Misplaced Pages

Revision as of 20:30, 5 July 2018

This article may be too long to read and navigate comfortably. Consider splitting content into sub-articles, condensing it, or adding subheadings. Please discuss this issue on the article's talk page. (March 2018)

The duodecimal system (also known as base 12 or dozenal) is a positional notation numeral system using twelve as its base. In this system, the number ten may be written by a rotated "2" (2) and the number eleven by a rotated "3" (3). This notation was introduced by Sir Isaac Pitman. These digit forms are available as Unicode characters on computerized systems since June 2015 as ↊ (Code point 218A) and ↋ (Code point 218B), respectively. Other notations use "A", "T", or "X" for ten and "B" or "E" for eleven. The number twelve (that is, the number written as "12" in the base ten numerical system) is instead written as "10" in duodecimal (meaning "1 dozen and 0 units", instead of "1 ten and 0 units"), whereas the digit string "12" means "1 dozen and 2 units" (i.e. the same number that in decimal is written as "14"). Similarly, in duodecimal "100" means "1 gross", "1000" means "1 great gross", and "0.1" means "1 twelfth" (instead of their decimal meanings "1 hundred", "1 thousand", and "1 tenth").

The number twelve, a superior highly composite number, is the smallest number with four non-trivial factors (2, 3, 4, 6), and the smallest to include as factors all four numbers (1 to 4) within the subitizing range, and the smallest abundant number. As a result of this increased factorability of the radix and its divisibility by a wide range of the most elemental numbers (whereas ten has only two non-trivial factors: 2 and 5, and not 3, 4, or 6), duodecimal representations fit more easily than decimal ones into many common patterns, as evidenced by the higher regularity observable in the duodecimal multiplication table. As a result, duodecimal has been described as the optimal number system. Of its factors, 2 and 3 are prime, which means the reciprocals of all 3-smooth numbers (such as 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 27, 32, 36, ...) have a terminating representation in duodecimal. In particular, the five most elementary fractions (+1⁄2, +1⁄3, +2⁄3, +1⁄4 and +3⁄4) all have a short terminating representation in duodecimal (0.6, 0.4, 0.8, 0.3 and 0.9, respectively), and twelve is the smallest radix with this feature (because it is the least common multiple of 3 and 4). This all makes it a more convenient number system for computing fractions than most other number systems in common use, such as the decimal, vigesimal, binary, octal and hexadecimal systems. Although the trigesimal and sexagesimal systems (where the reciprocals of all 5-smooth numbers terminate) do even better in this respect, this is at the cost of unwieldy multiplication tables and a much larger number of symbols to memorize.

Origin

In this section, numerals are based on decimal places. For example, 10 means ten, 12 means twelve.

Languages using duodecimal number systems are uncommon. Languages in the Nigerian Middle Belt such as Janji, Gbiri-Niragu (Gure-Kahugu), Piti, and the Nimbia dialect of Gwandara; the Chepang language of Nepal and the Mahl language of Minicoy Island in India are known to use duodecimal numerals.

Germanic languages have special words for 11 and 12, such as eleven and twelve in English. However, they are considered to come from Proto-Germanic *ainlif and *twalif (respectively one left and two left), both of which were decimal.

Historically, units of time in many civilizations are duodecimal. There are twelve signs of the zodiac, twelve months in a year, and the Babylonians had twelve hours in a day (although at some point this was changed to 24, which is twice as 12). Traditional Chinese calendars, clocks, and compasses are based on the twelve Earthly Branches. There are 12 inches in an imperial foot, 12 troy ounces in a troy pound, 12 old British pence in a shilling, 24 (12×2) hours in a day, and many other items counted by the dozen, gross (144, square of 12) or great gross (1728, cube of 12). The Romans used a fraction system based on 12, including the uncia which became both the English words ounce and inch. Pre-decimalisation, Ireland and the United Kingdom used a mixed duodecimal-vigesimal currency system (12 pence = 1 shilling, 20 shillings or 240 pence to the pound sterling or Irish pound), and Charlemagne established a monetary system that also had a mixed base of twelve and twenty, the remnants of which persist in many places.

The importance of 12 has been attributed to the number of lunar cycles in a year, and also to the fact that humans have 12 finger bones (phalanges) on one hand (three on each of four fingers). It is possible to count to 12 with the thumb acting as a pointer, touching each finger bone in turn. A traditional finger counting system still in use in many regions of Asia works in this way, and could help to explain the occurrence of numeral systems based on 12 and 60 besides those based on 10, 20 and 5. In this system, the one (usually right) hand counts repeatedly to 12, displaying the number of iterations on the other (usually left), until five dozens, i. e. the 60, are full.

Notations and pronunciations

Transdecimal symbols

In a duodecimal place system twelve is written as 10, but there are numerous proposals for how to write ten and eleven. The simplified notations use only basic and easy to access letters such as T and E (for ten and eleven), X and Z, t and e, d and k, others use A and B or a and b as in the hexadecimal system. Some employ Greek letters such as δ (standing for Greek δέκα 'ten') and ε (for Greek ένδεκα 'eleven'), or τ and ε. Frank Emerson Andrews, an early American advocate for duodecimal, suggested and used in his book New Numbers an X (from the Roman numeral for ten) and a script E (ℰ, U+2130).

The Dozenal Society of Great Britain proposes a rotated digit two 2 (↊, U+218A) for ten and a reversed or rotated digit three 3 (↋, U+218B) for eleven. This notation was introduced by Sir Isaac Pitman.

Until 2015, the Dozenal Society of America (DSA) used and , the symbols devised by William Addison Dwiggins. After the Pitman digits (↊, U+218A and ↋, U+218B) were added to Unicode in 2015, the DSA took a vote and then began publishing content using the Pitman digits instead. They still use the letters X and E as the equivalent in ASCII text.

Other proposals are more creative or aesthetic, for example, Edna Kramer in her 1951 book The Main Stream of Mathematics used a six-pointed asterisk (sextile) ⚹ for ten and a hash (or octothorpe) # for eleven. The symbols were chosen because they are available in typewriters and already present in telephone dials. This notation was used in publications of the Dozenal Society of America in the period 1974–2008. Many don't use any of the Hindu-Arabic symbols, under the principle of "separate identity."

Base notation

There are also varying proposals of how to distinguish a duodecimal number from a decimal one, or one in a different base. They include italicizing duodecimal numbers ( 54 = 64 ), adding a "Humphrey point" (a semicolon ";" instead of a decimal point "." ) to duodecimal numbers ( 54; = 64. ) ( 54;0 = 64.0 ), or some combination of the two. More also add extra marking to one or more bases. Others use subscript or affixed labels to indicate the base, allowing for more than decimal and duodecimal to be represented:

This allows one to write "54_z = 64_d," "54_twelve = 64_ten" or "doz 54 = dec 64." In programming, binary, octal, and hexadecimal often use a similar scheme: a binary number starts with 0b, octal with 0o, and hexadecimal with 0x.

Pronunciation

The Dozenal Society of America suggests the pronunciation of ten and eleven as "dek" and "el", each order has its own name and the prefix e- is added for fractions. The symbol corresponding to the decimal point or decimal comma, separating the whole number part from the fractional part, is the semicolon ";". The overall system is:

Multiple digits in this are pronounced differently. 12 is "one do two", 30 is "three do", 100 is "one gro", BA9 (ET9) is "el gro dek do nine", B8,65A,300 (E8,65T,300) is "el do eight bi-mo, six gro five do dek mo, three gro", and so on.

Advocacy and "dozenalism"

The case for the duodecimal system was put forth at length in F. Emerson Andrews' 1935 book New Numbers: How Acceptance of a Duodecimal Base Would Simplify Mathematics. Emerson noted that, due to the prevalence of factors of twelve in many traditional units of weight and measure, many of the computational advantages claimed for the metric system could be realized either by the adoption of ten-based weights and measure or by the adoption of the duodecimal number system.

Both the Dozenal Society of America and the Dozenal Society of Great Britain promote widespread adoption of the base-twelve system. They use the word "dozenal" instead of "duodecimal" to avoid the more overtly base-ten terminology. It should be noted that the etymology of 'dozenal' is itself also an expression based on base-ten terminology since 'dozen' is a direct derivation of the French word 'douzaine' which is a derivative of the French word for twelve, douze which is related to the old French word 'doze' from Latin 'duodecim'.

It has been suggested by some members of the Dozenal Society of America and Duodecimal Society of Great Britain that a more apt word would be 'uncial'. Uncial is a derivation of the Latin word 'one-twelfth' which is 'uncia' and also the base-twelve analogue of the Latin word 'one-tenth' which is 'decima'. In the same manner as decimal comes from the Latin word for one-tenth decima, (Latin for ten was decem), the direct analogue for a base-twelve system is uncial. An early use of this word can be found in Vol 1 Issue 2 of The Duodecimal Bulletin of the DSA dated June 1945 in which a submission on page 9 by a Pvt William S. Crosby titled "The Uncial Jottings of a Harried Infantryman", he includes the same argument for the word 'uncial'. Although not accepted by either of these two 'Uncial' societies, the use is beginning to grow.

