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24 December

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Today is the final day of the Chalkdust advent calendar.

Click here to open today’s door.

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23 December

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Welcome to the twenty third day of the 2016 Chalkdust Advent Calendar. Today, we have another puzzle for you to enjoy, plus the answer to the puzzles from 12 December.

Here’s today’s puzzle.

Finding Santa

It’s Christmas Eve Eve and Santa is so nervous that he decides to hide in the top floor of a nearby hotel. Santa starts in a random room and every hour moves to one of the rooms next door to his current room. There are 17 rooms on the top floor of the hotel.

The hotel manager is a very busy lady and only has time to allow you to look into one room every hour. It’s 2pm on Christmas Eve Eve, so you have 34 hours to find Santa or Christmas is ruined. Can you find him in time? 

Now for the solution to the puzzle from 12 December.

Odd factors

Pick a number. Call it $n$. Write down all the numbers from $n+1$ to $2n$ (inclusive). Under each of these, write its largest odd factor. What is the sum of these odd factors? 

Incredibly, the result will always be $n^2$.

To see why, imagine writing every number, $n+1\leq k\leq 2n$, in the form $$k=2^ab$$ where $b$ is an odd number and also the $k$’s largest odd factor. The next largest number whose largest odd factor is $b$ will be $2^{a+1}b=2k$. But this will be larger than $2n$, so outside the range. Therefore each number in the range has a different largest odd factor.

Each of the largest odd factors must be one of $1, 3, 5, …, 2n-1$, as they cannot be larger than $2n$. But there are $n$ odd numbers here and $n$ numbers in the range, so each number $1, 3, 5, …, 2n-1$ is the highest odd factor of one of the numbers (as the highest odd factors are all different).

Therefore, the sum of the odd factors is the sum of the first $n$ odd numbers, which is $n^2$.

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The Indisputable Existence of Santa Claus – A Review

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Welcome to the 22nd day of the 2016 Chalkdust Advent Calendar. Today we have for you, a book review!

Christmas, a time of celebration, joy and meeting family you never knew you had. Regardless of how joyous Christmas can be it also is undoubtedly stressful for some. What present could I get person $x$? Why does my Christmas tree look so ugly even though I spent 2 weeks decorating it? The simple answer is because you probably drenched it in tinsel and epileptic seizure inducing lights. The mathematical answer?

The Indisputable Existence of Santa Claus” is a recent release by Dr Hannah Fry and Dr Thomas Oléron Evans, Dr Hannah Fry who has previously written “The Mathematics of Love”. This book arranges itself as a step-by-step guide on how to prepare for the (mathematically) perfect Christmas, covering every detail from how to wrap presents according to their surface area to volume ratio, to using Markov chains as a means of perfecting the Queen’s speech. 

We shall start with the big question of whether Father Christmas himself actually exists. The first chapter gives a ‘seemingly’ valid proof of Santa’s existence, from another ‘seemingly’ valid proof of how 1+1=0. You must be thinking, what? Obviously there must be a flaw in the progression of the proof somewhere, which there is, but you can discover it for yourself by reading the book. 

Moving on, let’s have a look at the inconspicuous game of Secret Santa. For those unfamiliar with the concept, the classic approach to this game is simply, you write your name on a small piece of paper, fold it up and throw it into a hat. After a bit of shaking, someone picks out a piece and takes on the enormous responsibility of finding their victim, *coughs* I mean colleague, a present usually around the price range of £5. So why is this a rather inadequate way of organising this game? 

Well first, you risk the chance of picking out your own name. You might think that’s easy enough a problem to solve, just put it back in the hat…But what if you were the last person? Or even the second last person? Everybody knows your name is back in the hat, and they have a greater chance of picking your name, and in some case, 2 people will have each other’s name, Secret Santa is ruined. Goodnight. The book proposes another method of making sure that no one has their own name, and no one else knows who has their name. A clever yet simple solution involving derangements, alas, Christmas is saved, now lets hope your secret Santa isn’t a Scrooge

My favourite chapter in this book has to be the one on the Queen’s Speech. The beginning of the chapter is an analysis of the Queen’s vocabulary score (based solely on the number of unique words in the first 35K words of her speech. Surprisingly, poor old Lizzy scored lower than her counterparts Jay-Z and Shakespeare. Well, it seems as though maths might be able to give her a hand with that, with something special called a Markov chain, that determines the next word to place in a sentence given the word before. I won’t go into too much detail as to give it away; it just so happens that earlier this week I was reading about how Markov chains are used in determining the probability of flipping a coin and getting a certain outcome which is equivalent to another outcome, so this chapter peaked my interest even further. 

What is particularly good about this book is how accessible it is; at the end of every chapter there are endnotes that explain some of the maths mentioned and includes some extra reading material. The only questionable thing in this book might be the chapter about cooking turkey. They used a chicken instead. Enough said.

