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 2016-12-28 06:49:35 

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:
1Jack Jiang
2Steve Paget
3Joe Gage
4Tony Mann
5Stephen Cappella
6Cheng Wai Koo
7Demi Xin
9David Fox
10Bob 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 suspectMotive
1Dr. Uno (uno = Spanish 1)Obeying nameless entity
2Mr. Zwei (zwei = German 2)To worry others
3Ms. Trois (trois = French 3)To help really evil elephant
4Mrs. Quattro (quattro = Italian 4)For old unknown reasons
5Prof. Pum (pum = Welsh 5)For individual violent end
6Miss. Zes (zes = Dutch 6)Stopping idiotic xenophobia
7Lord Seacht (seacht = Irish 7)Suspect espied victim eating newlyweds
8Lady Oito (oito = Portuguese 8)Epic insanity got him today
9Rev. Novem (novem = Latin 9)Nobody in newsroom expected

1Throne roomWrench (1 vowel)
2Network roomRope (2 vowels)
3Beneath reedsRevolver (3 vowels)
4Edge of our gardenLead pipe (4 vowels)
5Fives courtNeighbour's sword (5 vowels)
6On the sixth floorSuper banana bomb (6 vowels)
7Sparse venueAntique candlestick (7 vowels)
8Weightlifting roomA foul tasting poison (8 vowels)
9Mathematics mezzanineRun 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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Thanks for the prizes. Fascinating books!
Steve Paget
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!
Thanks, Matthew! The puzzles were really fun, and piecing the clues was very interesting too!
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 2016-12-20 09:21:56 

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. 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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Thank you for the prompt response! It makes sense now and perhaps I should have read a little closer!
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 \(a\)...
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 15:57:54 

Christmas (2016) is Coming!

This year, the front page of 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 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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Ten correct submissions have been made. Just updating the pages to reflect this...
Have 10 correct submissions not been made yet?
Another Matthew
Thank you, Matthew!
Really enjoyed the extra bit at the end this year! Looking forward to 2017's calendar.
I'll email you if you are one of the winners to get the rest of your address!
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 2016-07-04 18:35:45 

Solving the Cross Diagonal Cover Problem

In five blog posts (1, 2, 3, 4, 5) Gaurish Korpal present the Cross diagonal cover problem, some ideas about how to solve it and some conjectures. In this post, I will present my solution to this problem. But first, the problem itself:
Draw with an \(m\times n\) rectangle, split into unit squares. Starting in the top left corner, move at 45° across the rectangle. When you reach the side, bounce off. Continue until you reach another corner of the rectangle:
How many squares will be coloured in when the process ends?

Restating the Problem

When I first saw this problem, it reminded me of Rebounds, a puzzle I posted here in 2014. To restate the problem in a similar way, we place a point in the centre of each unit square, then creating a second grid. I will call this the dual grid. The original problem is equivalent to asking, if a line bounces around the dual grid, how many corners will it pass through.
It it worth noting here that the dual grid is \(n-1\times m-1\): each side is one shorter than the original grid.
A corner cannot be travelled over more than twice: otherwise, the line would be retracing its past path; to do this requires it to have already hit a corner of the rectangle. Therefore we can calculate the number of distinct corners travelled to using:
$$\text{Distinct corners visited} = \text{Corners visited}-\text{Corners visited twice}$$

Introducing Mirrors

When I solved Rebounds, I imagined the line passing through mirror images of the rectange, rather then bouncing. For our example above, it would look like this:
Looking at the puzzle in this way, it can be seen that the line will travel through \(\mathrm{lcm}(n-1,m-1)\) squares, and so hit \(\mathrm{lcm}(n-1,m-1)+1\) corners (the \(+1\) appears due to fence panels and fence posts). We have shown that:
$$\text{Corners visited}=\mathrm{lcm}(n-1,m-1)+1$$

Collisions in the Mirror

To solve the problem, we need to work out how many corners are visited twice; or the number of times the line crosses itself in the rectangle.
To do this, imagine the mirror images of the red line in the mirrors. Ignoring the images parallel to the red line, and terminating the lines when they hit the red line gives the following diagram:
I have added extra rectangles to the diagram so that all the reflections that hit the red line and their starting points can be seen. The diagonal black line has been added because all lines outside that clearly cannot intersect the red line. We now need to justify two claims:
The first claim can be seen by reflecting the green lines and the parts of the red line they hit back into the top left rectangle.
The second claim can be shown in two parts:
First, each line starting from the border will meet the red line on the edge of a rectangle: this is because the green lines all start a multiple of two rectangles away, and meet at half this distance away (and half a multiple of two is a whole number).
Conversely, if a red line meets a green line at the edge of a rectangle, then the reflection of the red line in the edge (ie. the green line) must go back to a starting point on the boundary.