The renowned mathematician and mental calculator Alexander Craig Aitken was an outspoken advocate of the advantages and superiority of duodecimal over decimal:

The duodecimal tables are easy to master, easier than the decimal ones; and in elementary teaching they would be so much more interesting, since young children would find more fascinating things to do with twelve rods or blocks than with ten. Anyone having these tables at command will do these calculations more than one-and-a-half times as fast in the duodecimal scale as in the decimal. This is my experience; I am certain that even more so it would be the experience of others.

— A. C. Aitken, "Twelves and Tens" in The Listener (January 25, 1962)

But the final quantitative advantage, in my own experience, is this: in varied and extensive calculations of an ordinary and not unduly complicated kind, carried out over many years, I come to the conclusion that the efficiency of the decimal system might be rated at about 65 or less, if we assign 100 to the duodecimal.

— A. C. Aitken, The Case Against Decimalisation (1962)

In Jorge Luis Borges' short story Tlön, Uqbar, Orbis Tertius Herbert Ashe, a melancholy English engineer, working for the Southern Argentine Railway company, is converting a duodecimal number system to a hexadecimal system. He leaves behind on his death in 1937 a manuscript Orbis Tertius that posthumously identifies him as one of the anonymous authors of the encyclopaedia of Tlön.

In Leo Frankowski's Conrad Stargard novels, Conrad introduces a duodecimal system of arithmetic at the suggestion of a merchant, who is accustomed to buying and selling goods in dozens and grosses, rather than tens or hundreds. He then invents an entire system of weights and measures in base twelve, including a clock with twelve hours in a day, rather than twenty-four hours.

In Lee Carroll's Kryon: Alchemy of the Human Spirit, a chapter is dedicated to the advantages of the duodecimal system. The duodecimal system is supposedly suggested by Kryon (a fictional entity believed in by New Age circles) for all-round use, aiming at better and more natural representation of nature of the Universe through mathematics. An individual article "Mathematica" by James D. Watt (included in the above publication) exposes a few of the unusual symmetry connections between the duodecimal system and the golden ratio, as well as provides numerous number symmetry-based arguments for the universal nature of the base-12 number system.

In "Little Twelvetoes", American television series Schoolhouse Rock! portrayed an alien child using base-twelve arithmetic, using "dek", "el" and "doh" as names for ten, eleven and twelve, and Andrews' script-X and script-E for the digit symbols.

In computing

In March 2013, a proposal was submitted to include the digit forms for ten and eleven propagated by the Dozenal Societies of Great Britain and America in the Unicode Standard. Of these, the British forms were accepted for encoding as characters at code points U+218A turned digit two (↊) and U+218B turned digit three (↋) They have been included in the Unicode 8.0 release in June 2015.

Unicode points U+218C and U+218D seem to be reserved for the Dwiggins digits (stylized X and E).

Few fonts support these new characters, but Abibas, EB Garamond, Everson Mono, Squarish Sans CT, and Symbola do.

Also, the turned digits two and three are available in LaTeX as \textturntwo and \textturnthree.

Duodecimal clock

• Dozenal Clock by Joshua Harkey
• Dozenal Clock with four hands and a digital display, in several variants, by Paul Rapoport

Duodecimal metric systems

Systems of measurement proposed by dozenalists include:

• Tom Pendlebury's TGM system
• Takashi Suga's Universal Unit System

Comparison to other numeral systems

The number 12 has six factors, which are 1, 2, 3, 4, 6, and 12, of which 2 and 3 are prime. The decimal system has only four factors, which are 1, 2, 5, and 10, of which 2 and 5 are prime. Vigesimal (base 20) adds two factors to those of ten, namely 4 and 20, but no additional prime factor. Although twenty has 6 factors, 2 of them prime, similarly to twelve, it is also a much larger base, and so the digit set and the multiplication table are much larger. Binary has only two factors, 1 and 2, the latter being prime. Hexadecimal (base 16) has five factors, adding 4, 8 and 16 to those of 2, but no additional prime. Trigesimal (base 30) is the smallest system that has three different prime factors (all of the three smallest primes: 2, 3 and 5) and it has eight factors in total (1, 2, 3, 5, 6, 10, 15, and 30). Sexagesimal—which the ancient Sumerians and Babylonians among others actually used—adds the four convenient factors 4, 12, 20, and 60 to this but no new prime factors. The smallest system that has four different prime factors is base 210 and the pattern follows the primorials. In all base systems, there are similarities to the representation of multiples of numbers which are one less than the base.

Conversion tables to and from decimal

To convert numbers between bases, one can use the general conversion algorithm (see the relevant section under positional notation). Alternatively, one can use digit-conversion tables. The ones provided below can be used to convert any duodecimal number between 0.01 and ƐƐƐ,ƐƐƐ.ƐƐ to decimal, or any decimal number between 0.01 and 999,999.99 to duodecimal. To use them, the given number must first be decomposed into a sum of numbers with only one significant digit each. For example:

123,456.78 = 100,000 + 20,000 + 3,000 + 400 + 50 + 6 + 0.7 + 0.08

This decomposition works the same no matter what base the number is expressed in. Just isolate each non-zero digit, padding them with as many zeros as necessary to preserve their respective place values. If the digits in the given number include zeroes (for example, 102,304.05), these are, of course, left out in the digit decomposition (102,304.05 = 100,000 + 2,000 + 300 + 4 + 0.05). Then the digit conversion tables can be used to obtain the equivalent value in the target base for each digit. If the given number is in duodecimal and the target base is decimal, we get:

(duodecimal) 100,000 + 20,000 + 3,000 + 400 + 50 + 6 + 0.7 + 0.08 = (decimal) 248,832 + 41,472 + 5,184 + 576 + 60 + 6 + 0.583333333333... + 0.055555555555...

Now, because the summands are already converted to base ten, the usual decimal arithmetic is used to perform the addition and recompose the number, arriving at the conversion result:

Duodecimal  ----->  Decimal
  100,000     =    248,832
   20,000     =     41,472
    3,000     =      5,184
      400     =        576
       50     =         60
 +      6     =   +      6
        0.7   =          0.583333333333...
        0.08  =          0.055555555555...
--------------------------------------------
  123,456.78  =    296,130.638888888888...

That is, (duodecimal) 123,456.78 equals (decimal) 296,130.638 ≈ 296,130.64

If the given number is in decimal and the target base is duodecimal, the method is basically same. Using the digit conversion tables:

(decimal) 100,000 + 20,000 + 3,000 + 400 + 50 + 6 + 0.7 + 0.08 = (duodecimal) 49,ᘔ54 + Ɛ,6ᘔ8 + 1,8ᘔ0 + 294 + 42 + 6 + 0.849724972497249724972497... + 0.0Ɛ62ᘔ68781Ɛ05915343ᘔ0Ɛ62...

However, in order to do this sum and recompose the number, now the addition tables for the duodecimal system have to be used, instead of the addition tables for decimal most people are already familiar with, because the summands are now in base twelve and so the arithmetic with them has to be in duodecimal as well. In decimal, 6 + 6 equals 12, but in duodecimal it equals 10; so, if using decimal arithmetic with duodecimal numbers one would arrive at an incorrect result. Doing the arithmetic properly in duodecimal, one gets the result:

  Decimal  ----->  Duodecimal
  100,000     =     49,ᘔ54
   20,000     =      Ɛ,6ᘔ8
    3,000     =      1,8ᘔ0
      400     =        294
       50     =         42
 +      6     =   +      6
        0.7   =          0.849724972497249724972497...
        0.08  =          0.0Ɛ62ᘔ68781Ɛ05915343ᘔ0Ɛ62...
--------------------------------------------------------
  123,456.78  =     5Ɛ,540.943ᘔ0Ɛ62ᘔ68781Ɛ05915343ᘔ...

That is, (decimal) 123,456.78 equals (duodecimal) 5Ɛ,540.943ᘔ0Ɛ62ᘔ68781Ɛ059153... ≈ 5Ɛ,540.94

Duodecimal to decimal digit conversion

Decimal to duodecimal digit conversion

Conversion of powers

Some properties

(In this section, all numbers are written with duodecimal, using ᘔ for ten and Ɛ for eleven, unless other bases mentioned)

All squares end with square digits (i.e. end with 0, 1, 4 or 9), if n is divisible by both 2 and 3, then n ends with 0, if n is not divisible by 2 or 3, then n ends with 1, if n is divisible by 2 but not by 3, then n ends with 4, if n is not divisible by 2 but by 3, then n ends with 9. If the unit digit of n is 0, then the dozens digit of n is either 0 or 3, if the unit digit of n is 1, then the dozens digit of n is even, if the unit digit of n is 4, then the dozen digit of n is 0, 1, 4, 5, 8 or 9, if the unit digit of n is 9, then the dozen digit of n is either 0 or 6. (More specially, all squares of (primes ≥ 5) end with 1)

The digital root of a square is 1, 3, 4, 5 or Ɛ.

