Even XKCD is on the Car Talk bandwagon! (Click the image to go to the source)

Most significant to our present discussion, however, is Car Talk’s weekly diversion known as the Puzzler.  Each week, the brothers read a Puzzler (i.e. a brain teaser) to their listeners and request solutions.  The Puzzler’s solution is revealed the following week, and a new Puzzler is then presented; moreover, one of the correct listener submissions is chosen at random, and the winner receives a gift certificate for some Car Talk schwag.  While these puzzles are sometimes car-related, this is not a prerequisite, and indeed the puzzler I would now like to discuss makes no mention of cars.

Here’s the puzzle, with wording modified slightly from Car Talk’s website: Suppose there are 20,000 lights in a row, all turned off.  One person comes through and pulls the cord on every light, turning each of them on.  A second person then comes through and pulls the cord on every second light. A third person then pulls the cord on every third light, and so on. After the 20,000th person has gone through the room, which lights are turned on?

This problem also goes by the name of the Locker Problem, with open and shut lockers substituting for on and off lights.  But no matter how you contextualize the problem, the solution is the same.  To get some idea of what the answer should be, let’s consider what happens just for the first few (say, 10) light bulbs.  Feel free to think about this problem on your own before continuing on!

After the first person walks through the room, all the light bulbs are on.  So, the first 10 light bulbs will look like this:

After every second cord is pulled, half the lights will be on, and half will be off:

Now here’s where the fun begins.  When every third cord is pulled, off lights will turn on, and on lights will turn off.  This means that the arrangement of the first ten lights will look like this:

And, when every fourth cord is pulled, we get the following picture:

You can probably fill in the rest.  Note that when every 6th, 7th, 8th, 9th, or 10th cord is pulled, only one light in first 10 switches, so most of the work is already done.  In the end, the string of the first 10 lights will look like this:

Here we see that three of the first ten lights remain lit: the 1st, 4th, and 9th.  The astute reader will note that 1, 4, and 9 all share a common property; they are all perfect squares (1 = 1², 4 = 2², 9 = 3²).  The question to be asked, of course, is whether this pattern continues, and if it does, why?

To answer these questions we’ll need to think a bit more deeply about what’s going on.  First, when does a given light bulb in our string of 20,000 get its cord pulled?  Extrapolating from the first few cases, we see that if a person goes through the chain and pulls every dth cord, then the light bulbs with their cords pulled are precisely the ones whose numbers are multiples of d.  Therefore, the nth bulb’s cord is pulled once for every divisor of n.  More concretely, we see for example that the 12th light bulb will have its cord pulled 6 times: by the 1st, 2nd, 3rd, 4th, 6th, and 12th person, since 1, 2, 3, 4, 6, and 12 are precisely the divisors of 12.

What does this divisibility information tell us about the status of a given light bulb?  Since all the lights start in the off position, we see that a light will be off at the end if its cord is pulled an even number of times, and it will be on at the end if its cord is pulled an odd number of times.  In other words, the nth light will be on at the end of this process precisely when n has an odd number of divisors.

All we need to do now is convince ourselves that the numbers with an odd number of divisors are precisely the squares.  There are a couple of ways to do this.

Way 1: Pick your favorite whole number n.  For any divisor d of n, n/d is also a divisor (for example, if n = 30, the fact that 5 is a divisor immediately tells us that 30/5 = 6 is a divisor too).  In this way, we can naturally count up all the divisors of n in pairs.  Because of this, the only way we can have an odd number of divisors is if one of the pairs has the same number repeated twice, i.e. if for some divisor d, d and n/d are equal.  But if they are equal, this means that n = d², in other words, n is a square.

