Someone Else’s Homework: A Probability Question


My friend’s finished the last of the exams and been happy with the results. And I’m stuck thinking harder about a little thing that came across my Twitter feed last night. So let me share a different problem that we had discussed over the term.

It’s a probability question. Probability’s a great subject. So much of what people actually do involves estimating probabilities and making judgements based on them. In real life, yes, but also for fun. Like a lot of probability questions, this one is abstracted into a puzzle that’s nothing like anything anybody does for fun. But that makes it practical, anyway.

So. You have a bowl with fifteen balls inside. Five of the balls are labelled ‘1’. Five of the balls are labelled ‘2’. Five of the balls are labelled ‘3’. The balls are well-mixed, which is how mathematicians say that all of the balls are equally likely to be drawn out. Three balls are picked out, without being put back in. What’s the probability that the three balls have values which, together, add up to 6?

My friend’s instincts about this were right, knowing what things to calculate. There was part of actually doing one of these calculations that went wrong. And was complicated by my making a dumb mistake in my arithmetic. Fortunately my friend wasn’t shaken by my authority, and we got to what we’re pretty sure is the right answer.

Reading the Comics, May 20, 2017: Major Computer Malfunction Week Edition


I was hit by a massive computer malfunction this week, the kind that forced me to buy a new computer and spend half a week copying stuff over from a limping hard drive and hoping it would maybe work if I held things just right. Mercifully, Comic Strip Master Command gave me a relatively easy week. No huge rush of mathematically-themed comic strips and none that are going to take a thousand words of writing to describe. Let’s go.

Sam Hepburn’s Questionable Quotebook for the 14th includes this week’s anthropomorphic geometry sketch underneath its big text block.

Eric the Circle for the 15th, this one by “Claire the Square”, is the rare Eric the Circle to show off properties of circles. So maybe that’s the second anthropomorphic geometry sketch for the week. If the week hadn’t been dominated by my computer woes that might have formed the title for this edition.

Werner Wejp-Olsen’s Inspector Danger’s Crime Quiz for the 15th puts a mathematician in mortal peril and leaves him there to die. As is traditional for this sort of puzzle the mathematician left a dying clue. (Mathematicians were similarly kind to their investigators on the 4th of July, 2016 and the 9th of July, 2012. I was expecting the answer to be someone with a four-letter and an eight-letter name, none of which anybody here had. Doesn’t matter. It’ll never stand up in court.

John Graziano’s Ripley’s Believe It Or Not for the 17th features one of those astounding claims that grows out of number theory. Graziano asserts that there are an astounding 50,613,244,155,051,856 ways to score exactly 100 points in (ten-pin) bowling. I won’t deny that this seems high to me. But partitioning a number — that is, taking a (positive) whole number and writing down the different ways one can add up (positive) whole numbers to get that sum — often turns up a lot of possibilities. That there should be many ways to get a score of 100 by adding between ten and twenty numbers that could be between zero and ten each, plus the possibility of adding pairs of the numbers (for spares) or trios of numbers (for strikes) makes this less astonishing.

Wikipedia led me to this page, from Balmoral Software, about all the different ways there are to score different numbers. The most surprising thing it reveals to me is that 100 isn’t even the score with the greatest number of possible scores. 77 is. There are 172,542,309,343,731,946 ways to score exactly 77 points. I agree this ought to make me feel better about my game. It doesn’t. It turns out there are, altogether, something like 5,726,805,883,325,784,576 possible different outcomes for a bowling game. And how we can tell that, given there’s no practical way to go and list all of them, is described at the end of the page.

The technique is called “divide and conquer”. There’s no way to list all the outcomes of ten frames of bowling, but there’s certainly a way to list all the outcomes of one. Or two. Or three. So, work out how many possible scores there would be in few enough frames you can handle that. Then combine these shortened games into one that’s the full ten frames. There’s some trouble in matching up the ends of the short games. A spare or a strike in the last frame of a shortened game means one has to account for the first or first two frames of the next one. But this is still an easier problem than the one we started with.

Bill Amend’s FoxTrot Classics for the 18th (rerun from the 25th of May, 2006) is your standard percentages and infinities joke. Really would have expected Paige’s mother to be wise to this game by now, but this sort of thing happens.

Reading the Comics, July 6, 2016: Another Busy Week Edition


It’s supposed to be the summer vacation. I don’t know why Comic Strip Master Command is so eager to send me stuff. Maybe my standards are too loose. This doesn’t even cover all of last week’s mathematically-themed comics. I’ll need another that I’ve got set for Tuesday. I don’t mind.

Corey Pandolph and Phil Frank and Joe Troise’s The Elderberries rerun for the 3rd features one of my favorite examples of applied probability. The game show Deal or No Deal offered contestants the prize within a suitcase they picked, or a dealer’s offer. The offer would vary up or down as non-selected suitcases were picked, giving the chance for people to second-guess themselves. It also makes a good redemption game. The banker’s offer would typically be less than the expectation value, what you’d get on average from all the available suitcases. But now and then the dealer offered more than the expectation value and I got all ready to yell at the contestants.