No repdigits with more than one digit are squares, in fact, a square cannot end with three same digits except 000. (In contrast, in the decimal (base ᘔ) system squares may end in 444, the smallest example is 32 = ᘔ04 = 1444_ᘔ)

No four-digit palindromic numbers are squares. (we can easily to prove it, since all four-digit palindromic number are divisible by 11, and since they are squares, thus they must be divisible by 11 = 121, and the only four-digit palindromic number divisible by 121 are 1331, 2662, 3993, 5225, 6556, 7887, 8ƐƐ8, 9119, ᘔ44ᘔ and Ɛ77Ɛ, but none of them are squares)

R_n (where R_n is the repunit with length n) is a palindromic number for n ≤ Ɛ, but not for n ≥ 10, besides, 11 (also 1{0}1, i.e. 101, 1001, 10001, etc.) is a palindromic number for n ≤ 5, but not for n ≥ 6, and it is conjectured that no palindromic numbers are n-th powers if n ≥ 6.

A cube can end with all digits except 2, 6 and ᘔ (in fact, no perfect powers end with 2, 6 or ᘔ), if n is not congruent to 2 mod 4, then n ends with the same digit as n; if n is congruent to 2 mod 4, then n ends with the digit (the last digit of n +− 6).

The digital root of a cube can be any number.

If k≥2, then n ends with the same digit as n, thus, if i≥2, j≥2 and i and j have the same parity, then n and n end with the same digit.

Squares (and every powers) of 0, 1, 4, 9, 54, 69, 369, 854, 3854, 8369, Ɛ3854, 1Ɛ3854, ᘔ08369, ... end with the same digits as the number itself. (since they are automorphic numbers, from the only four solutions of x−x=0 in the ring of 10-adic numbers, these solutions are 0, 1, ...2Ɛ21Ɛ61Ɛ3854 and ...909ᘔ05ᘔ08369, since 10 is neither a prime not a prime power, the ring of the 10-adic numbers is not a field, thus there are solutions other than 0 and 1 for this equation in 10-adic numbers)

Except for 6 and 24, all even perfect numbers end with 54. Additionally, except for 6, 24 and 354, all even perfect numbers end with 054 or 854. Besides, if any odd perfect number exists, then it must end with 1, 09, 39, 69 or 99.

The digital root of an even perfect number is 1, 4, 6 or ᘔ.

Since 10 is the smallest abundant number, all numbers end with 0 are abundant numbers, besides, all numbers end with 6 except 6 itself are also abundant numbers.

The period of the unit digits of powers of a number must be a divisor of 2 (= λ(10), where λ is the Carmichael function).

The period of the final two digits of powers of a number must be a divisor of 10 (= λ(100)).

More generally, for every n≥2, the period of the final n digits of powers of a number must be a divisor of 10 (= λ(10)).

The period of the digital roots of powers of a number must be a divisor of ᘔ (= λ(Ɛ)).

The unit digit of a Fibonacci number can be any digit except 6 (if the unit digit of a Fibonacci number is 0, then the dozens digit of this number must also be 0, thus, all Fibonacci numbers divisible by 6 are also divisible by 100), and the unit digit of a Lucas number cannot be 0 or 9 (thus, no Lucas number is divisible by 10), besides, if a Lucas number ends with 2, then it must end with 0002, i.e., this number is congruent to 2 mod 10.

(Note that F_2ᘔ begins with L_1ᘔ, and F_2Ɛ begins with L_1Ɛ)

The period of the digit root of Fibonacci numbers is ᘔ.

The period of the unit digit of Fibonacci numbers is 20, the final two digits is also 20, the final three digits is 200, the final four digits is 2000, ..., the final n digits is 2×10 (n≥2). (see Pisano period)

There are only 13 possible values (of the totally 100 values, thus only 13%) of the final two digits of a Fibonacci number (see OEIS: A066853).

Except 0 = F₀ and 1 = F₁ = F₂, the only square Fibonacci number is 100 = F₁₀ (100 is the square of 10), thus, 10 is the only base such that 100 is a Fibonacci number (since 100 in a base is just the square of this base, and 0 and 1 cannot be the base of numeral system), and thus we can make the near value of the golden ratio: F₁₁/F₁₀ = 175/100 = 1.75 (since the ratio of two connected Fibonacci numbers is close to the golden ratio, as the numbers get large). Besides, the only cube Fibonacci number is 8 = F₆.

The smallest power of 2 starts with the digit Ɛ is 2 = Ɛ2ᘔ20ᘔ8. For all digits 1≤d≤Ɛ, there exists 0≤n≤21 such that 2 starts with the digit d.

2 = 59Ɛ18922Ɛ81631ᘔ39875663Ɛ89ᘔ853ᘔ91Ɛ595336ᘔ6114815ᘔ5ᘔ6929933ᘔ288Ɛ774Ɛ479575ᘔ628 may be the largest power of 2 not contain the digit 0, it has 65 digits. (Note that in the decimal (base ᘔ) system, the largest power of 2 not contain the digit 0 is 2 = Ɛ8196855ᘔ752550Ɛ455ᘔ1594 = 77371252455336267181195264_ᘔ, it has only 22 digits in base ᘔ)

The number 2 = 2 (see power of 2#Powers of two whose exponents are powers of two) is very close to googol (10), since it has ƐƐ digits. (thus, the Fermat number F₉ (=2+1) is very close to googol)

1001 is the first four-digit palindromic number, and it is also the smallest number expressible as the sum of two cubes in two different ways, i.e. 1001 = 1 + 1000 (=1 + 10) = 509 + 6Ɛ4 (=9 + ᘔ) (see taxicab number for other numbers), and it is also the smallest absolute Euler pseudoprime, note that there is no absolute Euler-Jacobi pseudoprime and no absolute strong pseudoprime. Since 1001 = 7×11×17, we can use the divisibility rule of 1001 (i.e. form the alternating sum of blocks of three from right to left) for the divisibility rule of 7, 11 and 17. Besides, if 6k+1, 10k+1 and 16k+1 are all primes, then the product of them must be a Carmichael number (absolute Fermat pseudoprime), the smallest case is indeed 1001 (for k = 1), but 1001 is not the smallest Carmichael number (the smallest Carmichael number is 3ᘔ9).

${\displaystyle {\sqrt {2}}}$ is very close to 1.5, since a near-value for ${\displaystyle {\sqrt {2}}}$ is 15/10 (=N₄/P₄, where N_n is nth NSW number, and P_n is nth Pell number, N_n/P_n is very close to ${\displaystyle {\sqrt {2}}}$ when n is large). Besides, ${\displaystyle {\sqrt {5}}}$ is very close to 2.2ᘔ, since a near-value for ${\displaystyle {\sqrt {5}}}$ is 22ᘔ/100 (=L₁₀/F₁₀, where L_n is nth Lucas number, and F_n is nth Fibonacci number, L_n/F_n is very close to ${\displaystyle {\sqrt {5}}}$ when n is large).

The reciprocal of n is terminating number if and only if n is 3-smooth, the 3-smooth numbers up to 1000 are 1, 2, 3, 4, 6, 8, 9, 10, 14, 16, 20, 23, 28, 30, 40, 46, 54, 60, 69, 80, 90, ᘔ8, 100, 116, 140, 160, 183, 194, 200, 230, 280, 300, 346, 368, 400, 460, 509, 540, 600, 690, 714, 800, 900, ᘔ16, ᘔ80, 1000.

If and only if n is a divisor of 20, then m = 1 mod n for every integer m coprime to n.

If and only if n is a divisor of 20, then all Dirichlet characters for n are all real.

If and only if n is a divisor of 20, then n is divisible by all numbers less than or equal to the square root of n.

If and only if n+1 is a divisor of 20, then ${\displaystyle C_{k}^{n}={\frac {n!}{k!(n-k)!}}}$ is squarefree for all 0 ≤ k ≤ n.

All prime numbers end with prime digits or 1 (i.e. end with 1, 2, 3, 5, 7 or Ɛ), more generally, except for 2 and 3, all prime numbers end with 1, 5, 7 or Ɛ (1 and all prime digits that do not divide 10), since all prime numbers other than 2 and 3 are coprime to 10.

