Tanya Khovanova's Math Blog

I ran an experiment. I copied multiple choices from the 2007 AMC 8 into a file and asked my son Sergei to try to guess the answers, looking only at the choices. I allowed him to keep several choices. The score I assigned depended on the number of leftover choices. If the leftover choices didn’t contain the right answer, the score for the problem was zero. Otherwise, it was scaled according to the number of choices he left. For example, if he had four choices left and the right answer was among them he got 1/4 of a point. Here are the choices:

1. 9, 10, 11, 12, 13.
2. 2/5, 1/2, 5/4, 5/3, 5/2.
3. 2, 5, 7, 10, 12.
4. 12, 15, 18, 30, 36.
5. 24, 25, 26, 27, 28.
6. 7, 17, 34, 41, 80.
7. 25, 26, 29, 33, 36.
8. 3, 4.5, 6, 9, 18.
9. 1, 2, 3, 4, cannot be determined.
10. 13, 20, 24, 28, 30.
11. Choose picture: I, II, III, IV, cannot be determined.
12. 1:1, 6:5, 3:2, 2:1, 3:1.
13. 503, 1006, 1504, 1507, 1510.
14. 5, 8, 13, 14, 18.
15. a+c < b, ab < c, a+b < c, ac < b, b/c = a.
16. Choose picture: 1, 2, 3, 4, 5.
17. 25, 35, 40, 45, 50.
18. 2, 5, 6, 8, 10.
19. 2, 64, 79, 96, 131.
20. 48, 50, 53, 54, 60.
21. 2/7, 3/8, 1/2, 4/7, 5/8.
22. 2, 4.5, 5, 6.2, 7.
23. 4, 6, 8, 10, 12.
24. 1/4, 1/3, 1/2, 2/3, 3/4.
25. 17/36, 35/72, 1/2, 37/72, 19/36.

It is clear that if you keep all choices, your score will be 5, which is the expected score for AMC if you are randomly guessing the answers. Sergei’s total score was 7.77, which is noticeably better than the expected 5.

There were two questions where Sergei felt that he knew the answer exactly. First was question number two with choices: 2/5, 1/2, 5/4, 5/3, 5/2. All but one of the choices has a 5 in it, so 1/2 must be wrong. Numbers 2/5 and 5/2 are inverses of each other, so if organizers expect you to make a mistake by inverting the right answer, then one of these choices must be the right answer. But 5/4 and 5/3 are better suited as a miscalculation of 5/2, than of 2/5. His choice was 5/2, and it was correct. The second question for which he was sure of the answer was question 19, with his answer 79. I still do not have a clue why.

Sergei’s result wasn’t too much better than just guessing. That means that AMC 8 organizers do a reasonably good job of hiding the real answer. You can try it at home and see if you can improve on Sergei’s result. I will publish the right answers as a comment to this essay in a week or so.

My knowledge about vampires comes mostly from the two TV series Buffy The Vampire Slayer and Angel. If you saw these series you would know that vampires can’t stand the sun. Therefore, they can’t get any tan at all and should be very pale. Angel doesn’t look pale but I never saw him going to a tanning spa. Nor did I ever see him taking vitamin D, as he should if he’s avoiding the sun.

But this is not why I’m confused about vampires. My biggest concerns are about vampires that are numbers.

Vampire numbers were invented by Clifford A. Pickover, who said:

If we are to believe best-selling novelist Anne Rice, vampires resemble humans in many respects, but live secret lives hidden among the rest of us mortals. Consider a numerical metaphor for vampires. I call numbers like 2187 vampire numbers because they’re formed when two progenitor numbers 27 and 81 are multiplied together (27 * 81 = 2187). Note that the vampire, 2187, contains the same digits as both parents, except that these digits are subtly hidden, scrambled in some fashion.

Some people call the parents of a vampire number fangs. Why would anyone call their parents fangs? I guess some parents are good at blood sucking and because they have all the power, they make the lives of their children a misery. So which name shall we use: parents or fangs?

Why should parents have the same number of digits? Maybe it’s a gesture of gender equality. But there is no mathematical reason to be politically correct, that is, for parents to have the same number of digits. For example, 126 is 61 times 2 and thus is the product of two numbers made from its digits. Pickover calls 126 a pseudovampire. So a pseudovampire with asymmetrical fangs, is a disfigured vampire, one whose fangs have a different number of digits. Have you ever seen fangs with digits?