This particular strip focuses on a smaller question: can you pick which of the many suitcases held the grand prize? And with the right setup, yes, you can pick it reliably.

Mac King and Bill King’s Magic in a Minute for the 3rd uses a bit of arithmetic to support a mind-reading magic trick. The instructions say to start with a number from 1 to 10 and do various bits of arithmetic which lead inevitably to 4. You can prove that for an arbitrary number, or you can just try it for all ten numbers. That’s tedious but not hard and it’ll prove the inevitability of 4 here. There aren’t many countries with names that start with ‘D’; Denmark’s surely the one any American (or European) reader is likeliest to name. But Dominica, the Dominican Republic, and Djibouti would also be answers. (List Of Countries Of The World.com also lists Dhekelia, which I never heard of either.) Anyway, with Denmark forced, ‘E’ almost begs for ‘elephant’. I suppose ’emu’ would do too, or ‘echidna’. And ‘elephant’ almost forces ‘grey’ for a color, although ‘white’ would be plausible too. A magician has to know how things like this work.

Werner Wejp-Olsen’s feature Inspector Danger’s Crime Quiz for the 4th features a mathematician as victim of the day’s puzzle murder. I admit I’m skeptical of deathbed identifications of murderers like this, but it would spoil a lot of puzzle mysteries if we disallowed them. (Does anyone know how often a deathbed identification actually happens?) I can’t make the alleged answer make any sense to me. Danger of the trade in murder puzzles.

Kris Straub’s Starship for the 4th uses mathematics as a stand-in for anything that’s hard to study and solve. I’m amused.

Edison Lee tells his grandfather how if the universe is infinitely large, then it's possible there's an exact duplicate to him. This makes his grandfather's head explode. Edison's father says 'I see you've been pondering incomprehensibles again'. Headless grandfather says 'I need a Twinkie.'
John Hambrock’s The Brilliant Mind of Edison lee for the 6th of July, 2016. I’m a little surprised the last panel wasn’t set on a duplicate Earth where things turned out a little differently.

John Hambrock’s The Brilliant Mind of Edison lee for the 6th is about the existentialist dread mathematics can inspire. Suppose there is a chance, within any given volume of space, of Earth being made. Well, it happened at least once, didn’t it? If the universe is vast enough, it seems hard to argue that there wouldn’t be two or three or, really, infinitely many versions of Earth. It’s a chilling thought. But it requires some big suppositions, most importantly that the universe actually is infinite. The observable universe, the one we can ever get a signal from, certainly isn’t. The entire universe including the stuff we can never get to? I don’t know that that’s infinite. I wouldn’t be surprised if it’s impossible to say, for good reason. Anyway, I’m not worried about it.

Jim Meddick’s Monty for the 6th is part of a storyline in which Monty is worshipped by tiny aliens who resemble him. They’re a bit nerdy, and calculate before they understand the relevant units. It’s a common mistake. Understand the problem before you start calculating.

Another Reasoning Puzzle From ChefMongoose


My friend ChefMongoose had another reasoning problem come to him, and I’m happy to share it further. It’s rather like that famous Singapore Birthday Problem that drove people crazy a couple of months ago. Here’s the problem:

I have a combination lock at work. There are three digits, all in the range 1 – 40; they’re all prime numbers. They’re X+Y, X+2Y, X+3Y — where X and Y are positive integers.

If I told you what X was but not Y, you wouldn’t be able to tell me the combination. If I told you what Y was but not X, you wouldn’t be able to tell me the combination. Now, what’s the combination?

I did work out the puzzle. It did make me notice a couple of strings of uniformly-spaced prime numbers I hadn’t done before, too, such as 3-13-23. (However, 3-13-23 isn’t one of the possible answers, because of the constraints of the problem. There aren’t positive X and Y for which X + Y = 3, X + 2Y = 13, and X + 3Y = 23.)

As with the Singapore Birthday Problem, this is a puzzle based on reasoning about the information we have. Mercifully there aren’t actually many prime numbers below 40, so if you want you can take the brute force approach and find all the strings of uniformly-spaced prime numbers. Then you can find what one matches the rules in ChefMongoose’s second paragraph.

I confess I wasn’t that systematic. I had a strong suspicion what the starting number of the sequence had to be, and then did some tests to be sure. I credit that to just having stared at lot at the smaller prime numbers in my life, so I’d had some intuitive feel for it. That’s a dangerous way to work. My intuitive feel, for example, hadn’t warned me about 3-13-23. But then there aren’t other trios of prime numbers spaced by ten, so that set would be ruled out by the “If I told you what Y was but not X” constraint. But now I know how to get stuff out of ChefMongoose’s work locker, you know, just in case.