The density of primes end with 1 is a relatively low, but the density of primes end with 5, 7 and Ɛ are nearly equal. (since all prime squares except 4 and 9 end with 1, no prime squares end with 5, 7 or Ɛ)

Except (3, 5), all twin primes end with (5, 7) or (Ɛ, 1).

If n ≥ 3 and n is not divisible by Ɛ, then there are infinitely many primes with digit sum n.

All palindromic primes except 11 has an odd number of digits, since all even-digit palindromic numbers are divisible by 11. The palindromic primes below 1000 are 2, 3, 5, 7, Ɛ, 11, 111, 131, 141, 171, 181, 1Ɛ1, 535, 545, 565, 575, 585, 5Ɛ5, 727, 737, 747, 767, 797, Ɛ1Ɛ, Ɛ2Ɛ, Ɛ6Ɛ.

All lucky numbers end with digit 1, 3, 7 or 9.

Except for 3, all Fermat primes end with 5.

Except for 3, all Mersenne primes end with 7.

Except for 2 and 3, all Sophie Germain primes end with 5 or Ɛ.

Except for 5 and 7, all safe primes end with Ɛ.

A prime p is Gaussian prime (prime in the ring ${\displaystyle Z}$, where ${\displaystyle i={\sqrt {-1}}}$) if and only if p ends with 7 or Ɛ (or p=3).

A prime p is Eisenstein prime (prime in the ring ${\displaystyle Z}$, where ${\displaystyle \omega ={\frac {-1+{\sqrt {3}}i}{2}}}$) if and only if p ends with 5 or Ɛ (or p=2).

A prime p can be written as x + y if and only if p ends with 1 or 5 (or p=2).

A prime p can be written as x + 3y if and only if p ends with 1 or 7 (or p=3).

All full reptend primes end with 5 or 7. (in fact, for all primes p ≥ 5, (p-1)/(the period length of 1/p) is odd if and only if p is end with 5 or 7, since 10 is a quadratic nonresidue mod p (i.e. ${\displaystyle \left({\frac {10}{p}}\right)=-1}$, where ${\displaystyle \left({\frac {m}{n}}\right)}$ is the Legendre symbol) if and only if p is end with 5 or 7, by quadratic reciprocity, and if 10 is a quadratic residue mod a prime, then 10 cannot be a primitive root mod this prime) However, the converse is not true, 17 is not a full reptend prime, since the recurring digits of 1/17 is 0.076Ɛ45076Ɛ45..., which has only period 6. If and only if p is a full reptend prime, then the recurring digits of 1/p is cyclic number, e.g. the recurring digits of 1/5 is the cyclic number 2497 (the cyclic permutations of the digits are this number multiplied by 1 to 4), and the recurring digits of 1/7 is the cyclic number 186ᘔ35 (the cyclic permutations of the digits are this number multiplied by 1 to 6). The full reptend primes below 1000 are 5, 7, 15, 27, 35, 37, 45, 57, 85, 87, 95, ᘔ7, Ɛ5, Ɛ7, 105, 107, 117, 125, 145, 167, 195, 1ᘔ5, 1Ɛ5, 1Ɛ7, 205, 225, 255, 267, 277, 285, 295, 315, 325, 365, 377, 397, 3ᘔ5, 3Ɛ5, 3Ɛ7, 415, 427, 435, 437, 447, 455, 465, 497, 4ᘔ5, 517, 527, 535, 545, 557, 565, 575, 585, 5Ɛ5, 615, 655, 675, 687, 695, 6ᘔ7, 705, 735, 737, 745, 767, 775, 785, 797, 817, 825, 835, 855, 865, 8Ɛ5, 8Ɛ7, 907, 927, 955, 965, 995, 9ᘔ7, 9Ɛ5, ᘔ07, ᘔ17, ᘔ35, ᘔ37, ᘔ45, ᘔ77, ᘔ87, ᘔ95, ᘔƐ7, Ɛ25, Ɛ37, Ɛ45, Ɛ95, Ɛ97, Ɛᘔ5, ƐƐ5, ƐƐ7. (Note that for the primes end with 5 or 7 below 30 (5, 7, 15, 17, 25 and 27, all numbers end with 5 or 7 below 30 are primes), 5, 7, 15 and 27 are full reptend primes, and since 5×25 = 101 = ${\displaystyle \Phi _{4}(10)}$, the period of 25 is 4, which is the same as the period of 5, and we can use the test of the divisiblity of 5 to test that of 25 (form the alternating sum of blocks of two from right to left), and since 7×17 = Ɛ1 = ${\displaystyle \Phi _{6}(10)}$, the period of 17 is 6, which is the same as the period of 7, and we can use the test of the divisiblity of 7 to test that of 17 (form the alternating sum of blocks of three from right to left), thus, 17 and 25 are not full reptend primes, and they are the only two non-full reptend primes end with 5 or 7 below 30)

By Midy theorem, if p is a prime with even period length (let its period length be n), then if we let ${\displaystyle {\frac {a}{p}}=0.{\overline {a_{1}a_{2}a_{3}...a_{n}}}}$, then a_i + a_i+n/2 = Ɛ for every 1 ≤ i ≤ n/2. e.g. 1/5 = 0.249724972497..., and 24 + 97 = ƐƐ, and 1/7 = 0.186ᘔ35186ᘔ35..., and 186 + ᘔ35 = ƐƐƐ, all primes (other than 2 and 3) ≤ 37 except Ɛ, 1Ɛ and 31 have even period length, thus they can use Midy theorem to get an Ɛ-repdigit number, the length of this number is the period length of this prime. (see below for the recurring digits for 1/n for all n ≤ 30)

The unique primes below 10 are Ɛ, 11, 111, Ɛ0Ɛ1, ƐƐ01, 11111, 24727225, Ɛ0Ɛ0Ɛ0Ɛ0Ɛ1, Ɛ00Ɛ00ƐƐ0ƐƐ1, 100ƐƐƐᘔƐᘔƐƐ000101, 1111111111111111111, ƐƐƐƐ0000ƐƐƐƐ0000ƐƐƐƐ0001, 100ƐƐƐᘔƐƐ0000ƐƐƐᘔƐƐ000101, 10ƐƐƐᘔᘔᘔƐ011110ƐᘔᘔᘔƐ00011, ƐƐƐƐƐƐƐƐ00000000ƐƐƐƐƐƐƐƐ00000001, ƐƐƐ000000ƐƐƐ000000ƐƐƐƐƐƐ000ƐƐƐƐƐƐ001, and the period length of their reciprocals are 1, 2, 3, ᘔ, 10, 5, 18, 1ᘔ, 19, 50, 17, 48, 70, 5ᘔ, 68, 53.

If p is a safe prime other than 5, 7 and Ɛ, then the period length of 1/p is (p-1)/2. (this is not true for all primes ends with Ɛ (other than Ɛ itself), the first counterexample is p = 2ƐƐ, where the period length of 1/p is only 37)

There is no full reptend prime ends with 1, since 10 is quadratic residue for all primes end with 1. (In contrast, in decimal (base ᘔ) system there are some such primes, and may be infinitely many such primes, the first few such primes in that base are 61_ᘔ = 51, 131_ᘔ = ᘔƐ, 181_ᘔ = 131, 461_ᘔ = 325, 491_ᘔ = 34Ɛ, see OEIS: A073761) (if so, then this prime p is a proper prime (i.e. for the reciprocal of such primes (1/p), each digit 0, 1, 2, ..., Ɛ appears in the repeating sequence the same number of times as does each other digit (namely, (p−1)/10 times)), see repeating decimal#Fractions with prime denominators) (In fact, not only for base 10 such primes do not exist, for all bases = 0 mod 4 (i.e. bases end with digit 0, 4 or 8), such primes do not exist)

5 and 7 are the only two safe primes which are also full reptend primes, since except 5 and 7, all safe primes end with Ɛ, and 10 is quadratic residue for all primes end with Ɛ. (In contrast, in decimal (base ᘔ) system there may be infinitely many such primes, the first few such primes in that base are 7_ᘔ = 7, 23_ᘔ = 1Ɛ, 47_ᘔ = 3Ɛ, 59_ᘔ = 4Ɛ, 167_ᘔ = 11Ɛ, see OEIS: A000353) (if so, then this prime p produces a stream of p−1 pseudo-random digits, see repeating decimal#Fractions with prime denominators) (In fact, not only for base 10 there are only finitely many such primes, of course for square bases (bases of the form k) only 2 may be full reptend prime (if the base is odd), and all odd primes are not full reptend primes, but since all safe primes are odd primes, for these bases such primes do not exist, besides, for the bases of the form 3k, only 5 and 7 can be such primes, the proof for these bases is completely the same as that for base 10)

For example, ${\displaystyle {\frac {1}{17}}}$ has period level 3, thus the numbers ${\displaystyle {\frac {a}{17}}}$ with integer 1 ≤ a ≤ 16 from 3 different cycles: 076Ɛ45 (for a = 1, 7, 8, Ɛ, 10, 16), 131ᘔ8ᘔ (for a = 2, 3, 5, 12, 14, 15) and 263958 (for a = 4, 6, 9, ᘔ, 11, 13). Besides, ${\displaystyle {\frac {1}{15}}}$ has period level 1, thus this number is a cyclic number and 15 is a full-reptend prime, and all of the numbers ${\displaystyle {\frac {a}{15}}}$ with integer 1 ≤ a ≤ 14 from the cycle 08579214Ɛ36429ᘔ7.

