Beth Malmskog

Math etc.

Month: March, 2012

What is the most important single-elimination tournament of March? The Rubik’s Cube-Off, of course.


Surely you’ve heard of the Rubik’s Cube-Off, the annual speed solving cube competition?  You know, the one that takes place on Pi Day, i.e. March 14, and that you qualify for by reciting as many digits of pi as you can? No?  Well, that’s because my frequent math mostly co-host, Jed Rendleman, hasn’t reported on it yet.  Or it could be because it doesn’t exist.  Jed is one of my many excellent students this semester at Wesleyan.  He’s been working with me on the radio versions of the last few puzzles and is really great at asking questions and being funny and all the other stuff that makes for a wildly successful math radio show.  Anyway, this weeks puzzle story is about Jed’s spring break and how I helped him cheat at the cube-off.  Which of course, just to be clear in case Jed is applying for grad school (or someone is background checking me!) didn’t actually happen.  No wire was worn into a no-communication-allowed tournament.  No windowless van was employed.  No digits of pi were recited.  But it does make for some kind of puzzle. 

So, the story is that Jed, being a journalist, wanted to go behind the scenes to report on the Cube-Off for the Wesleyan Argus.  He decided to enter as a contestant to get a real taste of the action at this, the bloodiest and most contentious Rubik’s Cube competition in the world.  The qualifying round, which consisted of reciting at least 120 digits of pi in under 1 minute, was not problem for Jed.  Being a bit of a savant, he had no problem memorizing the digits in a very short time, and handily qualified for the tournament.  He took his place among 10,040 other pi stars hoping to be the one winner of the Cube-Off.  So how does this competition work, you ask?  Well, first of all, beyond the qualifying round, it has nothing to do with pi.  Except they serve pie at the winner’s banquet, in celebration of the hallowed date.  The tournament itself is just your basic single elimination tournament, in which contestants go head-to-head in many rounds of cube-solving.  In each match, two contestants are given scrambled cubes. They compete to solve the cube as quickly as possible.   The one who solves it faster stays in the running, while the slower solver is out of the tournament.  


Early idea for choosing U.S. President--head-to-head Rubik's Cube competition.

Because of the nature of the qualifying round, every year a different number of people qualify to compete in the tournament.  That means the organizers can’t plan out the structure in advance.  They don’t even know how many matches to schedule.  So they have to figure it all out very quickly after the qualifying round, so that the Cube-Off can begin.  Alright, so the first question is, how many matches are there in the tournament?  The tournament judges decide to offer a cash prize to the first contestant who can tell them.  

Here’s where the ethics violation comes in.  You see, Jed was afraid that he would blow his cover at the Cube-Off because it turns out that he is not really nerdy enough to be at one of these things.  He doesn’t know any Star Trek jokes, has no idea how many sides the D&D dice have, can’t extensively discuss the merits of the old TI-85 versus the new TI-82, let alone take a side in the TI versus HP debate.  That last is all about graphing calculators, by the way.  Hopefully you, like Jed, have no idea what I’m talking about.  But the Cubers do!  Jed was afraid that the other contestants would find him out as a fake nerd and he would be a tournament outcast before he had a chance to gather enough material for his story.  So he’d asked me to be a sort of nerd coach, on call in case he needed some explanation or tips in passing for the kind of person who would be into all that stuff.  We decided that he should wear a wire and that I should be on call outside, in a van, ready to help in emergencies.  This was totally illegal, because outside help, like all calculating devices, are banned from the tournament.  They don’t want anybody programming some kind of cube solver into their TI-85s.  


Aw, just like my first graphing calculator! So cute!

Okay, so this is getting a little out of hand long, so I will get to the point.  Jed used his wire to ask me how many matches would need to be played in this years tournament of 10,041 people (including Jed) to get a Cube champion.  As a math puzzler, I immediately knew the answer.  I did one incredibly easy mathematical operation in my head and told him how many matches it would take.  He answered, and won the cash prize.  And immediately forgot all about the story he was writing.

So this week’s puzzle is:  How many matches are there in a single elimination tournament of 10,041 people, and what is the insight that let me figure it out incredibly quickly with a single mathematical operation?  If you know the answer, write me an email at  The best submission I get by Wednesday, April 4 will earn its writer an excellent t-shirt!