Way 2: By the fundamental theorem of arithmetic, Any positive whole number n > 1 can be written in an essentially unique way as a product of prime numbers, i.e. for any n > 1 we can write

,

where the numbers are primes, and the exponents are greater than zero (for example, ).  In particular, any divisor of n must be composed of the same primes as n itself, so that if d is a divisor, we can write

,

where each exponent is no larger than (for example, 12 is a divisor of 180 but 24 isn’t, since 2 goes into 24 three times but into 180 only twice).

How many divisors are there? Well, could divide d as few as 0 times, or as many as times – we have different ways to choose the exponent on , corresponding to the numbers 0, 1, 2, …, up to .  Similarly, we have different ways to choose the exponent on , and so on for each prime, so that there are different ways to choose the exponent on .  By the fundamental counting principle, this means that the number of divisors of n is equal to

Remember that our light will be on only if the number of divisors is odd, in other words, only if the above product is odd.  Notice that if any term in the above product is even, the product itself will be even – so in order for the product to be odd, each term in the product must be odd.  Since each must be odd, this means that each exponent must be even.  But if each is even, then is always a whole number, so

is a whole number whose square is equal to n.  Hence, once again we conclude n is a square.

There are plenty of variants to this problem worth pondering – for example, what if people only come in and pull every dth cord only for d prime?  Or, what if instead of the 2nd person pulling every 2nd cord, and the third person pulling every 3rd cord, the second person pulls every 2nd cord twice, the third person pulls every 3rd cord three times, and so on?  No doubt you can come up with some variants on your own as well.

If you prefer the locker version, here’s an interactive site where you can watch with satisfaction as lockers open and close.  No matter what model you use, though, this is a cute little problem on integers and their divisibility, and the result can be surprising for first time viewers.  So kudos to Car Talk for discussing this problem on a national stage! (Kudos also to Tom for drawing my attention to this particular episode of the program.)

Though smaller than the average cupcake, the macaron is also more labor-intensive, and is therefore frequently on the more expensive end of the confectionery spectrum.  The macarons at Bottega Louie, for example, will run you $1.75 each.

One of many delightful flavors

If you need a sweet fix, though, a single macaron may not be enough.  Anticipating such a first-world problem, Bottega Louie also offers boxes of macarons for purchase.  The boxes come in three sizes: the small holds five macarons, the medium holds thirteen, and for the true Francophiles, the largest box holds forty five.  If you buy a box, you can fill it with whatever flavors you like, and can then eat to your heart stomach’s content.

What does any of this have to do with mathematics?  As with so many things, a quantitative eye is useful when it comes time to look at the bottom line.  While the prices of most things decrease with scale – each donut in the purchase of a dozen is cheaper than the purchase of an individual donut, for example – in the case of these macarons, such scaling does not occur.

Let’s dig into some numbers.  The small box of macarons costs $10, or $2 per macaron.  This is more than a 10% increase in the price per macaroon; essentially, one is paying an extra $1.25 for a fancy little box.

What a tiny, fancy box.

Things are slightly better if you go bigger.  The medium box is $25, as compared to $22.75 for thirteen individual macarons.  That comes out to roughly $1.93 per macaroon, or a cost of $2.25 for the large box.  The largest box will run you $80, as compared to $78.75 for forty five individual macarons.  Equivalently, the cost is around $1.78 per macaron, and the additional cost of the box once again comes to $1.25.

In particular, it seems a little strange that the smallest and largest boxes both incur an additional $1.25 charge, while the one in the middle is an extra $2.25.  The mathematician in me would much rather see the box in the middle priced at $24, and the consumer in me would rather see all the boxes be cheaper per macaron than buying them individually.

Unfortunately, in the macaron game, this type of pricing is apparently not unheard of.  There is another small chain of macaron shops known as ‘lette, with several locations throughout the Los Angeles area.  I stopped in one this weekend and found the following prices displayed on the wall:

It’s not all bad news this time around.  While the smaller boxes cost more per macaron than buying individually ($2 each for the mini box, $1.75 for the box of six), the larger boxes are thankfully cheaper (around $1.63 each for the box of 12, around $1.58 for the box of 24).