There are only 9 repunit primes below R₁₀₀₀: R₂, R₃, R₅, R₁₇, R₈₁, R₉₁, R₂₂₅, R₂₅₅ and R_4ᘔ5 (R_n is the repunit with length n). If p is a Sophie Germain prime other than 2, 3 and 5, then R_p is divisible by 2p+1, thus R_p is not prime. (The length for the repunit (probable) primes are 2, 3, 5, 17, 81, 91, 225, 255, 4ᘔ5, 5777, 879Ɛ, 198Ɛ1, 23175, 311407, ..., note that 879Ɛ is the smallest (and the only known) such number ends with Ɛ)

By Fermat's little theorem, if p is a prime other than 2, 3 and Ɛ, then p divides the repunit with length p−1. (The converse is not true, the first counterexample is 55, which is composite (equals 5×11) but divides the repunit with length 54, the counterexamples up to 1000 are 55, 77, Ɛ1, 101, 187, 275, 4ᘔ7, 777, 781, Ɛ55, they are exactly the Fermat pseudoprimes for base 10 (composite numbers c such that 10 = 1 mod c) which are not divisible by Ɛ, they are called "deceptive primes", if n is deceptive prime, then R_n is also deceptive prime, thus there are infinitely may deceptive primes) Thus, we can prove that every positive integer coprime to 10 has a repunit multiple, and every positive integer has a multiple uses only 0's and 1's.

(this k is usually not prime, in fact, this k is not prime for all numbers n < 100 which are coprime to 10 except n = 55, and for n < 1000 which is coprime to 10, this k is prime only for n = 55, 101, 19Ɛ, 275 and 46Ɛ, and only 19Ɛ and 46Ɛ are itself prime, other 3 numbers are 5×11, 5×25 and 11×25, and this k for these n are successively 25, 11 and 5, which makes k×n = R₄ = 1111 = 5×11×25, besides, this k for n = 46Ɛ is 2ᘔ3Ɛ, which makes k×n = R₇ = 1111111, a repunit semiprime, and this k for n = 19Ɛ is a ᘔ8-digit prime number, with k×n = R_ᘔƐ, another repunit semiprime)

For every prime p except Ɛ, the repunit with length p is congruent to 1 mod p. (The converse is also not true, the counterexamples up to 1000 are 4, 6, 10, 33, 55, 77, Ɛ1, 101, 187, 1Ɛ0, 275, 444, 4ᘔ7, 777, 781, Ɛ55, they are called "repunit pseudoprimes" (or weak deceptive primes), all deceptive primes are also repunit pseudoprimes, if n is repunit pseudoprime, then R_n is also repunit pseudoprime, thus there are infinitely may repunit pseudoprimes. No repunit pseudoprimes are divisible by 8, 9 or Ɛ. (in fact, the repunit pseudoprimes are exactly the weak pseudoprimes for base 10 (composite numbers c such that 10 = 10 mod c) which are not divisible by Ɛ) Besides, the deceptive primes are exactly the repunit pseudoprimes which are coprime to 10)

Smallest multiple of n with digit sum 2 are: (0 if not exist)

2, 2, 20, 20, 101, 20, 1001, 20, 200, 1010, 0, 20, 11, 10010, 1010, 200, 100000001, 200, 1001, 1010, 10010, 0, 0, 20, 10000000001, 110, 2000, 10010, 101, 1010, 1000000000000001, 200, 0, 1000000010, 1000000000001, 200, ..., if and only if n is divisible by some prime p with 1/p odd period length, then such number does not exist.

Smallest multiple of n with digit sum 3 are: (0 if not exist)

3, 12, 3, 30, 21, 30, 12, 120, 30, 210, 0, 30, 0, 12, 210, 300, 201, 30, 10101, 210, 120, 0, 1010001, 120, 21, 0, 300, 120, 0, 210, 1010001, 1200, 0, 2010, 200001, 30, ..., such number does not exist for n divisible by Ɛ, 11 or 25.

Smallest multiple of n with digit sum 4 are: (0 if not exist)

4, 4, 13, 4, 13, 40, 103, 40, 130, 130, 0, 40, 22, 1030, 13, 40, 3001, 130, 2002, 130, 103, 0, 11101, 40, 10012, 22, 1300, 1030, 202, 130, 10003, 400, 0, 30010, 101101, 130, ..., such number is conjectured to exist for all n not divisible by Ɛ (of course, if n is divisible by Ɛ, then such number does not exist).

Smallest multiple of n with digit sum n are:

1, 2, 3, 4, 5, 6, 7, 8, 9, ᘔ, Ɛ, 1Ɛ0, 20Ɛ, 22ᘔ, 249, 268, 287, 2ᘔ6, 45ᘔ, 488, 4Ɛ6, 1Ɛᘔ, 8Ɛ4, 3Ɛᘔ0, 3ƐƐ, 23Ɛᘔ, 1899, ᘔᘔ8, 2Ɛ79, 4Ɛ96, 1Ɛᘔ9, 4ᘔᘔ8, 2ƐƐ9, 3ᘔƐᘔ, 799ᘔ, 5ƐƐ90, ..., such number is conjectured to exist for all n.

Write the recurring digits of 1/45 (=0.2872Ɛ3ᘔ23205525ᘔ784640ᘔᘔ4Ɛ9349081989Ɛ6696143757Ɛ117, which has period 44) to 44/45, we get a 44×44 prime reciprocal magic square (its magic number is 1Ɛᘔ), it is conjectured that there are infinitely many such primes, but 45 is the only such prime below 1000, all such primes are full reptend primes, i.e. the reciprocal of them are cyclic numbers, and 10 is a primitive root modulo these primes.

All numbers of the form 34{1} are composite (proof: 34{1_n} = 34×10+(10−1)/Ɛ = (309×10−1)/Ɛ and it can be factored to ((19×10−1)/Ɛ) × (19×10+1) for even n and divisible by 11 for odd n). Besides, 34 was proven to be the smallest n such that all numbers of the form n{1} are composite. However, the smallest prime of the form 23{1} is 23{1_Ɛ78}, it has Ɛ7ᘔ digits. The only other two n≤100 such that all numbers of the form n{1} are composite are 89 and 99 (the reason of 89 is the same as 34, and the reason of 99 is 99{1_n} is divisible by 5, 11 or 25).

The only known of the form 1{0}1 is 11 (see generalized Fermat prime), these are the primes obtained as the concatenation of a power of 10 followed by a 1. If n = 1 mod 11, then all numbers obtained as the concatenation of a power of n (>1) followed by a 1 are divisible by 11 and thus composite. Except 10, the smallest n not = 1 mod 11 such that all numbers obtained as the concatenation of a power of n (>1) followed by a 1 are composite was proven by Ɛᘔ, since all numbers obtained as the concatenation of a power of Ɛᘔ (>1) followed by a 1 are divisible by either Ɛ or 11 and thus composite. However, the smallest prime obtained as the concatenation of a power of 58 (>1) followed by a 1 is 10×58+1, it has 459655 digits.

All numbers of the form 1{5}1 are composite (proof: 1{5_n}1 = (14×10−41)/Ɛ and it can be factored to (4×10−7) × ((4×10−7)/Ɛ) for odd n and divisible by 11 for even n).

The emirps below 1000 are 15, 51, 57, 5Ɛ, 75, Ɛ5, 107, 117, 11Ɛ, 12Ɛ, 13Ɛ, 145, 157, 16Ɛ, 17Ɛ, 195, 19Ɛ, 1ᘔ7, 1Ɛ5, 507, 51Ɛ, 541, 577, 587, 591, 59Ɛ, 5Ɛ1, 5ƐƐ, 701, 705, 711, 751, 76Ɛ, 775, 785, 7ᘔ1, 7ƐƐ, Ɛ11, Ɛ15, Ɛ21, Ɛ31, Ɛ61, Ɛ67, Ɛ71, Ɛ91, Ɛ95, ƐƐ5, ƐƐ7.