Solution to the planes puzzle:  Ah!  So, this puzzle got a few answers despite the spring break lull.  Here’s one solution.  You can do it with 3 planes.  Call them A, B, and C.  Airplane A will be the one to fly all the way around the world.  Let’s say it takes 8 hours to fly around the world, just to keep it straight.  A, B, and C take off to fly “clockwise” around the world.  1/8th of the way (or after 1 hour), plane C gives planes A and B each enough fuel to go an additional 1/8th of the way.  Then C turns around and goes back to base.  1 hour later, 1/4 of the way around the world, plane B gives plane A enough gas to go an additional 1/8th of the way around the world, then turns around to go back to base. Plane A is now 1/4 of the way around the world, has a full tank of gas, and 2 hours have passed.  Now, after 2 hours later, when B arrives back at base, B and C fuel up and take off to go around the world in the opposite direction.  They go 1/8th of the way around the world that way, taking one hour, then C gives enough gas to go an additional 1/8th of the way around.  C turns around and goes back to base.  5 hours total have passed.  After another hour passes and B has gone another 1/8th of the way around the world, A meets B.  B gives A enough gas to go 1/8th of the way around (I should have named a unit for this!!), then turns around and proceeds with A back towards base.  6 hours have passed.  At this moment, C leaves base with a full tank of fuel.  After an hour, when 7 hours have passed (and A has made it 7/8ths of the way around the world), C meets A and B and gives them each enough fuel to get back to base.  Everybody gets home safe, at the same time!  


Hopefully this doesn't make you more confused...


How many hipster airplanes does it take to fly around the world?

Yeah, you probably don’t know, because, um, you’ve probably never heard of these airplanes anyway, because besides being really obscure I also made them up.

It doesn't take this many planes, I'll tell you that.

This week’s puzzle is about airplanes.  The story starts with a team of scientists discovering an amazing new fuel source.  This fuel produces no greenhouse gasses or pollutants of any type.  It is renewable.  One catch is that the only known source of this fuel is a tiny, very obscure island in the middle of the Pacific Ocean.  So the scientists have made their base on this island.  They are getting ready to announce their findings to the world.  Hooray! All of the world’s energy problems are solved.

However, these scientists are so out of touch with reality that they think they actually need a publicity stunt to go along with this announcement.  They decide that they will fly an airplane around the world using this amazing fuel.   One airplane must make a complete, uninterrupted circuit around the earth.  Yes, the long way around.  The pilot must fly a great circle, like the equator (though it could be a different great circle, for example including both the north and south poles).  Anyway, everybody agrees that this is a great idea (heh), but the problem is that when the scientists retrofit the airplanes they’ve got to run on this fuel, each plane can only hold enough fuel to take it half way around the world.

So it seems that the scientists are out of luck.  However, one scientist notices that there may be some hope if they use another plane as a helper.  Perhaps one plane could transfer fuel to another in flight.  They try this out and find that it works!  In fact, the pilots, planes, and fuel are all so amazing that fuel can be transferred from one plane to another in midflight, instantaneously and without any complications.  The two planes don’t even have to be going the same direction.  And another thing about these planes is that they change direction instantaneously, too, so you don’t even need to worry about that.

So everything seems to be back on.  However, the scientists soon realize that one helper plane won’t quite do the job.  To see why, lets consider what happens in the most basic scenario.  We’ll call the plane that is going to hopefully make the whole trip around the world plane A.  The helper plane will be plane B.  Let’s call the units of fuel worlds (Ws).  So we say that each plane can hold at most ½ W of fuel at a time. Considering the situation for a second, you see that our only possible hope is if both planes take off at the same time going the same direction with ½ W each of fuel.  The best that we can do here (without plane B running out of fuel on the way back and crashing into the ocean, which we don’t want) is to have B go 1/6 of the way around the world with A, then transfer 1/6 W of fuel to A, then go back to base with its remaining 1/6 W of fuel.  This will get plane A 4/6 of the way around the world, but if B refuels and takes off to meet A on the other side, B won’t be able to carry enough fuel to get them both home safely.  Ugh.

So the scientists clearly need to get some more planes involved.  The question is, how many planes to they need and how can they do it?


