Dork Scratchings

Rolling marbles  Here's a puzzle: Suppose you have N identical marbles rolling along a one dimensional table-top.  Each marble is randomly rolling to the left or to the right, all with the same speed.  Collisions are elastic, which means the marbles just change direction.  What is the maximum amount of time before all the marbles have rolled off the table? 8 marbles with speed 1 on a table of length 1 Answering this question is really difficult if you simply pick an individual marble and try to work out how long it might stay on the table as it bounces back and forth.  But there's a simpler way to look at it. Prior to each collision you have one marble rolling to the left and one to the right, and afterwards you still have one rolling to the left and one to the right.  If we swap labels following each collision then the labels never change direction.  Now it's easy to see that the answer is the same whether there are 100 marbles or just one.  And ...

 When I was a kid I pulled a Christmas Cracker and 6 cards with numbers fell out: It's a mind reading magic trick.  Have a go against my virtual assistant, and see if you can work out how it's done:

About 18 years ago a friend patiently explained to me how to solve a Rubik's Cube.  I memorized the instructions, but realized sooner or later I'd forget them.  So maybe a week later I wrote up my notes using dia , and printed them out .  Since then I've carried this same slip of paper around in my wallet, for those occasional opportunities when you're 'round someone's house and you spot a cube exhibiting a frustratingly high degree of entropy. It's still just about legible Each face of the cube is given a letter U - up D - down F - front B - back L - left R - right A single letter on it's own represents rotating the face 90$^\circ$ clockwise (looking at the face), and a letter followed by an apostrophe means rotate anti-clockwise.  Thus LL' is the same as doing nothing.  An exponent of 2 simply means do the preceeding action twice.  Below, I've split the sequences into subblocks with dashes to make them easier to memorize$^\dagger$. HOW TO SOLV...

Can you cover the 34 squares with 17 2x1 dominoes? Here's a puzzle I was shown by my lecturer Professor Schofield back in Bristol, last century: Make a 6x6 grid and remove two diagonally opposite squares so that you're left with 34 squares.  Given 17 two by one dominoes, can you cover the remaining area?  The illustration above shows one failed attempt. S C R O L L D O W N F O R T H E S O L U T I O N Sorry peeps.  It's an impossipuzzle.  To see why apply a checker pattern to the squares: 18 black squares but only 16 white squares The two removed squares were both white.  So there are fewer white squares than black ones.  But each domino placed covers exactly one white and one black square.  If you could cover it perfectly with 17 dominoes there would be the same number of black and white squares.  There isn't, so you can't!

In this post I will attempt to show that sitting down and having a nice cup of tea is the best approach to solving any new mathematical problem.  So, here's a problem: A plane has 100 seats and 100 passengers each with an allocated seat number.  The first passenger to get on the plane is blind and chooses a seat at random.  Each subsequent passenger to board chooses their own seat if still available, or a seat at random if not.  What is the probability that the 100th passenger gets their allocated seat? This is more subtle than it seems at first.  The last passenger could get their own seat because the blind passenger chooses the correct seat, or because the 2nd passenger chooses the blind person's seat, or because the first 87 passengers occupy the first 87 seats.  In fact there's a huge number of ways in which it could happen. Knuckleheaded Compsci solution Suppose we've forgotten to have a cup of tea.  Then we might just dive in an...

I found this during a terribly terribly boring day at work (before I discovered the light and vowed never to work in an open plan office again). https://plus.maths.org/content/magic-19 A lovely little problem.   Can you assign the numbers 1 to 19 to the nodes in the diagram such that the three numbers in every line segment always add up to 22?  I figured it out after about a day (mostly in the office) using logic alone.  But what I couldn't understand is how the problem setter knew a solution would exist?  I posed the question on the site but never got an answer.