These two claims show that the points where the line crosses itself in the dual rectangle match up one-to-one with the interior points at which the green lines start. So if we can count these points, we can solve the problem.

Counting the Interior Points

The green lines will start from all points that are a multiple of two rectangles (in both directions) away from the top left. The reason for this can be seen by reflecting the first little bit of the red line in all the mirrors:
We see that the perpendicular green lines that we are interested in, plus many other irrelevant lines, start from a grid of points on the corner of every other rectangle. To count these points, we will extend them into the full square:
To make for clearer counting, I have not drawn the unit squares.
Alternatively, this square can be thought of as being made up from two copies of the triangle.
We next notice that these points can never lie on the diagonal (the diagonal drawn in the diagram): a green point lying on the diagonal would imply that the red line met a corner before it did. Taking out the lines on which the green dots never lie, we get:
We can now count the green dots. In the square above, there are \(\displaystyle\frac{\mathrm{lcm}(m-1,n-1)}{m-1}\) rectangles vertically and \(\displaystyle\frac{\mathrm{lcm}(m-1,n-1)}{n-1}\) rectangles horizontally. Therefore there are \(\displaystyle\frac{\mathrm{lcm}(m-1,n-1)}{m-1}-1\) columns of green dots and \(\displaystyle\frac{\mathrm{lcm}(m-1,n-1)}{n-1}-1\) rows of green dots. (Again, we take one due to fence posts and fence panels.)
Of these green dots, half are in the triangle of interest, so:
$$\text{Corners visited twice} = \frac12\left(\frac{\mathrm{lcm}(m-1,n-1)}{m-1}-1\right)\left(\frac{\mathrm{lcm}(m-1,n-1)}{n-1}-1\right)$$

Putting it together

We can now put the two parts together to get:
$$\text{Distinct corners visited} = \mathrm{lcm}(m-1,n-1)+1 - \frac12\left(\frac{\mathrm{lcm}(m-1,n-1)}{m-1}-1\right)\left(\frac{\mathrm{lcm}(m-1,n-1)}{n-1}-1\right)$$
And we have solved the problem.


For the \(4\times6\) rectangle given, our formula gives:
$$\text{Distinct corners visited} = \mathrm{lcm}(3,5)+1 - \frac12\left(\frac{\mathrm{lcm}(3,5)}{3}-1\right)\left(\frac{\mathrm{lcm}(3,5)}{5}-1\right)$$ $$= 15+1 - \frac12(5-1)(3-1)$$ $$= 16 - 4 = 12$$
This is correct:

Disproving the Conjecture

In Gaurish's most recent post, he gave the following conjecture: The highest common factor (or greatest common divisor) of \(m\) and \(n\) always divides the number of coloured squares.
After trying to prove this for a while, I found that me attempted proof required that \(\mathrm{hcf}(n-1,m-1)^2-1\) is a multiple of \(\mathrm{hcf}(n,m)\). However this is not in general true (eg. 15,5).
In fact, 15,5 provides us with a counterexample to the conjecture:
In this diagram, 26 squares are coloured. However \(\mathrm{hcf}(15,5)=5\) and 5 is not a factor of 26.
Tags: puzzles