Reading the Comics, April 15, 2015: Tax Day Edition


Since it is mid-April, and most of the comic strips at Comics Kingdom and GoComics.com are based in the United States, Comic Strip Master Command ordered quite a few comics about taxes. Most of those are simple grumbling, but the subject naturally comes around to arithmetic and calculation and sometimes even logic. Thus, this is a Tax Day edition, though it’s bookended with Mutt and Jeff.

Bud Fisher’s Mutt And Jeff (April 11) — a rerun rom goodness only knows when, and almost certainly neither written nor drawn by Bud Fisher at that point — recounts a joke that has the form of a word problem in which a person’s age is deduced from information about the age. It’s an old form, but jokes about cutting the Gordion knot are probably always going to be reliable. I’m reminded there’s a story of Thomas Edison giving a new hire, mathematician, the problem of working out the volume of a light bulb. Edison got impatient with the mathematician treating it as a calculus problem — the volume of a rotationally symmetric object like a bulb is the sort of thing you can do by the end of Freshman Calculus — and instead filling a bulb with water, pouring the water into a graduated cylinder, and reading it off that.

Calculus under 50: vectors and stuff. Calculus over 50: diet and exercise problems.
Sandra Bell-Lundy’s Between Friends for the 12th of April, 2015. The link will likely expire around the 12th of May.

Sandra Bell-Lundy’s Between Friends (April 12) uses Calculus as the shorthand for “the hardest stuff you might have to deal with”. The symbols on the left-hand side are fair enough, although I’d think of them more as precalculus or linear algebra or physics, but they do parse well enough as long as I suppose that what sure looks like a couple of extraneous + signs are meant to refer to “t”. But “t” is a common enough variable in calculus problems, usually representing time, sometimes just representing “some parameter whose value we don’t really care about, but we don’t want it to be x”, and it looks an awful lot like a plus sign there too. On the right side, I have no idea what a root of forty minutes on a treadmill might be. It’s symbolic.

Continue reading “Reading the Comics, April 15, 2015: Tax Day Edition”

Denominated Mischief


I’ve finally got around to reading one of my Christmas presents, Alfred S Posamentier and Ingmar Lehman’s Magnificent Mistakes in Mathematics, which is about ways that mathematical reasoning can be led astray. A lot, at least in the early pages, is about the ways a calculation can be fowled by a bit of carelessness, especially things like dividing by zero, which seems like such an obvious mistake that who could make it once they’ve passed Algebra II?

They got to a most neat little erroneous calculation, though, and I wanted to share it since the flaw is not immediately obvious although the absurdity of the conclusion drives you to look for it. We begin with a straightforward problem that I think of as Algebra I-grade, though I admit my memories of taking Algebra I are pretty vague these days, so maybe I missed the target grade level by a year or two.

\frac{3x - 30}{11 - x} = \frac{x + 2}{x - 7} - 4

Multiply that 4 on the right-hand side by 1 — in this case, by \frac{x - 7}{x - 7} — and combine that into the numerator:

\frac{3x - 30}{11 - x} = \frac{x + 2 - 4(x - 7)}{x - 7}

Expand that parentheses and simplify the numerator on the right-hand side:

\frac{3x - 30}{11 - x} = \frac{3x - 30}{7 - x}

Since the fractions are equal, and the numerators are equal, therefore their denominators must be equal. Thus, 11 - x = 7 - x and therefore, 11 = 7.

Did you spot where the card got palmed there?

Another Reason Why It’s Got To Be 2


To circle back around that inscribed circle problem, about what the radius of the circle that just fits inside a right triangle with sides of length 5, 12, and 13: I’d had an approach for solving it different from HowardAt58’s geometric answer. This isn’t to imply that his answer’s wrong, I should point out: problems can often be solved by several different yet equally valid approaches. (It might almost be the definition of cutting-edge research if it’s a problem there’s only one approach for.)

A triangle with sides a, b, and c, with an inscribed circle, and three radial lines, one reaching to each side.
Figure 1. A triangle (meant to be a right triangle) with an inscribed circle of radius r. Three radial lines, perpendicular to the bases they touch, are included.

So here’s another geometry-based approach to finding what the radius of the circle that just fits inside the triangle has to be. We started off with the right triangle, and sides a and b and c; and there’s a circle inscribed in it. This is the biggest circle that’ll fit within the triangle. The circle has some radius, and we’ll just be a little daring and original and use the symbol “r” to stand for that radius. We can draw a line from the center of the circle to the point where the circle touches each of the legs, and that line is going to be of length r, because that’s the way circles work. My drawing, Figure 1, looks a little bit off because I was sketching this out on my iPad and being more exact about all this was just so, so much work.

The next step is to add three more lines to the figure, and this is going to make it easier to see what we want. What we’re adding are liens that go from the center of the circle to each of the corners of the original triangle. This divides the original triangle into three smaller ones, which I’ve lightly colored in as amber (on the upper left), green (on the upper right), and blue (on the bottom). The coloring is just to highlight the new triangles. I know the figure is looking even sketchier; take it up with how there’s no good mathematics-diagram sketching programs for a first-generation iPad, okay?