In the first book where vampires appeared Keys to Infinity the vampire numbers are called true vampire numbers as opposed to pseudovampire numbers.

We can add a zero at the end of a pseudovampire to get another pseudovampire, a trivial if obvious observation. To keep the parents equal, we can add two zeroes at the end of a vampire to get another vampire. Adding zeroes is not a very intellectual operation, but a vampire that can’t be created by adding zeroes to another vampire is more basic and, thus, more interesting. In the book Wonders of Numbers: Adventures in Mathematics, Mind, and Meaning a vampire where one of the multiplicands doesn’t have trailing zeroes is called a true vampire, as opposed to just a vampire. Thus, the trueness of vampires changes from book to book, adding some more confusion. It looks like the second definition of a true vampire is more widely adopted, so I will stick to it.

By analogy, we should call pseudovampires that do not end in zeroes, true pseudovampires. It’s interesting to note that by adding zeroes we can get a true vampire from any pseudovampire that is not a vampire. You see how easy it is to build equality? Just add zeroes.

A true vampire might not be true as a pseudovampire. For example, a vampire number 1260 = 20 * 61 is generated by adding a zero to a pseudovampire 126 = 2 * 61. In this case, the pseudovampire is truer than the vampire. Why does something more basic get a prefix “pseudo”?

Here’s another question. Why do vampires have to have two fangs? Can a vampire have three fangs? For example, 11439 = 9 * 31 * 41. This generalization of vampires should be called mutant vampires. Or multi-gender vampires.

To create more confusion, a mutant vampire can, at the same time, be a simple vampire: 1395 = 31 * 9 * 5 = 15 * 93.

Of course, nothing prevents a mutant vampire from being politically correct, that is, to have multiple and equal parents with the same number of digits, as in 197925 = 29 * 75 * 91.

People continue creating a mess with vampires. For example, a definition of a prime vampire number is floating around the Internet. When you look at this name, your first reaction is that a prime vampire is a prime number. But a vampire is never prime as it is always a product of numbers. By definition a prime vampire is a vampire with prime multiplicands, for example 124483 = 281 * 443. So “prime vampire number” is a very bad name. We should call these vampires prime-fanged vampires — this would be much more straightforward.

To eliminate some of this confusion, we mathematicians should go back and rename vampires consistently. But in the meantime, check out the illustration of vampire numbers shown above that I found at flickr.com with this description:

Like the count von Count in Sesame Street, there is a tradition that vampires suffer terribly from arithromania: the compulsion to count things. To keep vampires from wreaking murderous havoc at night, poppy seeds were strewn about their resting places. On waking, the vampire would be compelled to count the seeds. It would take him all night, and keep him from mischief.

My knowledge about vampires comes mostly from the two TV series Buffy The Vampire Slayer and Angel . If you saw these series you would know that vampires can’t stand the sun. Therefore, they can’t get any tan at all and should be very pale. Angel doesn’t look pale but I never saw him going to a tanning spa. Nor did I ever see him taking vitamin D, as he should if he’s avoiding the sun.

But this is not why I’m confused about vampires. My biggest concerns are about vampires that are numbers.

Vampire numbers were invented by Clifford A. Pickover, who said:

If we are to believe best-selling novelist Anne Rice , vampires resemble humans in many respects, but live secret lives hidden among the rest of us mortals. Consider a numerical metaphor for vampires. I call numbers like 2187 vampire numbers because they’re formed when two progenitor numbers 27 and 81 are multiplied together (27 * 81 = 2187). Note that the vampire, 2187, contains the same digits as both parents, except that these digits are subtly hidden, scrambled in some fashion.

Some people call the parents of a vampire number fangs. Why would anyone call their parents fangs? I guess some parents are good at blood sucking and because they have all the power, they make the lives of their children a misery. So which name shall we use: parents or fangs?

Why should parents have the same number of digits? Maybe it’s a gesture of gender equality. But there is no mathematical reason to be politically correct, that is, for parents to have the same number of digits. For example, 126 is 61 times 2 and thus is the product of two numbers made from its digits. Pickover calls 126 a pseudovampire. So a pseudovampire with asymmetrical fangs, is a disfigured vampire, one whose fangs have a different number of digits. Have you ever seen fangs with digits?

In the first book where vampires appeared Keys to Infinity the vampire numbers are called true vampire numbers as opposed to pseudovampire numbers.