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 2016-03-15 06:36:52 

The Mathematical Games of Martin Gardner

This article first appeared in issue 03 of Chalkdust. I highly recommend reading the rest of the magazine (and trying to solve the crossnumber I wrote for the issue).
It all began in December 1956, when an article about hexaflexagons was published in Scientific American. A hexaflexagon is a hexagonal paper toy which can be folded and then opened out to reveal hidden faces. If you have never made a hexaflexagon, then you should stop reading and make one right now. Once you've done so, you will understand why the article led to a craze in New York; you will probably even create your own mini-craze because you will just need to show it to everyone you know.
The author of the article was, of course, Martin Gardner.
A Christmas flexagon.
Martin Gardner was born in 1914 and grew up in Tulsa, Oklahoma. He earned a bachelor's degree in philosophy from the University of Chicago and after four years serving in the US Navy during the Second World War, he returned to Chicago and began writing. After a few years working on children's magazines and the occasional article for adults, Gardner was introduced to John Tukey, one of the students who had been involved in the creation of hexaflexagons.
Soon after the impact of the hexaflexagons article became clear, Gardner was asked if he had enough material to maintain a monthly column. This column, Mathematical Games, was written by Gardner every month from January 1956 for 26 years until December 1981. Throughout its run, the column introduced the world to a great number of mathematical ideas, including Penrose tiling, the Game of Life, public key encryption, the art of MC Escher, polyominoes and a matchbox machine learning robot called MENACE.


Gardner regularly received topics for the column directly from their inventors. His collaborators included Roger Penrose, Raymond Smullyan, Douglas Hofstadter, John Conway and many, many others. His closeness to researchers allowed him to write about ideas that the general public were previously unaware of and share newly researched ideas with the world.
In 1970, for example, John Conway invented the Game of Life, often simply referred to as Life. A few weeks later, Conway showed the game to Gardner, allowing him to write the first ever article about the now-popular game.
In Life, cells on a square lattice are either alive (black) or dead (white). The status of the cells in the next generation of the game is given by the following three rules:
For example, here is a starting configuration and its next two generations:
The first three generations of a game of Life.
The collection of blocks on the right of this game is called a glider, as it will glide to the right and upwards as the generations advance. If we start Life with a single glider, then the glider will glide across the board forever, always covering five squares: this starting position will not lead to the sad ending where everything is dead. It is not obvious, however, whether there is a starting configuration that will lead the number of occupied squares to increase without bound.
Gosper's glider gun.
Originally, Conway and Gardner thought that this was impossible, but after the article was published, a reader and mathematician called Bill Gosper discovered the glider gun: a starting arrangement in Life that fires a glider every 30 generations. As each of these gliders will go on to live forever, this starting configuration results in the number of live cells perpetually increasing!
This discovery allowed Conway to prove that any Turing machine can be built within Life: starting arrangements exist that can calculate the digits of pi, solve equations, or do any other calculation a computer is capable of (although very slowly)!

Encrypting with RSA

To encode the message \(809$, we will use the public key:
$$s=19\quad\text{and}\quad r=1769$$
The encoded message is the remainder when the message to the power of \(s\) is divided by \(r$:

Decrypting with RSA

To decode the message, we need the two prime factors of \(r\) (\(29\) and \(61\)). We multiply one less than each of these together:
\begin{align*} a&=(29-1)\times(61-1)\\[-2pt] &=1680. \end{align*}
We now need to find a number \(t\) such that \(st\equiv1\mod a\). Or in other words:
$$19t\equiv1\mod 1680$$
One solution of this equation is \(t=619\) (calculated via the extended Euclidean algorithm).
Then we calculate the remainder when the encoded message to the power of \(t\) is divided by \(r\):


Another concept that made it into Mathematical Games shortly after its discovery was public key cryptography. In mid-1977, mathematicians Ron Rivest, Adi Shamir and Leonard Adleman invented the method of encryption now known as RSA (the initials of their surnames). Here, messages are encoded using two publicly shared numbers, or keys. These numbers and the method used to encrypt messages can be publicly shared as knowing this information does not reveal how to decrypt the message. Rather, decryption of the message requires knowing the prime factors of one of the keys. If this key is the product of two very large prime numbers, then this is a very difficult task.

Something to think about

Gardner had no education in maths beyond high school, and at times had difficulty understanding the material he was writing about. He believed, however, that this was a strength and not a weakness: his struggle to understand led him to write in a way that other non-mathematicians could follow. This goes a long way to explaining the popularity of his column.
After Gardner finished working on the column, it was continued by Douglas Hofstadter and then AK Dewney before being passed down to Ian Stewart.
Gardner died in May 2010, leaving behind hundreds of books and articles. There could be no better way to end than with something for you to go away and think about. These of course all come from Martin Gardner's Mathematical Games:

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