Shunt Buster

Help! This one's killing me.

**Reminder YofX will be moving to www.yofx.org in 7 days on 7/20**

Sudoku Variation

My mother sent this sudoku variaton my way. She is a total puzzle junky and found this in the International Herald Tribune. All the normal sudoku rules apply, but you also have to fill the shaded areas with 1 through 9. Thanks, Mom!

Amoeba Fever Problem

Here's a very interesting problem sent in by Lee. Have fun playing with this.

Toby has a jar with one amoeba in it. Every minute, every amoeba turns into 0, 1, 2, or 3 amoebae with a probability of 1/4 for each case (dies, does nothing, splits into two, or splits into three). What is the probability that the amoeba population eventually dies out?

If you're wondering, Lee's brother Chip created the image above. I'm guessing it has something to do with the solution.

Math Donkeys

I love to give this problem to students.

Two Donkeys are walking down the road loaded down with sacks of potatoes. One looks over at the other and says, "hey, if you give me a sack, I'll have twice as many as you; and if I give you one we'll have the same number. Weird, huh?" The other donkey just says, "Geek" and keeps walking. How many potato sacks does each donkey have?

pic by kaysare

Breakout Sessions?

I remember being at the Barnes Seminar last year about this time and coming up with the following problem. There were probably 150 participants. We convened as a group in the morning and would then split up into breakout sessions of about 6 people. The break sessions were assigned not chosen, and organizers tried to have as little overlap as possible, giving you the opportunity to meet the most number of people. So here is the question, what is the maximum number of breakout sessions until at least two people find themselves in a group again? After you figure that out, what is the maximum number of non-overlapping breakout session of size p chosen from an overall group size of n?

pic by lululemonathletica

Gruesome, but Fun

The following riddle about disabled veterans is from Lewis Carroll's A Tangled Tale.

Say that 70% have lost an eye, 75% an ear, 80% an arm, 85% a leg. What percentage, at least, must have lost all four?

Found this one in Herstein's Abstract Algebra. pic wikimedia

Alfie's Children

This is one of my all time favorite problems to give to students.

Determine the ages of Alfie's children from analyzing the following conversation.

Alfie: Can you guess the age's of my children? The product of their ages is 36.

Bernice: That's not enough information.

Alfie: The sum of their ages is the same as our street address.

Bernice: I'll need a little more info.

Alfie: My oldest is a girl.

Bernice: Okay, I've got it.

pic by Alain Gree

Missing Algebra?

Coming down after your intermediate algebra final? Don't know how you are going to fill up the summer hours without all the homework problems? YofX is here to help. Here's a problem to both reinforce and extend your algebraic superpowers.

Given distinct positive real numbers, x,y,z such that is true. Write x,y,z in increasing order from left to right, and justify your answer algebraically.

Problem stolen from Awesome Math camp's application. Their camps look really fun. I wish I was 15 again!

A mathematical who-done-it?

Sue tells me her oldest has discovered the joy of logic problems. Here's a fun one from the exercises in Enderton's A Mathematical Introduction to Logic:

There are three suspects for a murder: Adams, Brown, and Clark. Adams says "I didn't do it. The victim was an old acquaintance of Brown's. But Clark hated him." Brown states "I didn't do it. I didn't even know the guy. Besides I was out of town all that week." Clark says "I didn't do it. I saw both Adams and Brown downtown with the victim that day; one of them must have done it." Assume that the two innocent men are telling the truth, but that the guilty man might not be. Who did it?

pic by martinmaters

May Issue of Wired

I really enjoyed the new issue of Wired. It's put together by guest editor, J.J. Abrams, who I'm ashamed to say I've never heard of, but whose issue as an homage to puzzling. He comes at it from various angles from a feature story of an artist who has been able to stump the best cryptographic minds in the world to an "enigmatrix" of the network connections between math, board games, plot, mysteries, magic code, and game theory. Best of all, it is full of puzzles. Enjoy the one above!

Mathematical Frenemies

This one came up on the problem widget (lower right) a while ago. I've been struggling with it.