A triangle with sides a, b, and c, and an inscribed circle. From the center of the circle are lines going to the vertices of the triangle, dividing the circle into three smaller triangles, with bases of lengths, a, b, and c respectively and all with the same height, r, the radius of the inscribed circle.
Figure 2. A triangle (meant to be a right triangle) with an inscribed circle of radius r. The triangle is divided into three smaller triangles meeting at the center of the inscribed circle.

If we can accept my drawings for what they are already, then, there’s the question of why I did all this subdividing, anyway? The good answer is: looking at this Figure 2, do you see what the areas of the amber, green, and blue triangles have to be? Well, the area of a triangle generally is half its base times its height. A base is the line connecting two of the vertices, and the height is the perpendicular distane between the third vertex and that base. So, for the amber triangle, “a” is obviously a base, and … say, now, isn’t “r” the height?

It is: the radius line is perpendicular to the triangle leg. That’s how inscribed circles work. You can prove this, although you might convince yourself of it more quickly by taking the lid of, say, a mayonnaise jar and a couple of straws. Try laying down the straws so they just touch the jar at one point, and so they cross one another (forming a triangle), and try to form a triangle where the straw isn’t perpendicular to the lid’s radius. That’s not proof, but, it’ll probably leave you confident it could be proven.

So coming back to this: the area of the amber triangle has to be one-half times a times r. And the area of the green triangle has to be one-half times b times r. The area of the blue triangle, yeah, one-half times c times r. This is great except that we have no idea what r is.

But we do know this: the amber triangle, green triangle, and blue triangle together make up the original triangle we started with. So the areas of the amber, green, and blue triangles added together have to equal the area of the original triangle, and we know that. Well, we can calculate that anyway. Call that area “A”. So we have this equation:

\frac12 ar + \frac12 br + \frac12 cr = A

Where a, b, and c we know because those are the legs of the triangle, and A we may not have offhand but we can calculate it right away. The radius has to be twice the area of the original triangle divided by the sum of a, b, and c. If it strikes you that this is twice the area of the circle divided by its perimeter, yeah, that it is.

Incidentally, we haven’t actually used the fact that this is a right triangle. All the reasoning done would work if the original triangle were anything — equilateral, isosceles, scalene, whatever you like. If the triangle is a right angle, the area is easy to work out — it’s one-half times a times b — but Heron’s Formula tells us the area of a triangle knowing nothing but the lengths of its three legs. So we have this:

(Right triangle)

r = \frac{1}{a + b + c} \cdot \left(a\cdot b\right)

(Arbitrary triangle)

r = \frac{1}{a + b + c} \cdot 2 \sqrt{p\cdot(p - a)\cdot(p - b)\cdot(p - c)} \mbox{ where }  p = \frac12\left(a + b + c\right) .

Since we started out with a Pythagorean right triangle, with sides 5, 12, and 13, then: a = 5, b = 12, c = 13; a times b is 60; a plus b plus c is 30; and therefore the radius of the inscribed circle is 60 divided by 30, or, 2.

Splitting a Cake with a Missing Piece


I wanted to offer something a little light today, as I’m in the midst of figuring out the next articles in a couple of my ongoing threads and getting ready for a guest posting. So here, from Mathematics Lounge, please enjoy this nice little puzzle about how to cut, into two even pieces, a cake that’s already had a piece cut out of it. It’s got a lovely answer and it’s worth pondering it and why that answer’s true before reading the solution. And there’s another, grin-worthy, solution offered in the comments.

Mathematics Lounge

Problem:
Jeremy and Jane would like to divide a rectangular cake in half, but their friend Bob (who can be a jerk sometimes) has already cut out a piece for himself. Bob’s slice is a rectangle of some arbitrary size and rotation. How can Jeremy and Jane divide the remaining cake into two equal portions, using a single cut with a sufficiently long knife?

cake_question

Description:
This is an interesting problem with a fairly elegant solution. It is the type of problem that can be posed as a math puzzle/riddle, and figured out on the spot with some ingenuity.

For this problem, we define a single cut as a separation of the area made by a straight line, viewed from above. For example, a cut that crosses a gap (like below) may intersect the cake in two separate places, but still counts as one cut. (This example, of course, clearly does…

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Reading the Comics, May 4, 2014: Summing the Series Edition


Before I get to today’s round of mathematics comics, a legend-or-joke, traditionally starring John Von Neumann as the mathematician.

The recreational word problem goes like this: two bicyclists, twenty miles apart, are pedaling toward each other, each at a steady ten miles an hour. A fly takes off from the first bicyclist, heading straight for the second at fifteen miles per hour (ground speed); when it touches the second bicyclist it instantly turns around and returns to the first at again fifteen miles per hour, at which point it turns around again and head for the second, and back to the first, and so on. By the time the bicyclists reach one another, the fly — having made, incidentally, infinitely many trips between them — has travelled some distance. What is it?