If you know the answer, send me an email at!  The best answers I receive by Wednesday, March 28, will win amazing math mostly/somewhat science t-shirts for their authors.


Congratulations to last week’s winners, Brian and Eli!  I got some great answers this week so it was a hard choice.  If you didn’t win last week, but you had the right answer, I’ll remember you and like you more next time.  So keep trying.  Thank you thank you to everyone who wrote in.

The solution to last week’s puzzle is simply this—the two kids go first, taking 2 minutes.  The fastest kid goes back, taking 1 minute.  Grandma and Mom cross, taking 10 minutes.  Pretty fast kid goes back, taking 2 minutes.  Both kids cross over, taking 2 more minutes.  Grand total: 2+1+10+2+2=17 minutes!  Awesome!

Reading the solution over, this may not seem challenging at all.  However, most people only come up with this solution after struggling for a long time trying out scenarios in which the fast kid always runs the flashlight back across the bridge.  This seems like the fastest way, right?  However, we know that this is doomed to failure because it means that each other person must cross with the fast kid, for a total of 17 minutes crossing time just going one way! That leaves the kid no time to ever cross back with the flashlight.

The old perilous-bridge-at-night-with-dying-flashlight puzzle



The bridge is like this, only, um, scarier. The internet didn't have any pictures as scary as the bridge in the story. And it is dark out.

A new puzzle!  A family of four is on one side of a rickety bridge across a deep and otherwise uncrossable chasm.  They desperately need to get to the other side.  Perhaps they are being pursued by monsters.  I don’t know.  But I do know that it’s night time–totally dark, no moon.  The bridge is a weak thing, which can only support the weight of two people at any given time.  And it doesn’t have a hand rail, or anything like that, so you could walk right off the edge if you can’t see where you’re going.  So anyone who is on the bridge had better have a flashlight with them.  Our family of four has a flashlight.  Just one, though, and the batteries are dying.  In fact, there are only 17 minutes of battery power left.  The family knows that it will take one of them (call her Grandma) 10 minutes to cross the bridge.  Another of them (Mom) will need 5 minutes, another (fast kid) will take 2 minutes, and another (really fast kid) will take 1 minute to cross.  If two people are on the bridge together, they will have to travel at the rate of the slower person, because they have to share the flashlight.  There is no funny stuff, like throwing the flashlight (or throwing Grandma) across.  The question, obviously, is how can the whole family get across in 17 minutes or less?  Can you save this family?

If this family lived in Connecticut, they would still be traumatized by the Halloween storm and would each have been carrying a hand-crank flashlight/radio with them at all times. Along with sleeping bags and some MREs. They could listen to WESU as they crossed the bridge, leisurely, one at a time if they felt like it. Luckily for us, they are poorly prepared.

Figure it out!  Send solutions or questions to by Tuesday March 6.  Best solutions will receive an incredible math mostly/somewhat science T-shirt!  Listen to Somewhat Science from 2:30-3:00 Friday afternoon (Eastern time) on WESU 88.1 FM ( for the solution.  And a lot of other great science related stories from Wesleyan students.  Speaking of Wesleyan students, big thanks to Jed Rendleman, my Math Mostly co-host this week.

Solution to last week’s puzzle:  Congratulations to James Ricci and Katherine Mullins, the two winners of last week’s Math Mostly write-in puzzle explosion! The puzzle was called the Trivium’s Checkerboard, but is commonly (and slightly gruesomely) known as the mutilated chessboard problem.  Their solutions are so good that I’ll just quote ’em.

James:  “Milo noticed that the checkerboard alternates between black and white squares. That is, adjacent to any black square there are only white squares, and vice versa.  Therefore each individual domino must cover exactly one white and one black square at once. However, there are exactly 32 black squares and 30 white squares! So no matter how he places the dominos down he will only ever be able to cover at most a set of 30 pairs of black and white squares leaving at least 2 black squares uncovered.”

Katherine: “White squares are only adjacent to black squares on Trivium’s board, and vice versa.  Thus, any domino that covers two adjacent squares must cover one white square and one black square.  Two white squares were removed from the checkerboard, so Trivium’s board has 2 more black squares than it has white squares.  If the only dominos available necessarily cover one white square and one black square, then at least 2 black squares remain uncovered at all times.”