A town creates a committee of six people to solve a new math problem. In the town, everyone is either a friend or an enemy. Prove that there are at least three friends or three enemies on the committee.

pic by Alana

Math Battle

If you've been reading along, you know I'm interested in finding a spectator friendly math competition. Kate left a provocative comment about the historical precedence of math battles on that post. I didn't run into any historical info, but when I googled 'math battle', I wound up with a Canadian group that has been running an interesting type of math competition it calls a math battle. What makes this more interesting than the usual paper and pencil test taken individually or in a group is that it is not graded. The answers are debated in front of a panel of judges and an audience. It starts in the normal manner with a problem set that each team works. Then the teams challenge each other's answers in a debate-style back and forth between the speakers. So not only can you score points by correctly solving the original problems, but you can also earn points by finding gaps or errors in your opponent's solution. I'm not sure this would work for us. I think John and Steve are thinking of something more immediate, physical, and faster moving, but I would love to see one of these competitions. It seems like an excellent idea. BTW, they also post their old problem sets so it is a great site to go to for a little mathematical recreation. I liked this one.

Two persons play a game on a board divided into 3 × 100 squares. They move in turn: the first places tiles of size 1 × 2 lengthwise (along the long axis of the board), the second, in the perpendicular direction. The loser is the one who cannot make a move. Which of the players can always win (no matter how his opponent plays), and what is the winning strategy?

pic by Bre!

Bonus problem for College Algebra

I thought this might be a nice bonus problem for a college algebra class.

Let A be some number. If c = log(A) and c = a+b where a is a non-negative integer and b is a decimal between zero and one, represent A in scientific notation.

Maybe a concrete form of the question would be better for students: If log(A) = 4.182, approximate A in scientific notation.

Prime Time

I was thinking about the possibility of defining a subset of the naturals containing numbers whose digits (appropriately grouped) are its own prime factorization. For example, take 32. It would be a number of this kind if 32 = (3)(2). (Of course it's not.) Are there any members of this set? Well, every prime number is trivially a member, 7 = 7, 11 = 11. The interesting question is, does this set have any composite members? I haven't been able to think of one yet.

I quickly checked the numbers less than 100 and there aren't any examples there. The only single digit numbers in this set are the primes. Consider the double digit numbers, 10 through 19. Let A be a digit, 0 through 9, if 1A = (1)(A), then 1A = A. Since no two digit number is equivalent to a one digit number, this isn't possible. (Note, 1A does not mean 1*A.) What about the numbers 20-29? If 2A = (2)(A), then the most (2)(A) could be is 14, because the most A could be is 7. But we know 2A is in the 20's so this is not possible. Any number in the 30's, 50's, and 70's will have the same problem. Any number 4A can only be grouped (4A) because 4 is not prime. So only the boring numbers (primes) in the 40's make it into the set. Same is true for the 60's, 80's, and 90's. What can we conclude? There is no composite example less than 100.

What about 3 digit numbers? We can start to make arguments by thinking of the grouping possibilities. If ABC is a three digit number, then we could group (A)(B)(C), (AB)(C), (A)(BC). Your mission: find a composite number in this set, or prove that there aren't any.

pic by kingfal

Math Contest

Friday was a big day for math at Tunxis. Both the Pi contest and CCC system-wide math contest were going down. It was a great day. The victor of the pi contest, Alex Bobman, memorized a whopping 302 digits! We are still waiting for the results of the math contest, but I'll post those as soon as I find out.
I was also able to get a copy of the math contest problems. I thought I'd pass along something fun.

A farmer had a daughter who spoke in riddles. One day the child was asked to count the number of goats and the number of ducks in the barnyard. She returned and said, "Twice the number of heads is 76 less than the number of of legs." How many goats were in the barnyard?

pic by ADoseofShipBoy

Probability Problem

Lee found this nice problem in the Minitab manual that accompanies "Elementary Statistics" by Mario F. Triola. The problem posed is:

Two points along a straight stick are randomly selected. The stick is then broken at these two points. Find the probability that the three pieces can be arranged to form a triangle.

pic by flowers & machinery

My New Favorite Puzzle: Tentaizu

I found a new puzzle. New to me anyway. It's called Tentaizu. I found it in the in-flight magazine on the way home from vacation. Btw, Southwest's magazine has the best puzzle section that I've ever seen. Not a reason to buy a ticket, but a definite perk. Anyway, I was completely charmed by this game. I thought we would play one this week. Maybe if we get excited we can try to automate a solution technique. I also tried to find a website that generates these like those I've found for Sudoku and KenKen. No luck.

Here are the instructions from Southwest's magazine:
In each of the grids on this page, stars are
hiding in 10 of the 49 squares. Your task is
to determine the positions of the stars. The
numbers in the grid provide clues: A number
in a square indicates how many stars lie next
to the square—in other words, how many
adjacent squares (including diagonally adjacent
squares) contain stars. No square with
a number in it contains a star, but a star may
appear in a square with no adjacent numbers.