We can add a zero at the end of a pseudovampire to get another pseudovampire, a trivial if obvious observation. To keep the parents equal, we can add two zeroes at the end of a vampire to get another vampire. Adding zeroes is not a very intellectual operation, but a vampire that can’t be created by adding zeroes to another vampire is more basic and, thus, more interesting. In the book Wonders of Numbers: Adventures in Mathematics, Mind, and Meaning a vampire where one of the multiplicands doesn’t have trailing zeroes is called a true vampire, as opposed to just a vampire. Thus, the trueness of vampires changes from book to book, adding some more confusion. It looks like the second definition of a true vampire is more widely adopted, so I will stick to it.

By analogy, we should call pseudovampires that do not end in zeroes, true pseudovampires. It’s interesting to note that by adding zeroes we can get a true vampire from any pseudovampire that is not a vampire. You see how easy it is to build equality? Just add zeroes.

A true vampire might not be true as a pseudovampire. For example, a vampire number 1260 = 20 * 61 is generated by adding a zero to a pseudovampire 126 = 2 * 61. In this case, the pseudovampire is truer than the vampire. Why does something more basic get a prefix “pseudo”?

Here’s another question. Why do vampires have to have two fangs? Can a vampire have three fangs? For example, 11439 = 9 * 31 * 41. This generalization of vampires should be called mutant vampires. Or multi-gender vampires.

To create more confusion, a mutant vampire can, at the same time, be a simple vampire: 1395 = 31 * 9 * 5 = 15 * 93.

Of course, nothing prevents a mutant vampire from being politically correct, that is, to have multiple and equal parents with the same number of digits, as in 197925 = 29 * 75 * 91.

People continue creating a mess with vampires. For example, a definition of a prime vampire number is floating around the Internet. When you look at this name, your first reaction is that a prime vampire is a prime number. But a vampire is never prime as it is always a product of numbers. By definition a prime vampire is a vampire with prime multiplicands, for example 124483 = 281 * 443. So “prime vampire number” is a very bad name. We should call these vampires prime-fanged vampires — this would be much more straightforward.

To eliminate some of this confusion, we mathematicians should go back and rename vampires consistently. But in the meantime, check out the illustration of vampire numbers shown above that I found at flickr.com with this description:

Like the count von Count in Sesame Street, there is a tradition that vampires suffer terribly from arithromania: the compulsion to count things. To keep vampires from wreaking murderous havoc at night, poppy seeds were strewn about their resting places. On waking, the vampire would be compelled to count the seeds. It would take him all night, and keep him from mischief.

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.

Once I witnessed John H. Conway factoring large numbers in his head. Impressed, I stared at him. Encouraged by my interest, he told me that if I ever want to be able to factor large numbers, I should know all the primes below one thousand.

The secret to knowing all such primes is to remember the composites, he continued. Obviously, we don’t need to remember trivial composites — the ones divisible by 2, 3, 5, or 11. Also, everyone knows all the squares below one thousand, so we can count squares as trivial composites. We only need to remember the non-trivial composites. There are not that many of them below one thousand — only 70. I mean, 70 is nothing compared to the number of primes: 168.

So, I need to remember the following seventy numbers:

91, 119, 133, 161, 203, 217, 221, 247, 259, 287, 299, 301, 323, 329, 343, 371, 377, 391, 403, 413, 427, 437, 469, 481, 493, 497, 511, 527, 533, 551, 553, 559, 581, 589, 611, 623, 629, 637, 667, 679, 689, 697, 703, 707, 713, 721, 731, 749, 763, 767, 779, 791, 793, 799, 817, 833, 851, 871, 889, 893, 899, 901, 917, 923, 931, 943, 949, 959, 973, 989.

If you are very ambitious and plan to learn the primes up to 50,000, then the trick of learning non-trivial composites instead of primes is of no use to you. Indeed, for larger numbers the density of primes goes down, while the density of non-trivial composites stays about the same or increases very slightly due to a smaller number of squares.

The turning point is around 11,625: the number of primes below 11,625 equals the number of non-trivial composites below it. So, compare your ambition to 11,625 and tailor your path of learning accordingly.

If you are lazy, you can learn primes only up to 100. In this case your path is clear; you should stick with remembering non-trivial composites, for you need to remember only one number: 91.

For his Master’s thesis, my son Alexey Radul wrote the Symmetriad — a computer program to compute and display highly symmetric objects.

