## The Puzzle Archives, II

This week, I’ll continue with some more problems from the contests for the 2014 conference of the International Group for Mathematical Creativity and Giftedness.  We’ll look at problems from the Intermediate Contest today.  Recall that the first three problems on all contests were the same; you can find them here.

The first problem I’ll share is a “ball and urn” problem.  These are a staple of mathematical contests everywhere.

You have 20 identical red balls and 14 identical green balls. You wish to put them into two baskets — one brown basket, and one yellow basket. In how many different ways can you do this if the number of green balls in either basket is less than the number of red balls?

Another popular puzzle idea is to write a problem or two which involve the year of the contest — in this case, 2014.

A positive integer is said to be fortunate if it is either divisible by 14, or contains the two adjacent digits “14” (in that order). How many fortunate integers n are there between 1 and 2014, inclusive?

The other two problems from the contest I’ll share with you today are from other contests shared with me by my colleagues.

In the figure below, the perimeters of three rectangles are given. You also know that the shaded rectangle is in fact a square. What is the perimeter of the rectangle in the lower left-hand corner?

I very much like this last problem.  It’s one of those problems that when you first look at it, it seems totally impossible — how could you consider all multiples of 23?  Nonetheless, there is a way to look at it and find the correct solution.  Can you find it?

Multiples of 23 have various digit sums. For example, 46 has digit sum 10, while 8 x 23 = 184 has digit sum 13. What is the smallest possible digit sum among all multiples of 23?

You can read more to see the solutions to these puzzles.  Enjoy!

## The Puzzle Archives, I

In going through some folders in my office the other day, I came across some sets of mathematics puzzles I wrote for a conference of the International Group for Mathematical Creativity and Giftedness in 2014.  Teachers of mathematics of all levels attended, from elementary school to university.  The organizing committee (which included me) thought it might be fun to have some mathematical activity that conference attendees could participate in.

So I and my colleagues created three levels of contests — Beginning, Intermediate, and Advanced — since it seemed that it would be difficult to create a single contest that everyone could enjoy.  But I did include three problems that were the same at every level, so all participants could talk about some aspect of the contests with each other.

Participants had a few days to get as many answers as they could, and we even had books for prizes!  Many remarked how much they enjoyed working out these puzzles.

Now this conference took place before I started writing my blog.   I have written several similar contests over the years for various audiences, and so I thought it would be nice to share some of my favorite puzzles from the contests with you.  And so The Puzzle Archives are born!

First, I’ll share the three puzzles common to all three contests.  I needed to create some puzzles which were fun, and didn’t require any specialized mathematical knowledge.  As I’m a fan of cryptarithms and the conference took place in Denver, I created the following puzzle.  Here, no letter stands for the digit “0.”

For the next puzzle, all you need to do is complete the magic square using the even numbers from 2 to 32.  Each row, column, and diagonal should add up to the same number.  There are two solutions to this puzzle — and so you need to find them both!

And of course, I had to include one of my favorite types of puzzles, a CrossNumber puzzle.  Remember, no entry in a CrossNumber puzzle can begin with “0.”

I also included a few geometry problems, staples of any math contest.  For the first one, you need to find the area of the smallest circle you could fit the following figure into.  Both triangles are equilateral; the smaller has side length 1 and the larger has side length 2.

And for the second one, you need to find the radius of the larger circle.  You are given that the smaller circle has a diameter of 2 units, and the sides of the square are 2 units long.  Moreover, the smaller circle is tangent to the square at the midpoint of its top edge, and is also tangent to the larger circle.

The last two problems I’ll share from this contest are number puzzles.  The first is a word problem, which I’ll include verbatim from the contest itself.

Tom and Jerry each have a bag of marbles. Tom says, “Hey, Jerry. I have four different colors of marbles in my bag. And the number of each is a different perfect square!” Jerry says, “Wow, Tom! I have four different colors of marbles, too, but the number of each of mine is a different perfect cube!”

If Tom and Jerry have the same total number of marbles, what is the least number of marbles they can have?

And finally, another cryptarithm, but with a twist.  In the following multiplication problem, F, I, N, and D represent different digits, and the x‘s can represent any digit.  Your job is to find the number F I N D. (And yes, you have enough information to solve the puzzle!)

Happy solving!  You can read more to see the solutions; I didn’t want to just put them at the bottom in case you accidentally saw any answers.  I hope you enjoy this new thread!