Overall, a great read, not too technical but with just enough maths to get you thinking. So even if your grumpy aunt Hilda despises anything to do with maths, there’s now a very slight chance she might enjoy it. 

Also the amount of cracker pulling rules I’ve never heard of is remarkable. I am now going to stop pulling both ends of my own cracker, despite my competitive nature.

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21 December

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Welcome to the twenty-first day of the 2016 Chalkdust Advent Calendar. Today, we bring you our final selection of fascinatingT&Cs apply facts, randomly generated by Santa’s elves.  Remember to send us your favourite scientific curiosities via Facebook, Twitter or email and we’ll feature the best in a blog next year.

The number 21

Setter of fiendish conjectures.

Today is the 21st day of the Chalkdust Advent calendar. The numbers 8 and 9 are the only powers of integers ($2^3$ and $3^2$ respectively) that are consecutive. This was conjectured by the Belgian mathematician Eugène Charles Catalan in 1844. It was proved in 2002 by Preda Mihailescu. Unlike Andrew Wiles, who proved Fermat’s last theorem, Mihailescu didn’t shoot to fame. This had absolutely nothing to do with the number 21.

Rudolph the red-nosed she-reindeer

Female. Female. Not male.

Rudolph the red-nosed reindeer is very confused as to why she has a male name. Male reindeers shed their antlers once they’ve finished using them as swords during the mating season in autumn, while females cast them in spring and regrow them in time for Christmas (when they use them as swords to fight other females over holes in the snow). I’m sure she is confused for many other reasons too. Such as why she has a red nose.

Father Christmas and the multiverse

If the multiverse theory is correct, and our universe is just one of an infinite number of universes, there may be one in which Father Christmas was not popularised by Coke adverts of the 1920s.

And that’s that from Santa’s elves. They’re off to find Rudelle.

[Pictures. Eugène Charles Catlan: Public domain; Rudelle: Tristan Ferne, CC BY-2.0]

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19 December

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Find your perfect partner with this wonderful tree diagram!


Attributions:

[Cauchy  – By Public domain – Library of Congress Prints and Photographs Division. From an illustration in: Das neunzehnte Jahrhundert in Bildnissen / Karl Werckmeister, ed. Berlin : Kunstverlag der photographische gesellschaft, 1901, vol. V, no. 581., Public Domain ; Knot – adapted from Flickr.com – knotted by Shelby Steward, CC-BY 2.0; Emmy Noether – By Unknown – Emmy Noether (1882-1935), Public Domain ; Python – adapted from Flickr.com – Python by Jonathan Kriz, CC-BY 2.0; Daniel Bernoulli – By Johann Jakob Haid – Here, Public Domain ; Scrooge – adapted from Flickr.com – Money by Tax Credits, CC-BY 2.0]

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16 December

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Behind today’s door… A quiz!

 


Attributions:

[Pictures: 1 – adapted from Flickr.com – Pi by Tom Magliery, CC-BY 2.0; 2 – adapted from Flickr.com – IMG_8118 by SAITOR, CC-BY 2.0; 3 – adapted from Flickr.com – Chichen Itza Pyramid by Shayne Bowman, CC-BY 2.0; 4 – adapted from Flickr.com – Chichen Itza Pyramid by Shayne Bowman, CC-BY 2.0; 5 – adapted from Flickr.com – Pac Man Board Game by Ian Crowfeather, CC-BY 2.0; 6 – adapted from Flickr.com – Keys grid with guide grid by Greg Borenstein, CC-BY 2.0; 7 – adapted from Flickr.com – Ellipses by Eric Wagner, CC-BY 2.0; 8 – adapted from Flickr.com – Equals by Mrs TeePot CC-BY 2.0; 9 – adapted from Flickr.com – Equals by Mrs TeePot CC-BY 2.0; other pictures by Chalkdust]

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The Chalkdust Christmas card

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Recently, some of you may have received a Chalkdust Christmas card. If not, it’s not because we hate you, it’s just that we couldn’t find your address… Unless we hate you, in which case it is because we hate you.

The card initially looks very boring: it is just a grid of squares with “Merry Christmas” written below it. Definitely NOT HOT… But there’s more. There’s a puzzle inside that leads you to add some colour to the squares to reveal a Christmassy picture.

Without giving any more away, here is the puzzle. If you’d like to give it to someone as a Christmas card (or just want to actually be able to colour it in), you can print and fold this lovely pdf.

Christmas Card 2016

The grid (click to enlarge)

Instructions

  1. Solve the puzzles below.
  2. Convert the answers to base 3.
  3. Write the answers in the boxes on the front cover.
  4. Colour squares containing a 1 green. Colour squares containing a 2 red. Leave squares containing a 0 unshaded.

Puzzles

  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 2756×312.
  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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