Bryan Bros and Units of Measurement

For the last couple years, one of my favorite sources of entertainment has been the wonderful world of YouTube Golf. Intending no disrespect to any other content creators, my favorite channels are the ones by Grant Horvat, the Bryan Bros (not to be confused with the twin tennis duo), Peter Finch, Bryson DeChambeau (of course), and Golf Girl Games (all of them absolutely, positively should have been in the Internet Invitational… but that’s another story for another day).

In a recent Bryan Bros video, my two interests collided. To make a long story short, a golf simulator projected that a tee shot on a par-3 ended 8 feet, 12 inches from the cup.

Co-host Wesley Bryan, to his great credit, immediately saw the computer glitch — this is an unusual way of saying the tee shot ended 9 feet from the cup. Hilarity ensued as the golfers held a stream-of-consciousness debate on the merits of metric and Imperial units. The video is below: the fun begins at the 21:41 mark and ends around 25:30.

Incoming Asteroid

Source: https://xkcd.com/3049/

Halloween Math Dad Joke

Q: “Why do mathematicians always confuse Halloween with Christmas?”

A: “Because 31-Oct = 25-Dec.”

My Mathematical Magic Show: Part 5e

As discussed earlier in this blog, here’s one of my favorite mathematical magic tricks. The trick works best when my audience has access to a calculator (including the calculator on a phone). The patter:

Write down any five-digit number you want. Just make sure that the same digit repeated (not something like 88,888).

(pause)

Now scramble the digits of your number, and write down the new number. Just be sure that any repeated digits appear the same number of times. (For example, if your first number was 14,232, your second number could be 24,231 or 13,422.)

(pause)

Is everyone done? Now subtract the smaller of the two numbers from the bigger, and write down the difference. Use a calculator if you wish.

(pause)

Has everyone written down the difference. Good. Now, pick any nonzero digit in the difference, and scratch it out.

(pause)

(I point to someone.) Which numbers did you not scratch out?

The audience member will say something like, “8, 2, 9, and 6.” To which I’ll reply in three seconds or less, “The number you scratched out was a 2.”

Then I’ll turn to someone else and ask which numbers were not scratched out. She’ll say something like, “3, 2, 0, and 7.” I’ll answer, “You scratched out a 6.”

As discussed in a previous post, the difference found by the audience member must be a multiple of 9. Since the sum of the digits of a multiple of 9 must also be a multiple of 9, the magician can quickly figure out the missing digit. In the previous example, $3+2+0+7=12$ . Since the next multiple of 9 after 12 is 18, the magician knows that the missing digit is $18-12 = 6$ .

To speed things up (and to reduce the possibility of a mental arithmetic mistake), the magician doesn’t actually have to add up all of the digits. If the audience member gives a digit of either 0 or 9, then the magician can ignore that digit for purposes of the trick. Likewise, if the magician notices that some subset of the given digits add up to 9, then those digits can be effectively ignored. In the current example, the magician could ignore the 0 and also the 2 and 7 (since $2+7=9$ ). That leaves only the 3, and clearly one needs to add 6 to 3 to get the next multiple of 9.

I was a little curious about how often this happens — how often the magician can get away with these shortcuts to find the missing digits. So I did some programming in Mathematica. Here’s what I found. If the audience starts with a 5-digit number, so that the difference must be some multiple of 9 between 9 and 99,999:

There are 690 multiples (out of 11,111, or about 6%) that do not reduce at all (for example, 57,888). So the magician can expect to do the full addition about one-sixth of the time.
There are 5535 multiples (about 50%) whose digits can be divided into subsets that sum to 9. So, about half the time, the magician can expect to quickly find the missing digit without having to add past 9.

I’ve put on my mathematical wish-list some kind of theorem about this splitting of digits of multiples of 9s.

Dimensional Lumber Tape Measure

Source: https://xkcd.com/3138/

Polynomial Long Division and Megan Moroney

A brief clip from Megan Moroney’s video “I’m Not Pretty” correctly uses polynomial long division to establish that $2x+3$ is a factor of $2x^4+5x^3+7x^2+16x+15$ . Even more amazingly, the fact that the remainder is $0$ actually fits artistically with the video.

And while I have her music on my mind, I can’t resist sharing her masterpiece “Tennessee Orange” and its playful commentary on the passion of college football fans.

Bracket Symbols

Source: https://xkcd.com/2954/

Mathematical Allusions in Shantaram: Index

I’m doing something that I should have done a long time ago: collecting a series of posts into one single post. The links below show my series on the mathematical allusions in the novel Shantaram.

Part 1: The Mandelbrot set

Part 2: Stubbornness

Part 3: Understanding a mathematical theorem for the first time

Part 4: Defining a meter

Track Meets and Floating Point Numbers

Here is one school’s results from a (relatively) recent track and field meet. Never mind the name of the school or the names of the athletes representing the school; this is a math blog and not a sports blog, even though I’m an avid sports fan. Furthermore, I have nothing but respect for young people who are both serious students and serious athletes. While I have no illusions about the global popularity of this blog, and while the information from the meet are in the public domain, I also have no desire to inadvertently subject these student-athletes to online abuse.