(Note:  The FIND puzzle was from a collection of problems shared by a colleague.  The first geometry problem may have come from elsewhere, but after four years, I can’t quite remember….)
Continue reading The Puzzle Archives, I

## More CrossNumber Puzzles

Last fall, I mentioned that while looking at the Puzzle Page of the FOCUS magazine published by the Mathematical Association of America, I thought to myself, “Hey, I write lots of puzzles.  Maybe some of mine can get published!”  So I submitted a few Number Search puzzles to the editor, and to my delight, she included them in the December/January issue.  Here’s the proof….

Incidentally, these puzzles are the same ones I wrote about almost two years ago — hard to believe I’ve been blogging that long!  So if you want to try them, you can look at Number Searches I and Number Searches II.

Since I had success with one round of puzzles, I thought I’d try again.  This time, I wanted to try a few CrossNumber puzzles (which I wrote about on my third blog post).  But as my audience was professional mathematicians and mathematics teachers, I wanted to try to come up with something a little more interesting than the puzzles in that post.

To my delight again, my new trio of puzzles was also accepted for publication!  So I thought I’d share them with you.  (And for those wondering, the editor does know I’m also blogging about these puzzles; very few of my followers are members of the MAA….)

Here is the first puzzle.

Answers are entered in the usual way, with the first digit of the number in the corresponding square, then going across or down as indicated.  In the completed puzzle, every square must be filled.

I thought this was an interesting twist, since every answer is a different power of an integer.  I included this as the “warmup” puzzle.  It is not terribly difficult if you have some software (like Mathematica) where you can just print out all the different powers and see which ones fit.  There are very few options, for example, for 3 Down.

The next puzzle is rather more challenging!

All the answers in this puzzle are perfect cubes with either three or four digits, and there are no empty squares in the completed puzzle.  But you might be wondering — where are the Across and Down clues?  Well, there aren’t any….

In this puzzle, the number of the clue tells you where the first digit of the number goes — or maybe the last digit.  And there’s more — the number can be written either horizontally or vertically — that’s for you to decide!  So, for example, if the answer to Clue 5 were “216,” there would be six different ways you could put it in the grid:  the “2” can go in the square labelled 5, and the number can be written up, down, or to the left.  Or the “6” can go in the square labelled 5, again with the same three options.

This makes for a more challenging puzzle.  If you want to try it, here is some help.  Let me give you a list of all the three- and four-digit cubes, along with their digit sum in parentheses:  125(8), 216(9), 343(10), 512(8), 729(18), 1000(1), 1331(8), 1728(18), 2197(19), 2744(17), 3375(18), 4096(19), 4913(17), 5832(18), 6859(28), 8000(8), 9261(18).  And in case you’re wondering, a number which is a palindrome reads the same forwards and backwards, like 343 or 1331.

The third puzzle is a bit open-ended.

To solve it, you have to fill each square with a digit so that you can circle (word search style) as many two- and three-digit perfect squares as possible. In the example above, you would count both 144 and 441, but you would only count 49 once. You could also count the 25 as well as the 625.

I don’t actually know the solution to this puzzle.  The best I could do was fill in the grid so I could circle 24 out of the 28 eligible perfect squares between 16 and 961.  In my submission to MAA FOCUS, I ask if any solver can do better.  Can you fit more than 24 perfect squares in the five-by-five grid?  I’d like to know!

I’m very excited about my puzzles appearing in a magazine for mathematicians.  I’m hoping to become a regular contributor to the Puzzle Page.  It is fortunate that the editor likes the style of my puzzles — when the magazine gets a new editor, things may change.  But until then, I’ll need to sharpen my wits to keep coming up with new puzzles!

## Beguiling Games IV: Scruffle.

We’ll start this installment of Beguiling Games by discussing who has a winning strategy in Splotch!  Recall the rules of the game:  players alternately color in squares in a 4 x 4 grid, with the goal of creating a specified splotch, shown below.  The splotch may be rotated and/or reflected as well.

For a more complete description together with an example of how the game is played, you can look at the previous installment of Beguiling Games.

OK, now for the solution!  It turns out that the second player can always win.  Let’s see how.  I found it easiest to think about a strategy by imagining the grid as being divided into quadrants, like this:

Now here is the important observation:  the first player who fills in a second square in any of the quadrants loses, regardless of whether the squares are adjacent or diagonally situated.  Of course there are other ways to lose, too — as with all these two-player games, there are multiple ways to analyze them.