The objects the Symmetriad is after are the 4D generalizations of Platonic and Archimedean solids. His thesis contains a picture gallery made with the Symmetriad, and these are my three favorite pictures.

The pictures have cute titles. The problem is that I still haven’t installed the WordPress plugin that allows me to put captions under the pictures. That is why you have to guess which title matches which picture. The titles are: “The Planets are Aligned,” “Great Jaws,” and “Diamonds are Forever.”

If you are wondering why one of the pictures shows a non-connected object, in fact the object is connected, but some of the edges are white, so that you can better see 3D cells of the object.

Subsequent to the publication of this thesis, Alexey enhanced his software to make even more dramatic pictures. The following pictures have no titles, so feel free to suggest some:

I think it would be nice to publish a book with all these pictures. As a book on symmetries should be published on a symmetric date, the next opportunity would be 01.02.2010.

I love Grand Tour puzzles more than I love Sudoku. You are given a graph, for example, a square grid like the one on the left. Some edges are highlighted. You need to find a closed path that visits each vertex exactly once and includes the highlighted edges as part of the path. Mathematically speaking you need to reconstruct a Hamiltonian circuit on a graph, once you are given a part of it. The highlighted edges are chosen to guarantee a unique solution to the puzzle.

On the left you can see a sample grand tour problem with its solution. This puzzle was designed by Glenn Iba. On Glenn’s Grand Tour Puzzle Page you can find many grand tour puzzles of varying levels of difficulty. The puzzles are playable. That is, you can click or unclick an edge. You can also branch out in a different color, which is especially useful for difficult puzzles when you want to test a hypothesis. I just want to warn you: these puzzles are addictive — I couldn’t stop playing until I solved all of them.

Below there is a simple grand tour puzzle from Glenn’s collection, but this time on a triangular grid:

You do not need a grid to construct a puzzle. But these puzzles look very natural on grids. I tried to analyze square grid puzzles a little bit. The first important point is that for square grids with odd number of vertices on each side of the square, Hamiltonian cycles do not exist. This point is easier to prove for directed Hamiltonian cycles. You can make a directed cycle from an undirected one by choosing a direction. If you have a directed cycle on a square grid, then the number of edges pointing up should be the same as the number of edges pointing down. We can say the same thing about edges on the cycle pointing left or right. Hence, the number of edges of a Hamiltonian cycle on a square grid should be even. At the same time, the number of edges of any Hamiltonian cycle equals the number of vertices.

I just proved that you need only consider square grids with an even number of vertices on each side. For square grids with two vertices on each side, there is only one Hamiltonian cycle, namely the border of the square. The only grand tour puzzle for this grid won’t have highlighted edges at all. For a square grid with four vertices on each side there are only two different Hamiltonian cycles up to isomorpshisms:

If we count all the reflections and rotations, we will get six Hamiltonian cycles. The next picture shows all 11 grand tour puzzles on this grid. If we count rotations and reflections, we will get 66 different grand tour puzzles.

Below are the sequences associated with this puzzle. Except for one case, I do not know if these sequences are in the Online Encyclopedia for Integer Sequences. I don’t know because I only counted two terms of each sequence, and this information is not sufficient to identify the sequence.

• A003763 Number of Hamiltonian cycles on 2n by 2n square grid of points. The sequence starts 1, 6, 1072, ….
• Number of Hamiltonian cycles up to isomorphism on 2n by 2n square grid of points. The sequence starts 1, 2, ….
• Number of Grand Tour puzzles on 2n by 2n square grid of points. The sequence starts 1, 11, ….
• Number of Grand Tour puzzles up to isomorphism on 2n by 2n square grid of points. The sequence starts 1, 66, ….
• The smallest number of edges in a Grand Tour puzzle on 2n by 2n square grid of points. The sequence starts 0, 2, ….
• The largest number of edges in a Grand Tour puzzle on 2n by 2n square grid of points. The sequence starts 0, 4, ….

If you look at Glenn Iba’s 6 by 6 square grid puzzles, you can see that the smallest number of edges is not more than 6. And the largest number of edges is no less than 12.

You can also make similar sequences for a triangular grid.

I invite you to calculate these sequences and submit them to the OEIS, if they are not already there.

