Ordering Puzzles

Ascertain the methodology by which the following lists are ordered.

100; 544; 702; 3; 51; 1392; 6; 41; 38
Zirconium; Berkelium; Lead=Krypton; Potassium; Tungsten; Roentgenium; Lithium
10; 225; 494; 207; 298; 205; 324
Seychelles; Singapore; Indonesia; Austria; Tonga; Morocco
66; 30; 74; 85; 29; 15; 70; 40; 90
484; 289; 149; 244; 59; 350; 154; 609; 213; 460
C; T; H; M; Y; B; W; F
W; H; U; I; X; G; L; Y; D; B; C

Calendars for Other Solar System Planets

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Mercury has resonant sidereal rotation to revolution resonance of 2:3, so 3 days equals 2 years on Mercury. The Mercurian calendar is trivially one day plus one leap day exhibited every second year per year.

We’ll get back to the Venusian calendar at the end, and you’ll see why there.

The Martian year is 686.97 Earth days long, and the Martian day is 24.62 Earth hours long. There are thus 669.67 Martian days in a Martian year. The Martian calendar should therefore have 670 days with one day deleted every three years. 670 has more factors (although only by a slight amount), so it should be used as the base amount for a calendar. For months, neither of Mars’ moons has a long enough orbital period to serve as a guide for month length, and thus year divisions would purely be optimized by mathematical philosophy. One reasonable division would be 10 months of 66 days plus 10 in-between month “special” days (or, 20 months of 33 days, or, if one doesn’t desire the in-between days, just 10 months of 67 days). In all cases, -1 day per three years.

Jupiter’s year is 4332.59 Earth days long, with a 9.925 Earth-hour rotational period, for 10476.79 Jovian days in a Jovian year, calling for 10476 days in a calendar with all but every fifth year having one day added. Jupiter has four highly significant moons, the largest of which is Ganymede, which orbits Jupiter in 17.30 Jovian days. If we put 18 days in a month, Jupiter’s calendar would therefore be 582 months of 18 days each, with +1 day per four out of five years.

Saturn’s year is 10759.22 Earth days long, with a 10.57 Earth-hour rotational period, for 24429.64 Saturnian days in a Saturnian year, calling for 24430 days in a calendar with every third year having one day deleted. Saturn’s most prominent moon, Titan, orbits Saturn once every 36.20 Saturnian days. Using 35 days for a month, Saturn’s calendar would 698 months of 35 days each, with -1 day per three years.

Uranus’ year is 30799.1 Earth days long, with a 17.24 Earth-hour rotational period, for 42875.8 Uranian days in a Uranian year, calling for 42876 days in a calendar with with every fifth year having one day deleted. Oberon orbits Uranus in 18.74 Uranian days. Using 18 days for a month, Uranus’ calendar would be 2382 months of 18 days each, with -1 day per five years.

Neptune’s year is 60190.0 Earth days long, with a 16.11 Earth-hour rotational period, for 89668.5 Neptunian days in a Neptunian year, calling for 89668 days in a calendar with an extra day every other year. Triton’s revolutionary period of 8.76 Neptunian days is rather short, and the next moon out, Nereid, orbits Neptune in 536.5 Neptunian days, which is too long and not close enough to any divisor of 89668, so let’s not consider moons and give Neptune a calendar of 773 months of 116 days each, with +1 day per two years.

Back to Venus. Venus is special because its revolutionary period is shorter than its rotational period. They are 224.70 Earth days and 243.02 Earth days, respectively. In addition to this, Venus rotates retrograde, but that doesn’t affect our calendar, although if anyone thinks of a way to incorporate this further abnormality, go ahead and mention it. There are 0.9246 Venusian days in a Venusian year, for one day, with +3 years every 37 years.

Mystery Hunts: Solving Times and Lengths of Winning Team Names, 2000-2013

After the conclusion of the record-longest Mystery Hunt in history, let’s take a look back.


In the chart above are the completion times in minutes (y-axis) for the given year (x-axis). I was unable to find accurate completion times for years before 2000. Note that 2010 and 2011 are notably close (the times for those two years were 41 hr 50 min and 41 hr 58 min, respectively). 2013, at 73 hr 18 min, not only was the year of the longest hunt but was also the year of the longest winning team name, <the entire text of Atlas Shrugged>. Here is a graph of solving time versus word length of the team name of the winning team, with a linear regression line attached.


Here is the same chart, but with ln(word length of team name of winning team) instead on the y-axis, hilariously taking a smaller correlation coefficient.


It is tempting to see how ln(ln(word length of team name of winning team)) goes, but unfortunately that involves a negative infinity since there were team names consisting of only one word in the original data.