And this is not hard problem to set up, inherently: each leg of the fly’s trip is going to be a certain ratio of the previous leg, which means that formulas for a geometric infinite series can be used. You just need to work out what the lengths of those legs are to start with, and what that ratio is, and then work out the formula in your head. This is a bit tedious and people given the problem may need some time and a couple sheets of paper to make it work.

Von Neumann, who was an expert in pretty much every field of mathematics and a good number of those in physics, allegedly heard the problem and immediately answered: 15 miles! And the problem-giver said, oh, he saw the trick. (Since the bicyclists will spend one hour pedaling before meeting, and the fly is travelling fifteen miles per hour all that time, it travels a total of a fifteen miles. Most people don’t think of that, and try to sum the infinite series instead.) And von Neumann said, “What trick? All I did was sum the infinite series.”

Did this charming story of a mathematician being all mathematicky happen? Wikipedia’s description of the event credits Paul Halmos’s recounting of Nicholas Metropolis’s recounting of the story, which as a source seems only marginally better than “I heard it on the Internet somewhere”. (Other versions of the story give different distances for the bicyclists and different speeds for the fly.) But it’s a wonderful legend and can be linked to a Herb and Jamaal comic strip from this past week.

Paul Trap’s Thatababy (April 29) has the baby “blame entropy”, which fits as a mathematical concept, it seems to me. Entropy as a concept was developed in the mid-19th century as a thermodynamical concept, and it’s one of those rare mathematical constructs which becomes a superstar of pop culture. It’s become something of a fancy word for disorder or chaos or just plain messes, and the notion that the entropy of a system is ever-increasing is probably the only bit of statistical mechanics an average person can be expected to know. (And the situation is more complicated than that; for example, it’s just more probable that the entropy is increasing in time.)

Entropy is a great concept, though, as besides capturing very well an idea that’s almost universally present, it also turns out to be meaningful in surprising new places. The most powerful of those is in information theory, which is just what the label suggests; the field grew out of the problem of making messages understandable even though the telegraph or telephone lines or radio beams on which they were sent would garble the messages some, even if people sent or received the messages perfectly, which they would not. The most captivating (to my mind) new place is in black holes: the event horizon of a black hole has a surface area which is (proportional to) its entropy, and consideration of such things as the conservation of energy and the link between entropy and surface area allow one to understand something of the way black holes ought to interact with matter and with one another, without the mathematics involved being nearly as complicated as I might have imagined a priori.

Meanwhile, Lincoln Pierce’s Big Nate (April 30) mentions how Nate’s Earned Run Average has changed over the course of two innings. Baseball is maybe the archetypical record-keeping statistics-driven sport; Alan Schwarz’s The Numbers Game: Baseball’s Lifelong Fascination With Statistics notes that the keeping of some statistical records were required at least as far back as 1837 (in the Constitution of the Olympic Ball Club of Philadelphia). Earned runs — along with nearly every other baseball statistic the non-stathead has heard of other than batting averages — were developed as a concept by the baseball evangelist and reporter Henry Chadwick, who presented them from 1867 as an attempt to measure the effectiveness of batting and fielding. (The idea of the pitcher as an active player, as opposed to a convenient way to get the ball into play, was still developing.) But — and isn’t this typical? — he would come to oppose the earned run average as a measure of pitching performance, because things that were really outside the pitcher’s control, such as stolen bases, contributed to it.

It seems to me there must be some connection between the record-keeping of baseball and the development of statistics as a concept in the 19th century. Granted the 19th century was a century of statistics, starting with nation-states measuring their populations, their demographics, their economies, and projecting what this would imply for future needs; and then with science, as statistical mechanics found it possible to quite well understand the behavior of millions of particles despite it being impossible to perfectly understand four; and in business, as manufacturing and money were made less individual and more standard. There was plenty to drive the field without an amusing game, but, I can’t help thinking of sports as a gateway into the field.

Creator.com's _Donald Duck_ for 2 May 2014: Ludwig von Drake orders his computer to stop with the thinking.

The Disney Company’s Donald Duck (May 2, rerun) suggests that Ludwig von Drake is continuing to have problems with his computing machine. Indeed, he’s apparently having the same problem yet. I’d like to know when these strips originally ran, but the host site of creators.com doesn’t give any hint.

Stephen Bentley’s Herb and Jamaal (May 3) has the kid whose name I don’t really know fret how he spent “so much time” on an equation which would’ve been easy if he’d used “common sense” instead. But that’s not a rare phenomenon mathematically: it’s quite possible to set up an equation, or a process, or a something which does indeed inevitably get you to a correct answer but which demands a lot of time and effort to finish, when a stroke of insight or recasting of the problem would remove that effort, as in the von Neumann legend. The commenter Dartpaw86, on the Comics Curmudgeon site, brought up another excellent example, from Katie Tiedrich’s Awkward Zombie web comic. (I didn’t use the insight shown in the comic to solve it, but I’m happy to say, I did get it right without going to pages of calculations, whether or not you believe me.)