In either case, though, the moral is the same: when it comes to macarons, make sure you do the math.  While we are all accustomed to the idea that larger purchases correspond to lower costs per unit, these examples show this is not necessarily the case.  If you’re only interested in stuffing your face, make sure to take a moment to crunch the numbers, as the best deal may not immediately present itself to you.

From a mathematical standpoint, though, there are a couple of inconsistencies. Michael makes no secret of the fact that he has worked for the company for 19 years.  His employees take this loyalty to heart, and in Michael Scott’s penultimate episode, “Michael’s Last Dundies,” they surprise their boss with a song parody of the Rent song “Seasons of Love,” which pays homage to such a long period of service.  Below is the relevant clip – if you don’t have access to Hulu, you can try this YouTube link, although I imagine it will get pulled eventually.

The song begins with a soulful rendition of the following lyrics, courtesy of Andy Bernard (Ed Helms):

9,986,000 minutes,
We actually sat down, and did the math
9,986,000 minutes,
That’s how may minutes that you’ve worked here.

Here’s a question: how does this number compare to Scott’s claim of 19 years?  Let’s make a few estimates.

For a crude estimate, we can simply take the length of a year to be 365 days, and count the number of minutes in 19 years given this assumption.  This isn’t hard; the answer is simply 19 (number of years) x 365 (number of days in a year) x 24 (number of hours in a day) x 60 (number of minutes in an hour) = 9,986,400.  This is exceptionally close to the value cited in the song – in fact, due to a syllable constraint needed to maintain faithfulness to the original song, it seems likely that this is how the number crunchers in the office actually came up with the value they sang.

Andy has the voice of an angel.

But how accurate is this number?  This depends on a few factors.  First, if Michael Scott really did begin 19 years ago, this estimate neglects a few leap years.  A better estimate would come from taking 365.25 days in a year, which in turn would give an estimate of 9,993,240 minutes (or 9,993,000 minutes, if one is interested in a number befitting of a Rent song parody).  What’s more, it’s unlikely that Michael worked exactly 19 years – the true length of his stay is probably somewhere between 19 and 20 years, or between 9,993,240 and 10,519,200 minutes.

Depending on your interpretation of the lyrics, though, any of the numbers given above may be much too large.  When Michael’s employees tell him “That’s how many minutes that you’ve worked here,” do they mean “that’s how many minutes since you’ve started working here,” or “that’s how many minutes you’ve spent working since you started working here?”  If the former, then these estimates seem more or less reasonable.  But if it’s the latter, we really should only be counting the minutes Scott spent at the office.  If we say that he spent 40 hours a week, roughly, in the office over the past 19 years, then with 365.25/7 weeks per year, this comes out to only 19 x 365.25/7 x 40 x 60, or roughly 2,379,343 minutes, a far cry from our earlier estimates.  To take things even further, based on the evidence provided by the show, the amount of work Michael actually does while in the office seems fairly minimal – if we’re talking strictly about the amount of time he’s spent working in those 19 years, I wouldn’t be surprised to find a much lower value.  Either way, there seem to be some accounting issues at work here that have been ignored.

Perhaps if accounting spent less time playing solitaire, these issues would have been ironed out.

Also, at the risk of sounding like comic book guy, just two years ago an episode of The Office focused on celebrations for Michael’s 15th anniversary with Dunder Mifflin; the subsequent cancellation of the festivities by new manager Stringer Bell Charles Miner drove Michael to quit and start a rival paper company.  One could argue that the time Michael spent not employed by Dunder Mifflin should not count towards his 19 years, but more importantly, 15 + 2 is only 17, not 19.  Before you try to argue that perhaps the show’s timeline is simply moving faster than the normal rate of one year per year (as TV shows are occasionally wont to do), one needs only to consider the timeline of Jim and Pam’s baby to see that this is not possible.  So in fact, all of the above discussion is fairly moot – indeed, we can’t even be sure how long Michael has worked at Dunder Mifflin.