I’m not kidding; there is such a paper. It is titled, “A Headache-causing Problem” and its authors are Conway (J.H.), Paterson (M.S.), and Moscow (U.S.S.R.). The acknowledgements in the paper shed some light on how Moscow became a mathematician:

The work described here was carried out when the first and second named authors enjoyed the hospitality of the third. The second and third authors are indebted to the first for expository details. The first and third authors gratefully remark that without the constant stimulation and witty encouragement of the second author this paper

[The next part was meant to be on the following page, Conway told me, but the editor missed the humor and just continued the sentence…]

was completed.

As a consequence of this joke, Moscow is envied by many mathematicians as it has an Erdős number of 2. Now wait for a couple of hundred years, and Moscow will be the only living mathematician with an Erdős number of 2. I can just imagine future mathematicians trying to persuade Moscow to coauthor papers with them, because this will be the only way for them to score an Erdős number of 3.

Even though I lived there for 30 years, I had no idea that Moscow had a talent for math. Of course, this talent only emerged when Moscow was more than 800 years old.

This wonderful paper by Moscow was very difficult to find. It was presented to Hendrik W. Lenstra on the occasion of his doctoral examination. It was published in 1977 in a book titled “Een pak met een korte broek,” which in Dutch means, “A Book in Short Trousers.”

I tried to find it on the Internet — it wasn’t there. I asked John Conway — it took him quite some time to find it. Here is the picture of John Conway searching for a headache-causing problem. Luckily for you and me, he found it. To save you from another headache, I am uploading the scan of it in pdf format here: A Headache-causing Problem by J.H. Conway, M.S. Paterson, and U.S.S.R. Moscow.

I hope that Moscow will not start complaining that I never asked its permission to post the paper. Some might argue that Moscow, U.S.S.R., doesn’t exist anymore, but I would counter that it exists, but with a changed name. If Moscow tries to sue me, I hope it’s not because it is still bitter that I left it behind in 1990.

Hey Moscow, it’s time we were friends again. Would you like to co-author a paper with me?

The Symmetries of Things by John H. Conway, Heidi Burgiel, and Chaim Goodman-Strauss is out. It is a beautiful edition with great pictures.

The first chapter is very nicely written and might be recommended to high school and undergraduate students. It covers symmetries of finite and infinite 2D objects.

The second chapter adds color to the theory. For beautiful colorful pictures with symmetry, there are two symmetry groups: the group that preserves the picture while ignoring its coloring and the group that preserves the picture while respecting its coloring. The latter group is a subgroup of a former group. This second chapter discusses all possible ways to symmetrically color a symmetric 2D picture. The chapter then continues with a discussion of group theory. This chapter is much more difficult to read than the first chapter, as it uses a lot of notations. The pictures are still beautiful, though.

The third chapter is even more difficult to read and the notations become even heavier. This chapter discusses hyperbolic groups and symmetries of objects in the hyperbolic space. Then the chapter moves into 3D and 4D. I guess that some parts of the second and the third chapters are not meant for light reading; they should be considered more as reference materials.

Here are pictures of a musnub cube (multiplied snub cube), built by John H. Conway. It is an infinite Archimedean polyhedron. The description of it is on page 338 and the diagram is on page 339 of the book. This object was glued together from stars and squares. Each corner of the square is glued to a point of one star and to an inside corner of another star. Mathematically, a star is not a regular polygon. If you look at stars with your mathematical eye, each star in the picture is not just a star, but rather the union of a regular hexagon with 6 regular triangles. That means the list of the face sizes around each vertex of a musnub cube officially is represented as 6.3.4.3.3.

The second picture shows a finished musnub cube. You can’t really finish building an infinite structure. Right? It is finished in the sense that John Conway finished doing what he was planning to do: to construct a part of a musnub cube inside a regular triangular pyramid.

Did I mention that I like the pictures in The Symmetries of Things?

Here I repeat the Conway’s Wizards Puzzle from a previous posting:

Last night I sat behind two wizards on a bus, and overheard the following:

— A: “I have a positive integral number of children, whose ages are positive integers, the sum of which is the number of this bus, while the product is my own age.”
— B: “How interesting! Perhaps if you told me your age and the number of your children, I could work out their individual ages?”
— A: “No.”
— B: “Aha! AT LAST I know how old you are!”

Now what was the number of the bus?

It is obvious that the first wizard has more than two children. If he had one child then his/her age would be the number of the bus and it would be the same as the father’s age. While it is unrealistic, in mathematics many strange things can happen. The important part is that if the wizard A had one child he couldn’t have said ‘No’. The same is true for two children: their age distribution is uniquely defined by the sum and the product of their ages.