However, having insights is hard. You can learn many of the tricks people use for different problems, but, say, no amount of studying the Awkward Zombie puzzle about a square inscribed in a circle inscribed in a square inscribed in a circle inscribed in a square will help you in working out the area left behind when a cylindrical tube is drilled out of a sphere. Setting up an approach that will, given enough work, get you a correct solution is worth knowing how to do, especially if you can give the boring part of actually doing the calculations to a computer, which is indefatigable and, certain duck-based operating systems aside, pretty reliable. That doesn’t mean you don’t feel dumb for missing the recasting.

Rick Detorie's _One Big Happy_ for 3 May 2014: Joe names the whole numbers.

Rick DeTorie’s One Big Happy (May 3) puns a little on the meaning of whole numbers. It might sound a little silly to have a name for only a handful of numbers, but, there’s no reason not to if the group is interesting enough. It’s possible (although I’d be surprised if it were the case) that there are only 47 Mersenne primes (a number, such as 7 or 31, that is one less than a whole power of 2), and we have the concept of the “odd perfect number”, when there might well not be any such thing.

The Math Blog Statistics, April 2014


Another month’s gone by and the statistics about viewership were pretty gratifying for April 2014. But I’m feeling awfully good about the place, because I’ve felt more gratified by the mathematics blog lately. It’s felt to me like there’ve been more comments and more interaction the past couple weeks, and it’s felt like it’s getting closer to supporting a community, which is thrilling, if not exactly measurable given what WordPress shares with me.

In March 2014, according to last month’s statistics survey, there were 453 views from 257 distinct viewers. That jumped pretty noticeably this month to 565 views, albeit from 235 distinct viewers, a views-per-visitor jump from 1.76 to 2.40. I suspect there’s some archive-bingers, and I’m happy to give anyone that thrill. It’s my greatest viewer count since June 2013, and the fourth-highest since December 2012 when WordPress started sharing statistics on unique visitors. I also noted at the start of April that while I’d reached 14,000 visitors in March I’d need a stroke of luck to reach 15,000 in April. I came close: the month topped out with my 14,931st view.

The most popular articles of the past thirty days were:

  1. How Dirac Made Every Number, the answer to that puzzle of how to construct any counting number using precisely four 2’s and ordinary operations (it’s a forehead slapper once you’ve seen it)
  2. Reading the Comics, April 27, 2014: The Poetry of Calculus Edition, as everyone wants to see some calculus poetry
  3. Can You Be As Clever As Dirac For A Little Bit in which that Dirac puzzle was laid out and the rules given
  4. How Many Trapezoids I Can Draw, always with the trapezoids
  5. Reading the Comics, April 21, 2014: Bill Amend In Name Only Edition, which includes a bundle of a lot of mathematics comics from Bill Amend’s FoxTrot in case you need some

The countries sending me the most viewers were the United States (294), Canada (65), Denmark (29), Austria (27), and the United kingdom (26), and I count nine countries sending me at least ten views each, which I think is a record but I haven’t been keeping track of that number. Sending me a single viewer each were Belgium, Brazil, Ecuador, Finland, Greece, Hungary, Malaysia, Morocco, Oman, Sweden, and Venezuela. Belgium, Brazil, Hungary, and Sweden were single-country viewers last month, and Hungary’s got a three-month single-viewer streak going. So, ah, hi, whoever that is in Hungary. Apparently nobody has ever visited me from Honduras.

Once again there’s a shortage of search term poetry, but there were some fair queries the past month, including:

How Dirac Made Every Number


A couple weeks back I offered a challenge taken from Graham Farmelo’s biography (The Strangest Man) of the physicist Paul Dirac. The physicist had been invited into a game to create whole numbers by using exactly four 2’s and the normal arithmetic operations, for example:

1 = \frac{2 + 2}{2 + 2}

2 = 2^{2 - \left(2 \div 2\right)}

4 = 2^2 \div 2 + 2

8 = 2^{2^{2}} \div 2

While four 2’s have to be used, and not any other numerals, it’s permitted to use the 2’s stupidly, as every one of my examples here does. Dirac went off and worked out a scheme for producing any positive integer from them. Now, if all goes well, Dirac’s answer should be behind this cut and it hasn’t been spoiled in the reader or the mails sent out to people reading it.

The answer made me slap my forehead and cry “of course”, if that helps you work it out before you look.