The non-repdigit permutable primes below 10 are 15, 57, 5Ɛ, 117, 11Ɛ, 5ƐƐƐ (the smallest representative prime of the permutation set).

The non-repdigit circular primes below 10 are 15, 57, 5Ɛ, 117, 11Ɛ, 175, 1Ɛ7, 157Ɛ, 555Ɛ, 115Ɛ77 (the smallest representative prime of the cycle).

The first few Smarandache primes are the concatenation of the first 5, 15, 4Ɛ, 151, ... positive integers.

The only known Smarandache–Wellin primes are 2 and 2357Ɛ11.

There are exactly 15 minimal primes, and they are 2, 3, 5, 7, Ɛ, 11, 61, 81, 91, 401, ᘔ41, 4441, ᘔ0ᘔ1, ᘔᘔᘔᘔ1, 44ᘔᘔᘔ1, ᘔᘔᘔ0001, ᘔᘔ000001.

The smallest weakly prime is 6Ɛ8ᘔƐ77.

The largest left-truncatable prime is 28-digit 471ᘔ34ᘔ164259Ɛᘔ16Ɛ324ᘔƐ8ᘔ32Ɛ7817, and the largest right-truncatable prime is ᘔ-digit 375ƐƐ5Ɛ515.

The only two base 10 Wieferich primes up to 10 are 1685 and 5Ɛ685, note that both of the numbers end with 685, and it is conjectured that all base 10 Wieferich primes end with 685. (there is also a note for the only two known base 2 Wieferich primes (771 and 2047) minus 1 written in base 2, 8 (= 2) and 14 (= 2), 770 = 010001000100₂ = 444₁₄ is a repdigit in base 14, and 2046 = 110110110110₂ = 6666₈ is also a repdigit in base 8, see Wieferich prime#Binary periodicity of p − 1)

There are 1, 2, 3, 5 and 6-digit (but not 4-digit) narcissistic numbers, there are totally 73 narcissistic numbers, the first few of which are 1, 2, 3, 4, 5, 6, 7, 8, 9, ᘔ, Ɛ, 25, ᘔ5, 577, 668, ᘔ83, 14765, 938ᘔ4, 369862, ᘔ2394ᘔ, ..., the largest of which is 43-digit 15079346ᘔ6Ɛ3Ɛ14ƐƐ56Ɛ395898Ɛ96629ᘔ8Ɛ01515344Ɛ4Ɛ0714Ɛ. (see OEIS: A161949)

The only two factorions are 1 and 2.

The only seven happy numbers below 1000 are 1, 10, 100, 222, 488, 848 and 884, almost all natural numbers are unhappy. All unhappy numbers get to one of these four cycles: {5, 21}, {8, 54, 35, 2ᘔ, 88, ᘔ8, 118, 56, 51, 22}, {18, 55, 42}, {68, 84}, or one of the only two fixed points other than 1: 25 and ᘔ5. (In contrast, in the decimal (base ᘔ) system there are Ɛᘔ happy numbers below 1000_ᘔ (=6Ɛ4), and all unhappy number get to this cycle: {4, 16, 37, 58, 89, 145, 42, 20}, there are no fixed points other than 1)

If we use the sum of the cubes (instead of squares) of the digits, then every natural numbers get to either 1 or the cycle {8, 368, 52Ɛ, ᘔ20, 700, 247, 2ᘔ7, 947, 7ᘔ8, 10ᘔ7, 940, 561, 246, 200}. (In contrast, in the decimal (base ᘔ) system all multiple of 3 get to 153_ᘔ (=109), and other numbers get to either one of these four fixed points: 1, 370, 371, 407, or one of these four cycles: {55, 250, 133}, {136, 244}, {160, 217, 352}, {919, 1459}) (for the example of the famous Hardy–Ramanujan number 1001 = 9 + ᘔ, we know that this sequence with initial term 9ᘔ is 9ᘔ, 1001, 2, 8, 368, 52Ɛ, ᘔ20, 700, 247, 2ᘔ7, 947, 7ᘔ8, 10ᘔ7, 940, 561, 246, 200, 8, 368, 52Ɛ, ᘔ20, 700, 247, 2ᘔ7, 947, 7ᘔ8, 10ᘔ7, 940, 561, 246, 200, 8, ...)

The harshad numbers up to 200 are 1, 2, 3, 4, 5, 6, 7, 8, 9, ᘔ, Ɛ, 10, 1ᘔ, 20, 29, 30, 38, 40, 47, 50, 56, 60, 65, 70, 74, 80, 83, 90, 92, ᘔ0, ᘔ1, Ɛ0, 100, 10ᘔ, 110, 115, 119, 120, 122, 128, 130, 134, 137, 146, 150, 153, 155, 164, 172, 173, 182, 191, 1ᘔ0, 1Ɛ0, 1Ɛᘔ, 200, although the sequence of factorials begins with harshad numbers, not all factorials are harshad numbers, after 7! (=2Ɛ00, with digit sum 11 but 11 does not divide 7!), 8ᘔ4! is the next that is not (8ᘔ4! has digit sum 8275 = Ɛ×8Ɛ7, thus not divide 8ᘔ4!). There are no 21 consecutive integers that are all harshad numbers, but there are infinitely many 20-tuples of consecutive integers that are all harshad numbers.

The Kaprekar numbers up to 10000 are 1, Ɛ, 56, 66, ƐƐ, 444, 778, ƐƐƐ, 12ᘔᘔ, 1640, 2046, 2929, 3333, 4973, 5Ɛ60, 6060, 7249, 8889, 9293, 9Ɛ76, ᘔ580, ᘔ912, ƐƐƐƐ.

The Kaprekar's routine of any four-digit number which is not repdigit converges to either the cycle {3ƐƐ8, 8284, 6376} or the cycle {4198, 8374, 5287, 6196, 7ƐƐ4, 7375}, and the Kaprekar map of any three-digit number which is not repdigit converges to the fixed point 5Ɛ6, and the Kaprekar map of any three-digit number which is not repdigit converges to the cycle {0Ɛ, ᘔ1, 83, 47, 29, 65}.

The self numbers up to 600 are 1, 3, 5, 7, 9, Ɛ, 20, 31, 42, 53, 64, 75, 86, 97, ᘔ8, Ɛ9, 10ᘔ, 110, 121, 132, 143, 154, 165, 176, 187, 198, 1ᘔ9, 1Ɛᘔ, 20Ɛ, 211, 222, 233, 244, 255, 266, 277, 288, 299, 2ᘔᘔ, 2ƐƐ, 310, 312, 323, 334, 345, 356, 367, 378, 389, 39ᘔ, 3ᘔƐ, 400, 411, 413, 424, 435, 446, 457, 468, 479, 48ᘔ, 49Ɛ, 4Ɛ0, 501, 512, 514, 525, 536, 547, 558, 569, 57ᘔ, 58Ɛ, 5ᘔ0, 5Ɛ1.

The Friedman numbers up to 1000 are 121=11, 127=7×21, 135=5×31, 144=4×41, 163=3×61, 368=8, 376=6×73, 441=(4+1), 445=5+4.

The Keith numbers up to 1000 are 11, 15, 1Ɛ, 22, 2ᘔ, 31, 33, 44, 49, 55, 62, 66, 77, 88, 93, 99, ᘔᘔ, ƐƐ, 125, 215, 24ᘔ, 405, 42ᘔ, 654, 80ᘔ, 8ᘔ3, ᘔ59.

There are totally 71822 polydivisible numbers, the largest of which is 24-digit 606890346850Ɛᘔ6800Ɛ036206464. However, there are no Ɛ-digit polydivisible numbers contain the digits 1 to Ɛ exactly once each.

The candidate Lychrel numbers up to 1000 are 179, 1Ɛ9, 278, 2Ɛ8, 377, 3Ɛ7, 476, 4Ɛ6, 575, 5Ɛ5, 674, 6Ɛ4, 773, 7Ɛ3, 872, 8Ɛ2, 971, 9Ɛ1, ᘔ2Ɛ, ᘔ3Ɛ, ᘔ5Ɛ, ᘔ70, ᘔᘔƐ, ᘔƐ0, Ɛ2ᘔ, Ɛ3ᘔ, Ɛ5ᘔ, Ɛᘔᘔ. The only suspected Lychrel seed numbers up to 1000 are 179, 1Ɛ9, ᘔ3Ɛ and ᘔ5Ɛ. However, it is unknown whether any Lychrel number exists. (Lychrel numbers only known to exist in these bases: Ɛ, 15, 18, 22 and all powers of 2)

Most numbers that end with 2 are nontotient (in fact, all nontotients < 58 except 2ᘔ end with 2), except 2 itself, the first counterexample is 92, which equals φ(ᘔ1) = φ(Ɛ) and φ(182) = φ(2×Ɛ), next counterexample is 362, which equals φ(381) = φ(1Ɛ) and φ(742) = φ(2×1Ɛ), there are only 9 such numbers ≤ 10000 (the number 2 itself is not counted), all such numbers (except the number 2 itself) are of the form φ(p) = p(p−1), where p is a prime ends with Ɛ.