Here is a generalization of this puzzle:

Last night I sat behind two wizards on a bus, and overheard the following:

— Wizard A: “I have a positive integral number of children, whose ages are positive integers, the sum of which is the number of this bus, while the product is my own age. Also, the sum of the squares of their ages is the number of dinosaurs in my collection.”
— Wizard B: “How interesting! Perhaps if you told me your age, the number of your children, and the number of dinosaurs, I could work out your children’s individual ages. ”
— Wizard A: “No.”
— Wizard B: “Aha! AT LAST I know how old you are!”

Now what was the number of the bus?

As usual with generalizations, they are drifting far from real life. For this puzzle, you have to open up your mind. In Conway’s original puzzle you do not need to assume that the wizard’s age is in a particular range, but once you solve it, you see that his age makes sense. In this generalized puzzle, you should assume that wizards can live thousands of years, and keep their libido that whole time. Wizards might spend so much of their youth thinking, that they postpone starting their families for a long time. The wizards’ wives are also generalized. They can produce children in great quantities and deliver multiple children at the same time in numbers exceeding the current world record.

Another difference with the original puzzle is that you can’t solve this one without a computer.

You can continue to the next step of generalization and create another puzzle by adding the next symmetric polynomial on the ages of the children, for example, the sum of cubes. In this case, I do not know if the puzzle works: that is, if there is an “AHA” moment there. I invite you, mighty geeks, to try it. Please, send me the answer.

In case you are wondering why the wizard is collecting dinosaurs, I need to point out to you that John H. Conway is a superb puzzle inventor. His puzzle includes a notation suggestion: a for the wizard’s age, b for the bus, c for the number of children. Hence, the dinosaurs.

Here is my simplified version of Conway’s wizards puzzle.

Last night when I was coming home from my writing class with Sue Katz, I sat behind two wizards on the bus, and overheard the following:

— Wizard A: “I have a positive integral number of children, whose ages are positive integers, the sum of which is the number of this bus, while the product is the amount of dollars I have in my pocket.”

At this point I interrupted the wizard. “Excuse me, professor, I overheard your conversation and can’t resist asking you a question. Usually when a father says ‘my children’ everyone assumes that he has at least two children. Can I assume that?”

— Wizard A: “No. I stated my assumptions up front. A positive integral number of children means one or more.”

I started thinking. If I were to explain this to a non-mathematician who assumes that ‘my children’ means more than one child, I would need to change the wizard’s statement into the following:

“I have at least one child. The ages of my one-or-more children are all positive integers. The sum of the ages of my children or the age of my only child is the number of this bus. The product of the ages of my children or my only child’s age is the amount of dollars I have in my pocket.”

Hmm. I like that mathematicians use ‘my children’ to indicate any number of children. Makes puzzles faster to type.

Anyway, the wizards continued their discussion:

— Wizard B: “How interesting! Perhaps if you told me the number of your children, I could work out their individual ages”
— Wizard A: “No.”
— Wizard B: “Aha! AT LAST I know how many children you have!”

If I were John Conway, I would have asked you next, “What is the number of the bus?” As I am not John Conway, I’ll ask you, “Why do we presume that Wizard A hasn’t cheated on his wife?”

The answer is that all wizards are notorious for making precise statements. If he cheats a lot, he would have started the conversation with, “The number of children I know about is a positive integer.” Or maybe, more discreetly, “My wife and I have a positive integral number of children.”

If you have already figured out the number of the bus, the bonus question is, “Why did I change the ‘age of the first wizard’ in Conway’s original puzzle into the ‘amount of dollars’ in my puzzle?”

When I left the bus, I started wondering why on earth anyone would ever want to sum up the ages of their children. And I remembered that I once did it myself. I was trying to persuade my sister to apply for U.S. citizenship. My argument was that by moving here the life expectancy of her children would increase by 30 years. Indeed, she has two sons and the male life expectancy in Russia and the U.S. has an astonishing 15-year difference. I have to admit that my argument is not very clean, as we do not know the causes for this difference and, besides, the data is for life expectancy at birth and it changes while our kids age. My sister dismissed my argument, saying that the low male life expectancy in Russia is due to alcoholism and that her family is not in the high-risk group.

So, there could be a reason to sum up the ages of your children, but why would anyone ever want to multiply the ages of their children? In any case, if the first wizard continues to keep an amount of dollars equaling the product of the ages of his children in his pocket, his pocket will do better than mutual funds for the next several years.