So let’s get specific.  Suppose the first player colors in square A1 (see the figure below).  The second player then colors in the square labelled B1.

At this point, the red squares indicate all the places the first player cannot play without losing the next turn.  So the first player must color in one of the two empty squares, after which the second player will color in the other one.  So after two turns for each player, the board now looks like this:

So no matter what square the first player colors in next, one of the quadrants will contain two filled-in squares, and so the second player will win on the next move.

A similar strategy may be used no matter where the first player begins.  Consider the first few moves in the following game.  The first player colors in the square A1, and the second player colors in B1.

Again, the first player must avoid the red squares, or else the second player would win on the next turn.  Whichever square the first player colors in next, the second player can always play “two away.”  The result will be 1) the first player will not be able to win on the next turn, and 2) one square in every quadrant will be colored in.  This means that the first player is forced to put a second square in one of the quadrants on the next move, meaning that the second player will win on the turn after that.

This is the simplest strategy I found for the second player.  I would be happy to hear if some reader found an even simpler way to describe a winning strategy!

What about using other splotches?  If the splotch contains too many squares, it is possible to force a draw.  For example, given the splotch below, either player may force a draw simply by coloring in the four corners on their first four moves.

Interestingly, it is difficult to come up with a splotch where the first player has a winning strategy (other than a splotch which is just a single square, of course).  The more squares included in the splotch, the more difficult the analysis.  But for simpler splotches, it seems a clever division of the board allows the second player to win.

For example, consider the following square splotch.

Now divide the board into the following 2 x 1 regions, or dominoes:

Player two has a simple winning strategy.  Whenever the first player fills in a square, the second player fills in the other square of the domino.  It should be clear that the second player can never lose this way.  The first player will eventually have to fill in a square directly above or below a filled-in domino, and when this happens, the second player wins on the next move.

A complete analysis of Splotch! is likely beyond reach.  Just counting the number of possible splotches (up to rotation and reflection) would be a challenging task unless you wrote a computer program to exhaustively find them.  Without rotations and reflections, there are 216 = 65,536 possible subsets of 16 squares, and hence 65,535 splotches (since a splotch must include at least one square).  So a computer program would be able to find them all relatively quickly.  The interested reader is welcome to undertake such a task….

Here is another simple two-player game for you to think about, which I call Scruffle.  It is played on a typical 3 x 3 Tic-Tac-Toe grid.  Players alternate playing either a 1, 2, or 3 anywhere in the grid.  A player wins when a number they place creates a column, row, or diagonal which contains a 1, 2, and 3 in any order.

There is one additional constraint:  only three of each number may be placed in the grid.  So once three 1’s (for example) are placed in the grid, no player may place another 1 anywhere in the grid.  This is not an arbitrary constraint — you can show that the game cannot end in a draw with this condition.  See if you can show this!

For the first puzzle, show that the first player has a winning strategy.  This is not difficult; the simplest strategy I found involves the first player’s second turn involving playing the same number they played on the first turn.

A slightly more challenging puzzle is to require the first player to play a different number than the number they played on their first move.  Does the first player still have a winning strategy?  I’ll give you the solution in the next installment of Beguiling Games!

## Bay Area Mathematical Artists, IV.

We had our last meeting of the Bay Area Mathematical Artists in 2017 this weekend!  We had a somewhat lower turnout than usual since we’re moving into the holiday season.  But it really wasn’t possible to move the seminar a week earlier, since many of us affiliated with universities were in the middle of Final Exams.

As we had been doing before, we began with a social half hour while waiting for everyone to show up.  We then moved on to the more formal part of the afternoon.

There were three speakers originally slated to give presentations, but one had to cancel due to illness.  Still, we had two very interesting talks.

The first talk, Squircular Calculations, was given by Chamberlain Fong.  Chamberlain did speak at the inaugural September meeting, but wanted a chance to practice a new talk he will be giving at the Joint Mathematics Meetings in San Diego this upcoming January.

So what is a squircle?  Let’s start with a well-known family of curves parameterized by p > 0:

$|x|^p+|y|^p=1.$

When p = 2, this gives the usual equation for a circle of radius 1 centered at the origin.  As p increases, this curve more and more closely approaches a square, and it is often said that “p = ∞” is in fact a square.

