heather

**2017-12-19**

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

Claudio

**Add a Comment**

**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**

Matthew

**2017-12-24**

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**

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**

Matthew

**2017-12-18**

Alex

**Add a Comment**

**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**

Steve Paget

**2017-01-19**

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**

Jack

**Add a Comment**

2018-05-03