All is not lost, though.  While we can’t accurately apply mathematics to this Office-inspired question, others out there have had more success with different questions.  For example, here is a computer science paper on automated recognition of phrases that can reasonably be followed up by the phrase “that’s what she said.”  What a neat problem!  It looks so hard, but they somehow manage to tame it.

Oh, Michael Scott.  You will be missed.

Last year, Professor Steven Strogatz of Cornell University wrote a series of op-eds for the New York Times that discussed the presence of mathematics in unlikely places. I discussed one of these columns here.  Now, either those articles were well-received, or Professor Strogatz is well-connected, because this year he’s back in the Times with a much more ambitious series of articles.  This time around, Strogatz is attempting to “[write] about the elements of mathematics, from preschool to grad school, for anyone out there who’d like to have a second chance at the subject.”

Preschool to grad school is a significant amount of ground to cover, but thus far Strogatz has used his articles to assault this goal with gusto.  To date, he has tackled counting, patterns in addition, negative numbers, division, and basic high school algebra.  This doesn’t really do justice to his content, though.  Along the way he gives the reader some Sesame Street, and discusses a number of tangential topics, including the inability of Verizon employees to do math, the half-your-age-plus-seven rule, and pre-WWI European history.  The latter comes about in a discussion of that old adage which is familiar to anyone who saw the first Alien vs. Predator movie: the enemy of my enemy is my friend.

Predators must be awesome at math.

While some of Professor Strogatz’s explanations are a bit hand wavy (in particular, his explanation of why (-1) x (-1) = 1 is a lacking), on the whole they are quite good.  In particular, he offers a nice explanation of what it is for a mathematical argument to be “elegant.”  But even more impressive than his writing is its location – to have a discussion of mathematics with as wide an audience as the New York Times readership is commendable.  Even if people are not inspired to learn more mathematics after reading these pieces, hopefully they will have at least learned something.  As with exercise, a little mathematics is better than no mathematics at all.

Moreover, these articles highlight aspects of math not usually seen in popular discourse.  Much like Paul Lockhart’s A Mathematician’s Lament (which Strogatz references), these snack-size essays are focused on simple mathematical ideas, and the beautiful (and sometimes unexpected) results that follow.  Nowhere here does Professor Strogatz multiply two really big numbers together; in fact, he’s quite sympathetic to the fact that for many people, there is nothing more tedious than calculation.  By leading the conversation in this way, he’s hopefully able to give a taste of what makes math beautiful to an audience for whom such a statement might otherwise be labeled heresy.

I don’t know where this series of articles is headed, but I look forward to finding out, and hope you do to.  Professor Strogatz’s articles are grouped together here.

(Hat tip to dad for sending me a few of these articles.)

Very recently I was with a group of friends, discussing all that would come to pass in this new year. One friend, whose anonymity I will preserve by referring to him only as “Smith,” was in the enviable position of being the only one among us whose age divided the current year (I won’t embarrass him by revealing his age, but given that it’s a divisor of 2010, this certainly restricts the possibilities). Once we realized this, it became natural to ask how common an occurrence this should be. In other words, how often can you expect your age to divide the current year? Of course, implicit in this is a choice of calendar – for our purposes, we will stick to commonly used Gregorian calendar, although the results would be equally valid under a different choice (e.g. the Hebrew calendar or Islamic calendar). For example, if you were 1, 7, 41, or 49 last year, your age divided the year (of 2009). Next year, only the one year olds will win out, since 2011 is a prime number.

Depending on the year you were born, you may find that this happens quite frequently, or not very frequently at all. For example, if you were born in the year 0, you’re in luck, because your age will divide the current year for at least part of the year for every subsequent year. The phrase “part of the year” is important, because in a given year you will be two different ages – the age before your birthday, and the age on and after your birthday. Of course, this isn’t an issue if you were born on January 1st or December 31st, but we will ignore this (simpler) case.