However, in Chamberlain’s opinion, the algebra becomes a bit unwieldy with this way of moving from a circle to a square.  He prefers the following parameterization:

$x^2+y^2-s^2x^2y^2=1,$

where s = 0 gives a circle, and the central portion of the curve when s = 1 is a square.  As s varies continuously from 0 to 1, the central portion of this curve continuously transforms from a circle to a square.  This parameterization was created by Manuel Fernandez Guasti; you can read his original paper here.

Chamberlain’s talk was about extending this idea in various ways into three dimensions.  He showed images of squircular cylinders, squircular cones, etc., and also gave equations in three-dimensional Cartesian coordinates for all these surfaces.  You can see some of the images in the title page of his presentation above.  It was quite fascinating, and there were lots of questions for Chamberlain when his talk was finished.  Feel free to email him at chamb3rlain@yahoo.com if you have further questions about squircles.

The second talk was given by Dan Bach (also a speaker at our inaugural meeting), entitled Making Curfaces with Mathematica.  Yes, “curfaces,” not “surfaces”!

Dan took us through a tour of his very extensive library of Mathematica-generated images.  He is fond of describing curves using parameters, and then changing the parameters over and over again to generate new images.

This is easy to do in Mathematica using the “Manipulate” command; below is a screen shot from Mathematica’s online documentation showing an example.

The parameter n is used in plotting a simple sine function — as you move the slider, the graph changes dynamically.  Note that any numerical parameter may be experimented with in this way.  Simply make a slider and watch how your image changes with the varying parameter.

So what are “curfaces”?  Dan uses the term for images create by a family of closely related curves which, when graphed together, suggest a surface.  As we see in the example above, the family of curves suggests a spiraling ribbon in which several brightly colored balls are nestled.  Dan showed several more examples of this and discussed the process he used to create them.  To see more examples, you can visit his website www.dansmath.com or email him at art@dansmath.com.

Once the talks were over, we had some time for puzzles!  Earlier in the week, when I knew we were not going to have an overabundance of talks, I asked participants to bring some of their favorite puzzles so we could all have some fun after the talks.  We were all intrigued with the wide variety of puzzles participants brought.

My dissection puzzle was actually quite popular — that is, until a few of the participants solved it!

You might recognize this from my recent blog post on geometrical dissections.  The pieces above are arranged to make a square, but they may also be rearranged to make an irregular dodecagon.  Some asked if I had any more copies of this puzzle, but unfortunately, I didn’t.  Maybe I’ll have to start making some….

As has been our tradition, many of us went out to dinner afterwards.  We went to our favorite nearby Indian buffet, and engaged in animated conversation.  Interestingly, after talking a bit about mathematics and art, Chamberlain began entertaining us with his wide repertoire of word puzzles.

To give just one example, he asked us to come up with what he calls “mismisnomers.”  Usually, the prefix “mis-” means to incorrectly take an action, as in “misspell.”  But some words, like “misnomer,” begin with “mis-,” while the remainder of the word, “nomer” is not even a word!  How many mismisnomers can you think of?  This and similar amusing puzzles kept us going for quite a while, until it was finally time to head home for the evening.

So that’s all for the Bay Area Mathematical Artists in 2017.  Stay tuned in 2018…our first meeting next year will be at the end of January, and I’ll be sure to let you know how it goes!

## Knights and Rogues

They were dark times in the Kingdom of Verdoon.  A band of fierce Rogues was terrorizing the local hamlets.  King Buford was very displeased — he had no other option but to send out a legion of Brave Knights and Loyal Knights to round up the Rogues and deliver them to the castle dungeons.

So he chose seven of his bravest Brave Knights, and seven of his most loyal Loyal Knights, and dispatched them across Mount Kerchoo to the hamlet of Dunken, where the Rogues were last reported to have been wreaking havoc.

The Knights were capable Knights, and soon rounded up seven of the fiercest Rogues, including their leader, Thorn Yackley.  Now they had the difficult task of escorting the Rogues back through the narrow, winding path over Mount Kerchoo.

The Rogues were so fierce, no single sword could subdue any of them, nor could a single mace.  But they were no match for both a sword and a mace.

Now the Brave Knights carried a single sword, and the Loyal Knights a single mace.  So on the trek back to the castle — where they had to walk single file, the path was so narrow — it was vital that each Rogue be flanked by both a Brave Knight and a Loyal Knight in the event the Rogue tried to escape.