List of Constellations by Accuracy of Bayer Designations

The brighter stars of constellations are designated by Greek letters according to Bayer Designations, for which the original intent was for the brightest star of a constellation to be designated Alpha (α), the second brightest Beta (β), the third brightest Gamma (γ), and so on through the Greek alphabet. Astronomers, including Bayer, however, hardly made an effort to make Bayer Designations actually indicate the brightest rank of stars in constellations, sometimes because it is hard to differentiate brightnesses (Bayer designated the stars with Greek letters before sufficient modern quantitative technology), sometimes because they wanted to have Greek letters outline patterns, and sometimes when it seems that they could only have been fooling around. In any case, here is a list of constellations ordered from most accurately Bayer-designated to most badly Bayer-designated. The quantity first stated is the sum of the positive differences between the apparent magnitude of the star actually Bayer-designated a Greek letter and the apparent magnitude of the star that should have been respectively Bayer-designated that letter. Divided by the number of Bayer-designated stars in the constellation, the final quantity shows the average amount by which a Bayer designation in that constellation is off by magnitude.

Here’s a picture of Orion with eight labeled stars, with magnitudes attached, for reference.


Note that with less stars Bayer-designated, it is naturally harder to make gigantic errors, and thus those most impressively Bayer-designated are those with many stars accurately Bayer-designated.

Canes Venatici: 0/2=0
Leo Minor: 0/1=0
Lynx: 0/1=0
Vulpecula: 0/1=0

Sextans: 0.02/5<0.01

Coma Berenices: 0.18/3=0.06
Scutum: 0.44/7=0.06

Dorado: 1.47/12=0.12
Vela: 0.92/7=0.13
Sculptor: 2.22/16=0.14
Lacerta: 0.28/2=0.14
Triangulum Australe: 1.44/10=0.14

Corona Australis: 1.77/11=0.16
Apus: 1.66/10=0.17
Volans: 1.66/10=0.17
Centaurus: 4.07/23=0.18
Monoceros: 1.09/6=0.18
Circinus: 1.49/8=0.19
Columba: 2.91/15=0.19
Mensa: 2.91/15=0.19
Cassiopeia: 4.79/24=0.20

Lyra: 2.73/13=0.21
Perseus: 4.65/22=0.21
Canis Minor: 1.52/7=0.22
Corona Borealis: 4.51/20=0.23
Delphinus: 2.26/10=0.23
Chamaeleon: 2.94/13=0.23
Caelum: 1.72/7=0.25
Camelopardalis: 0.74/3=0.25
Crux: 2.98/12=0.25

Grus: 5.25/21=0.25
Pictor: 3.05/12=0.25
Hydrus: 4.20/16=0.26
Reticulum: 2.69/10=0.27
Virgo: 6.69/24=0.28
Lupus: 6.83/24=0.28
Equuleus: 2.03/7=0.29
Triangulum: 1.75/6=0.29
Pyxis: 2.92/10=0.29

Horologium: 3.00/10=0.30
Gemini: 7.01/23=0.30
Andromena: 7.57/24=0.32
Cepheus: 5.70/18=0.32
Phoenix: 7.30/23=0.32
Indus: 5.07/15=0.34
Telescopium: 4.09/12=0.34
Cancer: 7.90/23=0.34
Lepus: 4.47/13=0.34
Pegasus: 7.65/22=0.35
Hercules: 8.01/23=0.35

Leo: 8.43/24=0.35
Ursa Minor: 3.53/10=0.35
Aries: 6.79/19=0.36
Serpens: 8.64/24=0.36
Crater: 4.34/12=0.36
Ara: 5.48/15=0.37
Musca: 4.04/11=0.37
Fornax: 7.5/20=0.38
Norma: 3.87/10=0.39
Boötes: 9.41/24=0.39
Aquarius: 9.42/24=0.39
Pavo: 8.75/22=0.40
Draco: 9.56/24=0.40

Ophiuchus: 9.22/23=0.40
Auriga: 9.24/23=0.40
Cygnus: 9.79/24=0.41
Pisces: 10.07/24=0.42
Tucana: 6.34/15=0.42
Hydra: 10.37/24=0.43
Microscopium: 4.47/10=0.45

Carina: 3.60/8=0.45
Octans: 10.80/24=0.45
Aquila: 10.37/23=0.45
Eridanus: 10.38/23=0.45
Capricornus: 10.92/24=0.46
Taurus: 10.98/24=0.46
Cetus: 10.22/22=0.46
Orion: 11.56/24=0.48

Ursa Major: 12.25/24=0.51
Piscis Austrinus: 7.34/14=0.52
Antlia: 3.73/7=0.53

Canis Major: 11.64/21=0.55
Puppis: 5.13/9=0.57

Sagitta: 5.13/8=0.64

Corvus: 4.62/7=0.66

Libra: 14.57/18=0.81
Sagittarius: 19.76/24=0.82

Scorpius: 20.12/22=0.91

Brightest stars in Libra: β (2.61), α (2.75), σ (3.25), υ (3.60), τ (3.66), γ (3.91), θ (4.13)
Brightest stars in Sagittarius: ε (1.79), σ (2.05), ζ (2.60), δ (2.72), λ (2.82), π (2.88), γ (2.98), η (3.10), φ (3.17), τ (3.32)
Brightest stars in Scorpius: α (1.06), λ (1.62), θ (1.86), δ (2.29), ε (2.29), κ (2.36), β (2.56), υ (2.70), τ (2.82), π (2.89), σ (2.90), ι (2.99)

In Scorpius, it can be seen that in α through λ, what is being designated is the backbone of the scorpion, traced from the heart upwards and then downwards. In Sagittarius, whatever pattern is being used is thoroughly mysterious, and α (3.96) literally comes 16th in the Greek letters. In these exceptional cases, Bayer Designations may show important patterns, but whatever they are, brightness is completely out of the question.

