Tanya Khovanova's Math Blog

I’ve got this puzzle from Nick Petry.

Captain Flint is dying. All his treasures are buried far away. He only has 99 pieces of gold with him. Filled with remorse at the last moments of his life, he decides that he only wants to take one piece of gold with him to his grave. The rest of the gold he will give to the families of two men that he had killed the day before.

Though Captain Flint is heavily drunk he notices that no matter which piece he takes for himself, he can divide the leftover 98 pieces into two piles of 49 pieces each of the same weight. Prove that all the gold pieces are of the same weight.

For an additional challenge, Sasha Shapovalov suggested the following generalization of the previous puzzle.

Captain Flint has N gold pieces and yesterday he killed not two but K men. He wants to take one piece with him to his grave and to divide the rest into equi-weighted piles, not necessarily of the same number of pieces. If he can choose any piece to take with him and is able to divide the rest, prove that N – 1 is divisible by K.

Both of these puzzles can be easily solved if the weight of every gold piece is an integer or even a rational number. If you don’t assume that the weights are rational numbers, then I do not know an elementary solution, but I do know a simple and beautiful solution using linear algebra for both puzzles.

Even pirates need linear algebra to divide their treasure. Hooray for linear algebra!

Here is a logic puzzle for kids:

— John has more than a thousand books, said Pete.
— No, he has less than one thousand books, said Ann.
— Surely, he has at least one book, said Mary.

If only one statement is true, are you sure you know how many books John has?

Here is a math problem from the 1977 USSR Math Olympiad:

Let A be a 2n-digit number. We call this number special if it is a square and a concatenation of two n-digits squares. Also, the first n-digit square can’t start with zero; the second n-digit square can start with zero, but can’t be equal to zero.

• Find all two- and four-digit special numbers.
• Prove that there exists a 20-digit special number.
• Prove that not more than ten 100-digit special numbers exist.
• Prove that there exists a 30-digit special number.

Obviously, these questions are divided into two groups: show the existence and estimate the bound. Furthermore, this problem can be naturally divided into two other groups. Do you see them? The puzzle about special numbers makes a special day today — you get a four-in-one puzzle.

David Bernstein gave me this puzzle. He says that the puzzle was given at a Moscow math Olympiad a long time ago. At that time there were no computer science olympiads yet. I do not know why this puzzle feels to me like computer science. Maybe because the trivial solution is of order N, the easy solution is of order square root of N and the requested solution is of order logarithm of N:

Can you cut a square into N convex pieces minimizing the number of possible intersections of any straight line with your pieces?

It is easy to maximize the number of intersections. If you cut your square with N-1 parallel cuts into N equal thin rectangles, then there exists a line with N intersections.

It is easy to cut a square to guarantee no more than 2√N intersections. Can you cut your square so that any line makes no more than 2log₂N intersections?

A puzzle from the 1977 USSR math Olympiad can be solved naturally with linear algebra:

Seven dwarfs are sitting at a round table. Each dwarf has a cup partially filled with milk. Each dwarf in turn divides all his milk evenly between the six other cups. After the seventh dwarf has done this, every cup happens to contain the initial amount of milk. What was the initial distribution of milk?

Can you use linear algebra to intelligently solve this puzzle?

This puzzle is a generalization of a problem from the 1977 USSR math Olympiad:

At the beginning of the game you are given a polynomial, which has 1 as its leading coefficient and 1 as its constant term. Two people play. On your turn you assign a real value to one of the unknown coefficients. The person that goes last wins if the polynomial has no real roots at the end. Who wins?

It is clear that if the last person’s goal is for the polynomial to have a root, then the game is trivial: in this case, he can always make 1 a root with the last move. Also, an odd degree polynomial always has a real root. Therefore, to make the game interesting we should assume that the degree of the polynomial is even.

Though I can’t imaging myself ever being interested in playing this game, figuring out the strategy is a lot of fun.

Here is a math puzzle for kids from a nice collection of ThinkFun Visual Brainstorms:

Wendy has cats. All but two of them are Siamese, all but two of them are Persian, and all but two of them are Maine Coon. How many cats does Wendy have altogether?

This puzzle has two answers: the expected answer and an unexpected answer. Can you find both?

Here is a puzzle that my ex-brother-in-law, Dodik, gave to me today:

Prove that every group with more than two elements has a non-trivial automorphism.

I usually love puzzles that are solved with a counter-intuitive brilliant idea. This puzzle is different — I didn’t solve it in one elegant swoop. But I still love the puzzle: it feels so natural, and it’s solution feels so natural, that I even decided to call this puzzle “organic.” Or, maybe, I am just in an organic mood today waiting for my organic bananas to be delivered from Boston Organics.

Here is a funny puzzle from Kvant (1996, vol 4), the best Russian journal of recreational math.

When Alice goes through the looking-glass, she might meet many multi-headed, multi-armed, multi-legged beings. A being with H heads, A arms and L legs is considered:

• smart, if H > A+L,
• strong, if A > H+L,
• fast, if L > H+A.

Can there exist a well-rounded personality behind the looking-glass: someone who is smart, strong and fast at the same time?

One of Microsoft’s biggest contributions to humanity is the popularization of manhole covers. The most famous question that Microsoft asks during job interviews of geeks is probably, “Why are manhole covers round?” Supposedly the right answer is that if a manhole cover is round it can’t be dropped into the hole. See, for example, How Would You Move Mount Fuji? Microsoft’s Cult of the Puzzle – How the World’s Smartest Company Selects the Most Creative Thinkers. This book by William Poundstone is dedicated exclusively to Microsoft’s interview puzzles.

All we need, actually, is for the cover not to fit into the hole. For example, if the hole is small, the cover could be almost any shape, as long as the diameter of the cover is bigger than any straight segment that fits into the hole.

Microsoft makes an implicit assumption that the cover is about the same shape and size as the hole; otherwise we would waste a lot of extra cover material.

Even with this assumption, there is a good deal of flexibility in possible shapes if our only concern is that the cover shouldn’t fit into the hole. It is sufficient for the cover to be any shape with a constant width.

Here are some additional answers to why the cover should be round. Microsoft accepts the answer that a round cover is easier to roll. I’m not sure why a cover would ever have to be moved away from its hole. But I agree that if kids try to steal a cover, it would be much easier to escape with a round one.

Another answer that Microsoft supposedly accepts is that you do not need to rotate a round cover to align it with the opening when you are putting it back. This way if there is a lane divider painted on the cover it will point in a new random direction.

You can find many other explanations at the wiki article devoted to this subject. The most reasonable is that manhole covers are round because manholes are round. Duh!

Thanks to Microsoft there are now many websites with pictures of and discussions on the shape of manhole covers. For example, Manhole Covers Etc. or Manhole.ca or Manhole Covers of the World. As you can see many manhole covers are square or rectangular. They say that New Hampshire had triangular covers at some point.

But my favorite answer to this interview question was sent to me by Jorge Tierno:

Since manhole covers are not necessarily round, but you are asking why they are round, you are probably asking why round manhole covers are round. Round manhole covers are round by definition.

Now I have my own favorite question to ask you during a job interview: “Why are some manhole covers square?”