Can You Be As Clever As Dirac For A Little Bit


I’ve been reading Graham Farmelo’s The Strangest Man: The Hidden Life of Paul Dirac, which is a quite good biography about a really interestingly odd man and important physicist. Among the things mentioned is that at one point Dirac was invited in to one of those number-challenge puzzles that even today sometimes make the rounds of the Internet. This one is to construct whole numbers using exactly four 2’s and the normal, non-exotic operations — addition, subtraction, exponentials, roots, the sort of thing you can learn without having to study calculus. For example:

1 = \left(2 \div 2\right) \cdot \left(2 \div 2\right)
2 = 2 \cdot 2^{\left(2 - 2\right)}
3 = 2 + \left(\frac{2}{2}\right)^2
4 = 2 + 2 + 2 - 2

Now these aren’t unique; for example, you could also form 2 by writing 2 \div 2 + 2 \div 2, or as 2^{\left(2 + 2\right)\div 2} . But the game is to form as many whole numbers as you can, and to find the highest number you can.

Dirac went to work and, complained his friends, broke the game because he found a formula that can any positive whole number, using exactly four 2’s.

I couldn’t think of it, and had to look to the endnotes to find what it was, but you might be smarter than me, and might have fun playing around with it before giving up and looking in the endnotes yourself. The important things are, it has to produce any positive integer, it has to use exactly four 2’s (although they may be used stupidly, as in the examples I gave above), and it has to use only common arithmetic operators (an ambiguous term, I admit, but, if you can find it on a non-scientific calculator or in a high school algebra textbook outside the chapter warming you up to calculus you’re probably fine). Good luck.

January 2014’s Statistics


So how does the first month of 2014 compare to the last month of 2013, in terms of popularity? The raw numbers are looking up: I went from 176 unique visitors looking at 352 pages in December up to 283 unique visitors looking at 498 pages. If WordPress’s statistics are to be believed that’s my greatest number of page views since June of 2013, and the greatest number of visitors since February. This hurt the ratio of views per visitor a little, which dropped from 2.00 to 1.76, but we can’t have everything unless I write stuff that lots of people want to read and they figure they want to read a lot more based on that, which is just crazy talk. The most popular articles, though, were:

  1. Something Neat About Triangles, this delightful thing about forming an equilateral triangle starting from any old triangle.
  2. How Many Trapezoids I Can Draw, with my best guess for how many different kinds of trapezoids there are (and despite its popularity I haven’t seen a kind not listed here, which surprises me).
  3. Factor Finding, linking over to IvaSallay’s quite interesting blog with a great recreational mathematics puzzle (or educational puzzle, depending on how you came into it) that drove me and a friend crazy with this week’s puzzles.
  4. What’s The Worst Way To Pack? in which I go looking for the least-efficient packing of spheres and show off these neat Mystery Science Theater 3000 foam balls I got.
  5. Reading The Comics, December 29, 2013, the old year’s last bunch of mathematics-themed comic strips.

The countries sending me readers the most often were the United States (281), Canada (52), the United Kingdom (25), and Austria (23). Sending me just a single reader each this past month were a pretty good list:
Bulgaria, France, Greece, Israel, Morocco, the Netherlands, Norway, Portugal, Romania, Russia, Serbia, Singapore, South Korea, Spain, and Viet Nam. Returning on that list from last month are Norway, Romania, Spain, and Viet Nam, and none of those were single-country viewers back in November 2013.

Factor Finding


I imagine everyone in the world has seen this already, but, over on the Find The factors blog is a string of mathematics puzzles. The one to which I link amounts to writing out a multiplication table, where the rows and columns have been scrambled, and you have to work out which row is which based on the select handful of numbers in the table. That is, the first row might be the multiples of 6, the next row the multiples of 9, the next row the multiples of 4; and the first column the multiples of 4, the second column multiples of 5, the third column multiples of 2, and so on.

I think this is a fun exercise. It’s more challenging than the day’s Jumble (which this year has had a disturbing number of ones that can be solved on sight, without any unscrambling of words), without being so time-consuming as Sudoku, and if you’re trying to learn the times tables (which I admit probably few readers around here are trying to do) there’s a lot of chance to think about what the multiplication tables are to work out the puzzle. There’s a fresh puzzle every week, as well as a good number of tools for people learning multiplication.

The Intersecting Lines


I haven’t had much chance to sit and think about this, but that’s no reason to keep my readers away from it. Elke Stangl has been pondering a probability problem regarding three intersecting lines on a plane, a spinoff of a physics problem about finding the center of mass of an object by the method of pinning it up from a couple different points and dropping the plumb line. My first impulse, of turning this into a matrix equation, flopped for what were as soon as I worked out a determinant obvious reasons, but that hardly means I’m stuck just yet.

Some Difficult Math Problems That You Understand


What the heck, it’s Friday, I might as well put up a little fun thing here. Over on the Maths In A Minute blog is a quartet of mathematics problems which anyone can understand on their first reading of them, but which are unsolved and which have been sitting there unsolved for generations.