Let’s take a more detailed example. Suppose you were born in 1982. In 1983, after your first birthday, your age will divide the year (since 1 divides everything). Similarly, in 1984, your age will divide the year after your 2nd birthday, since 1984 is even. And in 1986 your age will divide the year until your 4th birthday, since 1986 ÷ 3 = 662. Unfortunately, you will be too young to appreciate this arithmetic coincidence at any of these opportunities, and unless you live to be 661, you’ll never again be able to say that your age divides the year.

However, if you were born just a few years earlier, in 1979, you’ll find that your age divides the year quite frequently. In fact, by the year 2000, the only years in which your age wouldn’t have divided the year at all would have been 1987, 1988, 1993, 1994, 1996, 1997, and 1999.

Why is it that some years allow for one’s age to be divisible by the year quite frequently, while other years do not? The answer is quite simple. Suppose we let b denote the birth year, and we let a denote a person’s age. That person will be a years of age from their birthday in year b + a until their birthday in year b + a + 1. Therefore, your age will divide the year from your birthday until the end of the year if a divides b + a, or from the first of the year until your birthday if divides b + a + 1. So, the question becomes: when does a divide b + a, and when does it divide b + a + 1?

In the first case, since a always divides a, we know that a divides b + a if and only if a divides b. By the second same argument, we see that a divides b + a + 1 if and only if a divides b + 1. In other words, we conclude the following:

Your age will divide the current year if, and only if, either (i) it is between January 1st and your birthday, and your age divides the year after you were born, or (ii) it is between your birthday and December 31st, and your age divides the year you were born. To put it more simply, your age will divide the year for at least part of the time you are at that age if and only if that age divides the year of your birth or the year after your birth.

With this knowledge, it’s easy to see why people born in 1979 will have their age divide the current year more frequently than people born in 1982. In the former case, determining the set of ages which will divide the current year is equivalent to finding the divisors of 1979 and 1980. 1979 is a prime number, so it will never be the case that your age will divide the year between your birthday and December 31st (except after your 1st birthday); on the other hand, 1980 has a prime factorization of 2 x 2 x 3 x 3 x 5 x 11, which gives it a large number of small factors, and consequently a large number of solutions to the problem.

By contrast, if you were born in 1982, you won’t get many factors either way: 1982 factors as 2 x 991, and 1983 factors as 3 x 661. This is why, if you are born in 1982, your age won’t divide the current year after you’re 3.

While it’s not often that mathematics comes up when I’m with my friends at home, I certainly relish every opportunity. I hope that this may serve as an example to all of you who would like to make mathematics more of a part of your everyday life, especially in social circles into which math rarely intrudes. Single guys looking for first date conversation material are especially urged to keep this sentiment in mind.

That’s right – Ghiradelli now makes salad.

It was my friend Jack who pointed out the placement of the decimal point. Apparently the people who work in cafeterias in Utah are the same people who work at Verizon call centers. If you ever want cheap salad, I guess this is the place to go – $0.0029 per ounce is a price that can’t be beat!

It’s unlikely that anyone will try to exploit this small misprint to score a pound of salad for just under 5 cents, but someone would certainly be within his or her rights to do so. The lesson I learned is that even when you are surrounded by mathematicians, you are never truly safe from the consequences of an insufficient math education. Of course, at the end of the day, the typo really is inconsequential, but as highlighted in the Verizon call I posted earlier, even simple misunderstandings such as these can have significant consequences.

Then again, maybe the salad really was that cheap, in which case I really should have stuffed my luggage with vegetables.

On a related note, I would encourage all of you to start writing the dollar amounts on your checks as more complicated mathematical expressions. Everyone could use a boost to their mathematical literacy, bankers included.

The audio clip is quite long, and the longer it goes on, the more depressing it gets.