Too Alive

Once I thought the only thing to live was life itself.
And that a heart that’s beating proudly was a heart in health.
But now I’m too alive, and life itself is killing me.
And my heart’s seeking a cage with which to set its plagued self free.

The First Set of Views/Visitors Stats

Starting 03 December 2012, WordPress statistics started releasing information on unique visitors to blogs per day in addition to total views. Here’s how my first set looked. Numbers in parentheses indicate views/visitors ratios.


On top of that, on 02 January 2013, I had a total of 246 views (from 87 unique visitors), one of the highest daily view counts there has been on this blog (the highest is 898, on 10 April 2012, but I don’t think there has been a second day when it was over 400), but more importantly, I had views from eight different countries, which I believe is a new record for daily country variety on this blog. Yay!

13 Math Problems to Start Off 2013

Because why not.

1) A particular alien warrior has eight hands, arranged generally in the directions of the vertices of a cube. When it goes to war, each hand must carry a sword, a shield or nothing, and it must carry at least one sword. This alien is rotationally symmetric over the sagittal-coronal axis, and the cube that the hands form has edges parallel to these and the transverse axis. Note that the alien is not reflectively symmetric over the transverse plane. In how many distinct ways can this alien arm?

2) Paula is usually a reliable painter, but once in a while, she does make a snafu. Suppose that on the nth day that Paula works, she accidentally spills φ(n)-2 gallons of paint, where φ(n) is the number of positive integers less than or equal to n that are relatively prime to n. Fortunately, due to the fact that her boss was impressed by her ability to spill antipaint that she demonstrated on her first two days at work, her boss decided to refrain from firing her until her work becomes unprofitable. Her work becomes unprofitable when the total number of gallons of paint she has spilled exceeds the number of days she has worked. After what day of her work does Paula get fired?

3) An RPG hero is on a quest to put an end to the rampant inflation plaguing the world. The first night, the stay at an inn costs 100 gollld. The second night, a stay at a commensurate inn costs 200 gollld. This rate of inflation persists until the end of the hero’s journey and for two days before its start. Before the thirty-second night of the journey, the source of inflation is finally vanquished. Assuming the prices immediately drop to their original values the next day, what is the annual rate of deflation of the value of gollld over that night?

4) In one of the journeys of the hero mentioned in the above problem, the hero encounters a locked door. The hero is a swordsman and can deal 200 HP of damage with one blow, which takes one second to execute. With the hero is an archer who can shoot an arrow which deals 250 HP of damage with one blow, which takes one and a half seconds to execute. Also in the party is a mage whose best spell deals 900 HP of damage and takes five seconds to execute. Assuming that if they don’t bring the door down within 60 seconds, they will get bored and actually look for the key, minimally how much HP must the door have to force the party to ascertain the location of its key?

5) Suppose a paper beats a rock if it can actually wrap all the way around a rock. Suppose paper has a density of 500 kilograms per cubic meter and rock has a density of 6000 kilograms per cubic meter. Also, suppose paper is 0.1 millimeters thick and rectangular and that rock is spherical. Is there a particular weight above or below which the rock requires a heavier amount of paper to be able to beat it, and what is this weight?

6) Two Asians have just finished dining, and have finally agreed to split the bill exactly in half and have each party pay their half. The bill was $27.83, so the agreement required the chopping of a penny. Although the one who was agreed on to do the chopping was pretty proficient at martial arts, he is unsuccessful at splitting the penny close enough to half three-fourth of the time, thus calling for the bringing out of a new penny to split in half. Each preparing and splitting of a penny takes ten seconds. Suppose that each minute is worth $1.00 to each of them. How long must they have been arguing over who pays the bill before they agreed on the method of paying for it for there to be as much money wasted then as is the expected amount of money wasted while attempting to split a penny in half?

7) Suppose in the above problem that one person plans to pay with a credit card as soon as the other is done successfully halving the penny, and the other person is repaying half of the bill over to the first person in cash. What is the expected number of minutes it must take for the restaurant to become cash only so that the first person will have to take back the credit card and find cash to pay with?

8) The author of these problems was too lazy to write thirteen math problems, and thus decided to call that the “13” in the title of this set of problems was written in another base. Fearing that his excuse was too base, he decided to neutralize it a bit. Suppose this excuse has a pH of 13 (base five) and is monoprotic. 20 mL of the excuse is present. One molar hydrochloric acid must be titrated with this excuse to reach neutrality. If it takes one second to dispense 5 milliliters of acid, how much shorter does it take to neutralize the excuse a bit rather than to neutralize it a byte, assuming that a byte is the full amount of neutralization?