The question:  In how many ways could the Brave Knights, Loyal Knights, and the Rogues be lined up so that on the treacherous way back, each Rogue was flanked by both a Brave Knight and a Loyal Knight?

(For this problem, consider the Brave Knights indistinguishable from each other; similarly for the Loyal Knights and Rogues.)

Recently, I had been challenged to create a lesson which involved problem-solving with binomial coefficients.  But I didn’t want to create just any lesson — I wanted something interesting and novel.

Now if you’ve ever taught combinatorics and counting, you know that there is a deluge of problems about binomial coefficients — from choosing committee members, sitting people around a circular table, lining up books on shelves, etc.  So it’s not an easy task to come up with something you think might be new.

That was the genesis of Knights and Rogues.  Of course it may very likely not be new, since it really is difficult to come up with a combinatorics problem no one has thought about before.  But I did ask a colleague who knows a lot about problem solving, and he hadn’t seen it before.

The story is just to get students interested.  Combinatorially, we could state the problem as follows:

Given p 0’s, q 1’s, and r 2’s, in how many ways can you line them up so that each 2 is surrounded by both a 0 and a 1?

It turns out that there is a lot going on in this problem.  With three parameters, the analysis is far from straightforward.

There are a few restrictions on the parameters.  For example, we need $p+q>r,$ or there are not enough Knights to flank the Rogues.  We also need

$p,q\ge\left\lceil\dfrac r2\right\rceil,$

or else there wouldn’t be enough Brave Knights and Loyal Knights to flank the Rogues.  For example,  if p = 2, q = 1, and r = 3 (so that q just falls short of the above inequality), you can see the problem — the one 1 can flank at most two 2’s, and so there is one 2 without an adjacent 1.  But if increase q to 2, we can have either the sequence 0212021 or 1202120.

Of course it is possible to consider specific examples, but it is also possible to consider infinite families of parameters — we’ll look at the case p = 2, qn, and r = 2 to give you an idea of the reasoning involved.

Now when n = 1, there is only one possibility:  02120.    But when n is larger than 1, we need to look at two separate cases:  when the 2’s are separated by more than one 0/1, so that we are looking at two blocks of 021 or 120.  When the 2’s are separated by exactly one 0/1, we only have the options 02120 or 12021.  We need to count each case separately.

Let’s look at the first case.  We have two blocks of either 021 or 120.  This uses up the two 1’s but there are n – 2 1’s left over.  So we have n things to arrange:  two blocks, and n – 2 1’s.  This may done in

$4\displaystyle{n \choose 2}$

ways, since we first choose the 2 places for the blocks, but then multiply by 4 since each block has two possibilities:  021 or 120.

In the second case, if the 2’s are only separated by one number, we may have a block of 02120 and n – 1 1’s to place, giving n possibilities. With a block of 12021, we have the block, one 0, and n – 2’s to place, which can be done in

$\displaystyle2{n\choose2}$

ways (since the order of the block and the one 0 matters). Adding these three cases together results in

$\displaystyle4{n\choose2}+n+2{n\choose2}=3n^2-2n$

possible lineups for the Knights and Rogues.

Interestingly, these are the octagonal numbers (A000567 in the OEIS).  I have no idea why, though….

What I like about this problem is that parameterizing p, q, and r in different ways produces seemingly very different results.  One interesting parameterization is p = qrn.  It turns out that this sequence — at least the first several values, which I used the computer to generate and didn’t work out by hand — is also in the OEIS (A141147), and the description is the

number of linear arrangements of n blue, n red and n green items such that the first item is blue and there are no adjacent items of the same color (first and last elements considered as adjacent).

I find this particularly intriguing since in Knights and Rogues, 0’s and 1’s can be adjacent — but even though 2’s are not adjacent, it is not possible to have any block be 020 or 121, although these are counted in A141147.  Finding a bijection between the two linear arrangements would be a very interesting problem.

I hope you enjoyed reading about Knights and Rogues!  Again, this is another example of mathematics as it happens — I came up with this problem within the past few weeks, and I’ve just scratched the surface of this interesting puzzle.  I hope it’s actually a new problem, but I won’t be surprised if I get a comment like, “Oh, I saw that problem in…..”  Generating new, novel mathematics problems is not an easy task…..

And incidentally, the answer to the initial problem is 80,096.  Did you get it?