With that preamble, here are the school’s results:

The unusual score for jumps caught my attention. Clearly a score of $16\frac{1}{3}$ was intended, but this isn’t displayed. (This is also a lesson about using unnecessary precision… unlike the total points field showing 152.33.) Before this unusual decimal expansion, however, I should generally describe how teams are scored at a track and field meet… or at least the high school and college meets that I’ve attended in the United States.

Each meet has multiple events, often categorized as sprints, hurdles, distance races, throwing events, jumping events, relay races, and multi-sport events (like the decathlon). At each event, first place gets 10 points, second place gets 8 points, third place gets 6 points, fourth place gets 5 points, fifth place gets 4 points, sixth place gets 3 points, seventh place gets 2 points, and eighth place gets 1 point.

Let’s explain the last two lines first. At this meet, athletes from this team finished second, sixth, and seventh in the one multi-sport event, earning $8+3+2=13$ points for the school. A relay team finished third in the 4×100 meter relay, earning another 6 points for the school.

The third-to-last line — jumps — requires some explanation. No athlete from the school finished in top eight in the long jump or the triple jump (0 points). One athlete won the high jump (10 points). And one athletic finished in a three-way tie for second place in the pole vault. In the case of such a tie, the points for second, third, and fourth place are averaged and given to all three competitors, for $(8+6+5)/3 = 6\frac{1}{3}$ points. In total, the school earned $16\frac{1}{3}$ points from the jumping events.

In jumping events, it is possible (but rare) for athletes to tie. The table above shows the results of the competition for the top eight finishers. The lingo: P means the competitor passed at that height (to save time and energy), O means a successful attempt, and X means a failed attempt. So, the winner passed at all heights up to and including 2.70 meters, succeeded on the first attempt at 2.85, 3.00, and 3.15 meters. This athlete was the only one who cleared 3.15 meters and thus won the competition. This athlete then failed three times at 3.30 meters: each athlete has three attempts at each height; three failures at one height means elimination from the competition.

The athletes in the next three lines had the exact same performance: success at 2.70 and 2.85 meters on the first attempt, and then three straight failed attempts at 3.00 meters. Because there is nothing to distinguish the three performances, the athletes are deemed to be tied.

The athletes in fifth and sixth place also cleared 2.85 meters, but on their second attempts. Therefore, they are behind the athletes who cleared 2.85 meters on the first attempt. Furthermore, the athlete in fifth place cleared 2.70 meters on the first attempt, while the athlete in sixth place needed two attempts. Similarly, the athletes in seventh and eighth place both cleared 2.70 meters; the tiebreaker is the number of attempts needed at 2.40 meters.

Ties can also happen in elite competition as well. This dramatically happened in the men’s high jump at the 2021 Summer Olympics and the women’s pole vault at the 2023 World Championships, where the top two competitors tied and decided to share the gold medal.

By contrast, at the 2024 Olympics, the top two competitors in the men’s high jump tied but decided to continue the competition with a jumpoff until there was one winner.

My apologies to any track and field experts if my description of the scoring wasn’t quite perfect.

Back to mathematics… and back to the scores. Why did the computer think that the number of points from jumps was 16.333333492279053 and not $16\frac{1}{3} = 16.3333\dots$ ?

There are two parts to the answer: (1) Computers store numbers in binary, and (2) they only store a finite number of binary digits.

Converting 16 into binary is easy: since $16=2^4$ , its representation in binary is 10000.

Converting $\displaystyle \frac{1}{3}$ into binary is more challenging, and perhaps I’ll write a separate post on this topic. This particular fraction can be found by using the formula for an infinite geometric series:

$a + ar + ar^2 + ar^3 + \dots = \displaystyle \frac{a}{1-r}$

If we let $a = r = \displaystyle \frac{1}{4}$ , then we find

$\displaystyle \frac{1}{4} + \frac{1}{16} + \frac{1}{64} + \dots = \displaystyle \frac{ \frac{1}{4} } { 1 - \frac{1}{4} } = \frac{ 1/4}{3/4} = \frac{1}{3}$ .

Said another way,

$\displaystyle \frac{1}{3} = \frac{0}{2^1} + \frac{1}{2^2} + \frac{0}{2^3} + \frac{1}{2^4} + \frac{0}{2^5} + \frac{1}{2^6} + \dots = 0.01010101\dots$

Combining the two results,

$\displaystyle 16\frac{1}{3} = 10000.010101010101010101010101010101\dots$

This is mathematically correct; however, computers use floating-point arithmetic only store a finite number of digits to represent any number. In this case, we can reverse-engineer to figure out how many digits are stored. In this case, after some trial and error, I found that 21 digits were apparently stored after the decimal point:

$\displaystyle 16\frac{1}{3} \approx 10000.010101010101010101011$

This is equivalent to the sum $2^4 + \displaystyle \frac{1}{2^2} + \frac{1}{2^4} + \frac{1}{2^6} + \dots + \frac{1}{2^{20}} + \frac{1}{2^{21}}$ ; notice that the last fraction is basically rounding up in binary. Mathematica confirms that this sum matches the sum shown in the school’s team score:

So the computer showed far too many decimal places in the “Jumps” field, and it probably should’ve been programmed to show only two decimal places, like in the “Points” field.

I close by linking to this previous post on the 1991 Gulf War, describing why a similarly small error in approximating $\displaystyle \frac{1}{10}$ in binary tragically led to a bigger computational error that caused the death of 28 soldiers.