If we let the musical notes in an octave be numbers in the cyclic group Z₁₀: C=0, C#=1, D=2, Eb=3, E=4, F=5, F#=6, G=7, Ab=8, A=9, Bb=ᘔ, B=Ɛ (see pitch class and music scale) (thus, if we let the middle C be 0, then the notes in a piano are -33 to 40), then x and x+3 are small 3-degree, x and x+4 are big 3-degree, x and x+7 are perfect 5 degree (thus, we can use 7x for x = 0 to Ɛ to get the five degree cycle), etc. (since an octave is 10 semitones, a small 3-degree is 3 semitones, a big 3-degree is 4 semitones, and a perfect 5 degree is 7 semitones, etc.) (if we let an octave be 1, then a semitone will be 0.1, and we can write all 10 notes on a cycle, the difference of two connected notes is 26 degrees or ${\displaystyle {\frac {\pi }{6}}}$ radians) Besides, the x major chord (x) is {x, x+4, x+7} in Z₁₀, and the x minor chord (xm) is {x, x+3, x+7} in Z₁₀, and the x major 7th chord (x) is {x, x+4, x+7, x+Ɛ}, and the x minor 7th chord (x) is {x, x+3, x+7, x+ᘔ}, and the x dominant 7th chord (x) is {x, x+4, x+7, x+ᘔ}, and the x diminished 7th triad (x) is {x, x+3, x+6, x+9}, since the frequency of x and x+6 is not simple integer fraction, they are not harmonic, and this diminished 7th triad is corresponding the beast number 666 (three 6's). Besides, x major scale uses the notes {x, x+2, x+4, x+5, x+7, x+9, x+Ɛ}, and x minor scale uses the notes {x, x+2, x+3, x+5, x+7, x+8, x+ᘔ}. Besides, the frequency of x+10 is twice as that of x, the frequency of x+7 is 1.6 (=3/2) times as that of x, and the frequency of x+5 is 1.4 (=4/3) times as that of x, they are all simple integer fractions (ratios of small integers), and they all have at most one digit after the duodecimal point, and we can found that 1.6 = ᘔ9.8Ɛ5809 is very close to 2 = ᘔ8, since 2 = 2134ᘔ8 is very close to 3 = 217669, the simple frequency fractions found for the scales are only 0.6, 0.8, 0.9, 1.4, 1.6 and 2, however, since the frequency of x+10 is twice as that of x, thus the frequency of x+1 (i.e. a semitone higher than x) is ${\displaystyle {\sqrt{2}}}$ (=2) times as that of x. Let f(x) be the frequency of x, then we have f(2)/f(0) = 9/8 (=1.16), f(4)/f(2) = ᘔ/9 (=1.14), and f(5)/f(4) = 14/13 (this number is very close to ${\displaystyle {\sqrt{2}}}$), and thus we have that f(5)/f(0) = (9/8) × (ᘔ/9) × (14/13) = 4/3. Also, we can found that 2 is very close to 1.4, and 2 is very close to 1.6.

All orders of non-cyclic simple group end with 0 (thus, all orders of unsolvable group end with 0), however, we can prove that no groups with order 10, 20, 30 or 40 are simple, thus 50 is the smallest order of non-cyclic simple group (thus, all groups with order < 50 are solvable), (50 is the order of the alternating group A₅, which is a non-cyclic simple group, and thus an unsolvable group) next three orders of non-cyclic simple group are 120, 260 and 360. (Edit: I found that this is not completely true (although this is true for all orders ≤ 14000), the smallest counterexample is 14ᘔ28, however, all such orders are divisible by 4 and either 3 or 5 (i.e. divisible by either 10 or 18), and all such orders have at least 3 distinct prime factors, by these conditions, the smallest possible such order is indeed 50 = 2 × 3 × 5, next possible such order is 70 = 2 × 3 × 7, however, by Sylow theorems, the number of Sylow 7-subgroups of all groups with order 70 (i.e. the number of subgroups with order 7 of all groups with order 70) is congruent to 1 mod 7 and divides 70, hence must be 1, thus the subgroup with order 7 is a normal subgroup of the group with order 70, thus all groups with order 70 have a nontrivial normal subgroup and cannot be simple groups)

Prime numbers

(In this section, all numbers are written with duodecimal)

A natural number (i.e. 1, 2, 3, 4, 5, 6, etc.) is called a prime number (or a prime) if it has exactly two positive divisors, 1 and the number itself. Natural numbers greater than 1 that are not prime are called composite.

The first 1ᘔ5 prime numbers (all the prime numbers less than 1000) are:

2, 3, 5, 7, Ɛ, 11, 15, 17, 1Ɛ, 25, 27, 31, 35, 37, 3Ɛ, 45, 4Ɛ, 51, 57, 5Ɛ, 61, 67, 6Ɛ, 75, 81, 85, 87, 8Ɛ, 91, 95, ᘔ7, ᘔƐ, Ɛ5, Ɛ7, 105, 107, 111, 117, 11Ɛ, 125, 12Ɛ, 131, 13Ɛ, 141, 145, 147, 157, 167, 16Ɛ, 171, 175, 17Ɛ, 181, 18Ɛ, 195, 19Ɛ, 1ᘔ5, 1ᘔ7, 1Ɛ1, 1Ɛ5, 1Ɛ7, 205, 217, 21Ɛ, 221, 225, 237, 241, 24Ɛ, 251, 255, 25Ɛ, 267, 271, 277, 27Ɛ, 285, 291, 295, 2ᘔ1, 2ᘔƐ, 2Ɛ1, 2ƐƐ, 301, 307, 30Ɛ, 315, 321, 325, 327, 32Ɛ, 33Ɛ, 347, 34Ɛ, 357, 35Ɛ, 365, 375, 377, 391, 397, 3ᘔ5, 3ᘔƐ, 3Ɛ5, 3Ɛ7, 401, 40Ɛ, 415, 41Ɛ, 421, 427, 431, 435, 437, 447, 455, 457, 45Ɛ, 465, 46Ɛ, 471, 481, 485, 48Ɛ, 497, 4ᘔ5, 4Ɛ1, 4ƐƐ, 507, 511, 517, 51Ɛ, 527, 531, 535, 541, 545, 557, 565, 575, 577, 585, 587, 58Ɛ, 591, 59Ɛ, 5Ɛ1, 5Ɛ5, 5Ɛ7, 5ƐƐ, 611, 615, 617, 61Ɛ, 637, 63Ɛ, 647, 655, 661, 665, 66Ɛ, 675, 687, 68Ɛ, 695, 69Ɛ, 6ᘔ7, 6Ɛ1, 701, 705, 70Ɛ, 711, 71Ɛ, 721, 727, 735, 737, 745, 747, 751, 767, 76Ɛ, 771, 775, 77Ɛ, 785, 791, 797, 7ᘔ1, 7ƐƐ, 801, 80Ɛ, 817, 825, 82Ɛ, 835, 841, 851, 855, 85Ɛ, 865, 867, 871, 881, 88Ɛ, 8ᘔ5, 8ᘔ7, 8ᘔƐ, 8Ɛ5, 8Ɛ7, 901, 905, 907, 90Ɛ, 91Ɛ, 921, 927, 955, 95Ɛ, 965, 971, 987, 995, 9ᘔ7, 9ᘔƐ, 9Ɛ1, 9Ɛ5, 9ƐƐ, ᘔ07, ᘔ0Ɛ, ᘔ11, ᘔ17, ᘔ27, ᘔ35, ᘔ37, ᘔ3Ɛ, ᘔ41, ᘔ45, ᘔ4Ɛ, ᘔ5Ɛ, ᘔ6Ɛ, ᘔ77, ᘔ87, ᘔ91, ᘔ95, ᘔ9Ɛ, ᘔᘔ7, ᘔᘔƐ, ᘔƐ7, ᘔƐƐ, Ɛ11, Ɛ15, Ɛ1Ɛ, Ɛ21, Ɛ25, Ɛ2Ɛ, Ɛ31, Ɛ37, Ɛ45, Ɛ61, Ɛ67, Ɛ6Ɛ, Ɛ71, Ɛ91, Ɛ95, Ɛ97, Ɛᘔ5, ƐƐ5, ƐƐ7

Except 2 and 3, all primes end with 1, 5, 7 or Ɛ. The first k such that all of 10k, 10k + 1, 10k + 2, ..., 10k + Ɛ are all composite is 38, i.e. all of 380, 381, 382, ..., 38Ɛ are composite.