There’s, realistically, no chance that you personally are going to solve them. While probably everyone has some mathematical talent, and anyone with the ability to reason is at least in principle endowed with the relevant tools, it’s not likely that you’ll find a path to solving them that hasn’t been tried and that’s failed before. But that doesn’t mean they’re not worth looking at and thinking about because you can nevertheless discover things you didn’t know before, and notice stuff about mathematics that you hadn’t realized. Even if your discovery only delights yourself, it’s still a delight to you.

(I didn’t run across this page myself, though could have; the Algebra Fact of the Day brought me to it.)

Geometry the Old-Fashioned Way


I failed to keep note of where I got this link from, so I apologize to whatever fine person sent me over here.

Science Vs Magic.net has a splendid page that lets one have fun doing geometry in the classical Greek format, with tools that are the equivalent of straightedge and compass. The compass is used properly, too: there’s no cheating and copying distance by lifting the compass carefully and setting it back down. You have to draw from a center point and a chosen radius each time, the way classical constructions are supposed to be done.

There’s a package of forty challenges offered, things like drawing squares or making rosettes of circles or the like, and if that’s not enough there’s the challenge in beating a par for the number of moves required. Meanwhile I’m gratified to learn that, years after I had to do this stuff for school, I’ve still remembered how to do bisections of lines and angles, and how to drop perpendiculars to a point.

(I haven’t figured yet how to draw a circle to an arbitrary point — sometimes you don’t need to connect anywhere particular — but imagine if I read the instructions maybe that would be obvious or something.)

Counting From 52 to 11,108


fluffy once again brings to my attention the work of Inder J Taneja, who got into the Annals of Improbable Research for a fun parlor-game sort of project a couple of months ago. This was for coming up with ways to (most of) the numbers from 44 up to 1,000 using the digits 1 through 9 in order (ascending and descending), in combinations of addition, multiplication, and exponentiation. Taneja got back in Improbable this weekend with a follow-up project, listing the numbers that can be formed all the way out to a pleasant 11,111.

Taneja’s paper, available at arxiv.org, is that rare mathematics paper that you don’t need to be a mathematician to read, although it isn’t going to strike anyone as very enlightening. The ingenuity involved in many of them is impressive, though, and Taneja lists some interesting things such as how many numbers in a given range can’t be made by the digits in ascending or descending order. (Remarkably, to me at least, everything from 1,001 to 2,000 can be done in ascending or descending order.)

Continue reading “Counting From 52 to 11,108”

How Big Is This Number? Answered


My little question about just how big a number 3^{3^{15}} was got answered just exactly right by John Friedrich, so if you wondered about how I could say a number took about seven million digits just to write out, there’s your answer. Friedrich gives it as a number with 6,846,169 digits, and I agree. Better, the calculator I found which was able to handle this (MatCalcLite, a free calculator app I have on my iPad) agrees too: it claims that 3^{3^{15}} is about 3.25 \times 10^{6 846 168} which has that magic 6,846,169 digits.

Friedrich uses logarithms to work it out, and this is one of the things logarithms are good for in these days when you don’t generally need them to do multiplications and divisions. You can look at logarithms as letting you evaluate the lengths of numbers — how many digits they need to work out — rather than the numbers themselves, and this brings to the field of accessibility numbers that would otherwise be too big to work with, even on the calculator. (Another thing logarithms are good for is that they’re quite nice to work with if you have to do calculus, so once you’re comfortable with them, you start looking for chances to slip them into analysis.)

One nagging little point about Friedrich’s work, though, is that you need to know the logarithm of 3 to work it out. (Also you need the logarithm of 10, or you could try using the common logarithm — the logarithm base ten — of 3 instead.) For finding the actual number that’s fine; trying to get this answer with any precision without looking up the logarithm of 3 is quirky if not crazy.

But what if you want to do this purely by the joys of mental arithmetic? Could you work out 3^{3^{15}} without finding a table of logarithms? Obviously you can’t if you want a really precise answer, and here 3.25 \times 10^{6 846 168} counts as precise, but could you at least get a good idea of how big a number it is?

Reblog: The Pigeon Hole Principle


Parker Glynn-Adey here speaks some about the Pigeon Hole Principle, which is one of those little corners of mathematics whose name alone brings a smile to people’s faces. There are a couple of ways of stating the principle. The version I remember from time immemorial is that if one has N pigeons and a smaller number M of pigeon-holes, then if we’ve put all the pigeons somewhere, there must be at least one pigeon-hole with more than one pigeon.

Glynn-Adey starts from a more general way of describing this situation, and goes through a couple of equivalent versions of the idea, before launching into some of the neat little puzzles that follow directly from this idea. Some of them are nicely surprising and I recommend any of the exercises as a fun pastime.

I admit that when I first learned of the Pigeon-Hole Principle it was in a class that also needed the idea of keeping pigeons on purpose explained to it. We’d have thought more naturally of cubby-holes, but hadn’t ever encountered a cubby.

Parker Glynn-Adey

Below the cut are some pigeon hole related questions I collected together for a Math Circle at the Fields Institute.

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