Jean-Marc's Angle of Fire

Jean-Marc passed on this problem to me. Here's the story. You're on one side of a pit of fire. You see two walls protruding from it. You must figure out how to build a wall so that if it were continued past the pit of fire it would bisect the angle. I solved this with trig the other day during a meeting and told Jean-Marc. He said the fun solution is purely geometric. So I'm back at work on this problem with my weak geometry. Help!

What Would a Spectator Friendly Math Competition Look Like?

John and Steve have given me the challenge of creating a math competition that is actually fun to watch. There are tons of competitions out there. Rob runs one every year at Tunxis. However, almost all of these involve students sitting at a desk solving problems. What would a spectator friendly math competition look like? I have been thinking in two directions on this.

1) Taking math problems and making them physical. Classic bucket problems could be fun to watch if you took away paper and pencil and gave competitors actual buckets. (Think Die Hard with a Vengeance minus the bomb.) Or, here's a pack of M&M's, add these two numbers in base 6. Compass and protractor constructions would be interesting, particularly because hardly anyone knows how to do them anymore.

2) Structure the competition so that the audience can evaluate or judge the participants. Math contests are usually only judged for correctness and speed by experts. John mentioned the example of the poetry slam. How could this happen for math? Here are some ideas to get started with. Contestants solve a problem at home before the contest, come to the contest, and explain their solutions in 3 minutes or less. Then the audience judges them on the clarity/creativity of their explanations. Or, contestants prepare a 3 minute lecture/demonstration about math and the audience judges. Another wilder idea is to have them perform a creative task that doesn't have a correct answer. I use an enrichment activity in Math for Liberal Arts that might work. I have students create a poem using the power set structure on a set of words. The audience would definitely be able to evaluate this.

I'm not sure exactly what John and Steve had in mind for the bigger picture. I think they plan to have contests like this in all disciplines. They definitely gave me an interesting thought problem though.

Solving The Rectangles Problem

I keep posting problems and you do all the solving. I figured I better walk the walk, so I worked on Lee's problem from last week's post. It was really a nice one. The idea was to give the dimensions of five rectangles such that
1) the sides have unique integer dimensions between 1 and 10
2) they can be arranged into a square.

The first thing I did was narrow the size of the possible squares. I did it originally with an area argument. The minimum area of 5 rectangles that meet criteria 1 would be 1*10+2*9+3*8+4*7+5*6 = 110. The max would be 10*9+8*7+6*5+4*3+2*1= 190. So, 110<11^2, 12^2, 13^2<190. That limits the squares I was considering to only three different dimensions. (Later I found another way to argue for these as the only three possibilities.)

I have to admit I had a little more information than you guys did on the blog. Lee is a puzzle maker and he brought in a wood version of one of the solutions to show me. I was playing around with it on my desk one day and found it surprisingly hard to arrange into a square. It took me around 5 minutes to arrange 5 rectangles into a square.



Seeing the correct configuration was the insight that broke the problem for me. The rectangles looked something like this when assembled.



Note that this isn't drawn to scale, but essentially you have one rectangle in the middle with the others spiraling around it. The dimensions of the middle depend on the dimensions you choose for your outer rectangles.


That's when I started playing with me three possible areas.

1) If the overall square is 11 by 11, then what are my options for the inner rectangle. 10 can't be one of the sides of the inner triangle because I have to add two numbers to it to get 11. 9 also can't be it because I'd have to add two one's to it to get 11, and I'm only allowed one one. 8 is the highest possible side of the inner rectangle, because the outer rectangles could have 2 and 1 as sides.

But then since the sides of the whole rectangle are 11, we can fill in...

Now only 3,4,5,6,7 are left, and the only pairs that add to 11 are (5,6) and (4,7). That means 3 has to be the other dimension of the center rectangle.

What two remaining numbers could you add to 3 to get 11? There aren't any. Conclusion: 8 can't be the largest dimension of the inner rectangle. Try 7. And so the process repeats through all possible cases. I really didn't do it that systemattically, but this makes for a better demonstration.

When I got through with the process I only found two solutions: An 11x11 rectangle like this

and a 13x13 like this

At this point, I freaked out. I thought my logic was impeccable but Lee promised 4 solutions. At some point I thought to go back and look at the solution he gave me and saw that the sides were identical to mine, but his solution had flipped two of my sides giving different dimensions to the surrounding four rectangles. I'll show you with the 13x13,

So that's it, Lee also has a solution page on his website check it out. What I want now is a proof that 5 rectangles with unique sides cannot be arranged into a square in any other way (than the way I described above). There has to be a simple topological proof of this. Help?