## Beguiling Games III: Splotch!

In this installment of Beguiling Games, we’ll learn how to play Splotch!

But first, I’ll start off by giving you the solution to the last puzzle I presented in the last installment.  The statement of the puzzle is too long to repeat, so you can refresh your memory at Beguiling Games II.

First, it is important to note that Lucas’s statement actually provides no information!   Suppose Ophelia’s card was Truthteller.  Then she would have told the truth in Round 1, passed her card to Lucas, and he would truthfully have stated that she told the truth in Round 1.

But what if Ophelia’s card was Liar?  Then her statement “I am a Truthteller” in Round 1 would in fact have been a lie.  Now she passes her card to Lucas.  He lies and says she told the truth in Round 1!  What this means is Lucas could have made his statement in Round 2 regardless of what card Ophelia passed him.

Note that the same logic applies to Mordecai’s statement.  He could have said “Lucas also told the truth in the first round” regardless of whether Lucas passed him a Truthteller or a Liar card.

Now let’s examine Nancy’s statement in some detail.  She said that there is at least one liar at the table.  Could she have lied?

Well, if the fact that the there is at least one liar at the table is false, that means everyone is a Truthteller.  But there is no way Nancy could have known this, since the only two cards she saw were hers and the card passed to her by Mordecai.

That means Nancy must have told the truth in Round 2, and Mordecai must have passed her a Truthteller card.  But in order for her to have sufficient information to say there is at least one Liar at the table, she must have been holding a Liar card in Round 1.

Now this information could be deduced by anyone at the table.  In other words, anyone would know that Nancy held a Liar card in Round 1, and Mordecai held a Truthteller card.  That leaves Lucas’s and Ophelia’s cards in Round 1.

The only person who could know both these cards would be Lucas — he knew his own card in Round 1, and he knew Ophelia’s card because she passed it to him in Round 2.  So he had enough information to declare after Nancy’s statement.

It is important to point out that there is no way to know what Lucas’s and Ophelia’s cards actually were.  All we need to know is that Lucas knew what both of them were.

Did you figure it out?  It takes a little bit of reasoning, but all the facts were there.

Now as I mentioned in the last installment, this time I’d give you a geometrical two-player game to work out.  I call it Splotch!

In the game of Splotch!, players alternate coloring in squares on a 4 x 4 grid.  The goal is to create a target shape, called a splotch.  A player wins when he or she colors in a square which completes a splotch, and then announces the win.

So if a player completes a splotch but doesn’t notice it and doesn’t announce the win, then play continues until someone completes another splotch.  You can only call Splotch! on your turn, so you cannot win by calling it if your opponent fails to.  It is also important to remember that the square you color in on your turn must be part of a splotch you announce.

In the version of Splotch! I’m sharing with you this week, the target splotch is:

Just so you know how it works, let’s look at a sample game, shown below.

There are two players, A and B.  A goes first, and plays in the lower left corner (labelled with A1).  B colors in a square in the top row (B1).  Then A’s second move is in the second row.  B wins on the next play by completing — and announcing! — a splotch.  The splotch may be rotated (as in this example), reflected, or even both!  So you’ve got to watch carefully.

Now admittedly, A was not a very clever player in this round of Splotch!  In fact, B could also have won by playing as follows:

But here is the question:  In this version of Splotch!, which player can force a win every time?  Remember:  you must be able to provide a response to every move by your opponent!  I’ll reveal which player can force a win in my next installment of Beguiling Games, and I’ll share my particular strategy for winning.

One final remark — a bit of a tangent, but related to puzzles and games.  If you’ve been following my blog for a long while, you might remember my (fiendishly diabolical) number searches.  (These were posted well over a year ago.)

Here, the numbers in the right are in base 10, but you have to convert them to another base to find them in the grid!  You can read more about these puzzles here and here.

I submitted these puzzles to MAA Focus, the newsmagazine of the Mathematical Association of America.  I am happy to report that they will be featured on the Puzzle Page in the December 2017/January 2018 issue!

Now I will admit that this doesn’t exactly make me famous, given the number of people who subscribe to MAA Focus.  But I composed these puzzles specifically for my blog — so if I hadn’t decided to write a blog, these puzzles might never have been created.  Perhaps another reason to write a blog!

Stay tuned for the next round of Beguiling Games, where you’ll learn (if you didn’t already figure it out for yourself) who has a winning strategy in the game of Splotch!