The density of primes end with 1 is a relatively low (< 1/4), but the density of primes end with 5, 7 and Ɛ are nearly equal (all are a little more than 1/4). (i.e. for a given natural number N, the number of primes end with 1 less than N is usually smaller than the number of primes end with 5 (or 7, or Ɛ) less than N) e.g. For all 1426 primes < 10000, there are 3Ɛ8 primes (2Ɛ.3%) end with 1, 410 primes (30.3%) end with 5, 412 primes (30.5%) end with 7, 406 primes (2Ɛ.Ɛ%) end with Ɛ. It is conjectured that for every natural number N ≥ 10, the number of primes end with 1 less than N is smaller than the number of primes end with 5 (or 7, or Ɛ) less than N. (Note: the percentage in this sequence are also in duodecimal, i.e. 20% means 0.2 or 20/100 = 1/6, 36% means 0.36 or 36/100 = 7/20, 58.7% means 0.587 or 587/1000)

13665 is the smallest prime p such that the number of primes end with 1 or 5 ≤ p is more than the number of primes end with 3, 7 or Ɛ ≤ p (see OEIS: A007350, of course, 3 is the only prime ends with 3). Besides, 9ᘔ03693ᘔ831 is the smallest prime p such that the number of primes end with 1 or 7 ≤ p is more than the number of primes end with 2, 5 or Ɛ ≤ p (see OEIS: A007352, of course, 2 is the only prime ends with 2). Question: What is the smallest prime p such that the number of primes end with 1 ≤ p is more than the number of primes end with d ≤ p for at least one of d = 5, 7 or Ɛ?

All squares of primes (except 2 and 3) end with 1.

There are 2ᘔ primes between 1 and 100, 23 primes between 101 and 200, 1ᘔ primes between 201 and 300, 1ᘔ primes between 301 and 400, 1Ɛ primes between 401 and 500, 1ᘔ primes between 501 and 600, 16 primes between 601 and 700, 1ᘔ primes between 701 and 800, 18 primes between 801 and 900, 16 primes between 901 and ᘔ00, 1ᘔ primes between ᘔ01 and Ɛ00, 17 primes between Ɛ01 and 1000.

There are about N/ln(N) primes less than N, where ln is the natural logarithm, i.e. the logarithm with base e = 2.875236069821... (see prime number theorem), thus there are about ${\displaystyle {\frac {10^{n}}{n\cdot ln10}}}$ primes less than 10 (i.e. with at most n digits), and ln(10) = 2.599Ɛ035Ɛ8169...

In the following table, numbers shaded in cyan are primes.

Divisibility rules

(In this section, all numbers are written with duodecimal)

This section is about the divisibility rules in duodecimal.

1

Any integer is divisible by 1.

2

If a number is divisible by 2 then the unit digit of that number will be 0, 2, 4, 6, 8 or ᘔ.

3

If a number is divisible by 3 then the unit digit of that number will be 0, 3, 6 or 9.

4

If a number is divisible by 4 then the unit digit of that number will be 0, 4 or 8.

5

To test for divisibility by 5, double the units digit and subtract the result from the number formed by the rest of the digits. If the result is divisible by 5 then the given number is divisible by 5.

This rule comes from 21(5*5)

Examples:
13     rule => |1-2*3| = 5 which is divisible by 5.
2Ɛᘔ5   rule => |2Ɛᘔ-2*5| = 2Ɛ0(5*70) which is divisible by 5(or apply the rule on 2Ɛ0).

OR

To test for divisibility by 5, subtract the units digit and triple of the result to the number formed by the rest of the digits. If the result is divisible by 5 then the given number is divisible by 5.

This rule comes from 13(5*3)

Examples:
13     rule => |3-3*1| = 0 which is divisible by 5.
2Ɛᘔ5   rule => |5-3*2Ɛᘔ| = 8Ɛ1(5*195) which is divisible by 5(or apply the rule on 8Ɛ1).

OR

Form the alternating sum of blocks of two from right to left. If the result is divisible by 5 then the given number is divisible by 5.

This rule comes from 101, since 101 = 5*25, thus this rule can be also tested for the divisibility by 25.

Example:

97,374,627 => 27-46+37-97 = -7Ɛ which is divisible by 5.

6

If a number is divisible by 6 then the unit digit of that number will be 0 or 6.

7

To test for divisibility by 7, triple the units digit and add the result to the number formed by the rest of the digits. If the result is divisible by 7 then the given number is divisible by 7.

This rule comes from 2Ɛ(7*5)

Examples:
12     rule => |3*2+1| = 7 which is divisible by 7.
271Ɛ    rule => |3*Ɛ+271| = 29ᘔ(7*4ᘔ) which is divisible by 7(or apply the rule on 29ᘔ).

OR

To test for divisibility by 7, subtract the units digit and double the result from the number formed by the rest of the digits. If the result is divisible by 7 then the given number is divisible by 7.

This rule comes from 12(7*2)

Examples:
12     rule => |2-2*1| = 0 which is divisible by 7.
271Ɛ    rule => |Ɛ-2*271| = 513(7*89) which is divisible by 7(or apply the rule on 513).

OR

To test for divisibility by 7, 4 times the units digit and subtract the result from the number formed by the rest of the digits. If the result is divisible by 7 then the given number is divisible by 7.

This rule comes from 41(7*7)

Examples:
12     rule => |4*2-1| = 7 which is divisible by 7.
271Ɛ    rule => |4*Ɛ-271| = 235(7*3Ɛ) which is divisible by 7(or apply the rule on 235).

OR

Form the alternating sum of blocks of three from right to left. If the result is divisible by 7 then the given number is divisible by 7.

This rule comes from 1001, since 1001 = 7*11*17, thus this rule can be also tested for the divisibility by 11 and 17.

Example:

386,967,443 => 443-967+386 = -168 which is divisible by 7.

8

If the 2-digit number formed by the last 2 digits of the given number is divisible by 8 then the given number is divisible by 8.

Example: 1Ɛ48, 4120

     rule => since 48(8*7) divisible by 8, then 1Ɛ48 is divisible by 8.
     rule => since 20(8*3) divisible by 8, then 4120 is divisible by 8.

9

If the 2-digit number formed by the last 2 digits of the given number is divisible by 9 then the given number is divisible by 9.

Example: 7423, 8330

     rule => since 23(9*3) divisible by 9, then 7423 is divisible by 9.
     rule => since 30(9*4) divisible by 9, then 8330 is divisible by 9.

ᘔ

If the number is divisible by 2 and 5 then the number is divisible by ᘔ.

Ɛ

If the sum of the digits of a number is divisible by Ɛ then the number is divisible by Ɛ (the equivalent of casting out nines in decimal).

Example: 29, 61Ɛ13

     rule => 2+9 = Ɛ which is divisible by Ɛ, then 29 is divisible by Ɛ.
     rule => 6+1+Ɛ+1+3 = 1ᘔ which is divisible by Ɛ, then 61Ɛ13 is divisible by Ɛ.

10

If a number is divisible by 10 then the unit digit of that number will be 0.

11

Sum the alternate digits and subtract the sums. If the result is divisible by 11 the number is divisible by 11 (the equivalent of divisibility by eleven in decimal).

Example: 66, 9427

     rule => |6-6| = 0 which is divisible by 11, then 66 is divisible by 11.
     rule => |(9+2)-(4+7)| = |ᘔ-ᘔ| = 0 which is divisible by 11, then 9427 is divisible by 11.

12

If the number is divisible by 2 and 7 then the number is divisible by 12.

13

If the number is divisible by 3 and 5 then the number is divisible by 13.

14

If the 2-digit number formed by the last 2 digits of the given number is divisible by 14 then the given number is divisible by 14.

Example: 1468, 7394

     rule => since 68(14*5) divisible by 14, then 1468 is divisible by 14.
     rule => since 94(14*7) divisible by 14, then 7394 is divisible by 14.

Fractions and irrational numbers

Fractions

Duodecimal fractions may be simple:

• ⁠1/2⁠ = 0.6
• ⁠1/3⁠ = 0.4
• ⁠1/4⁠ = 0.3
• ⁠1/6⁠ = 0.2
• ⁠1/8⁠ = 0.16
• ⁠1/9⁠ = 0.14
• ⁠1/10⁠ = 0.1 (note that this is a twelfth, ⁠1/ᘔ⁠ is a tenth)
• ⁠1/14⁠ = 0.09 (note that this is a sixteenth, ⁠1/12⁠ is a fourteenth)

or complicated:

• ⁠1/5⁠ = 0.249724972497... recurring (rounded to 0.24ᘔ)
• ⁠1/7⁠ = 0.186ᘔ35186ᘔ35... recurring (rounded to 0.187)
• ⁠1/ᘔ⁠ = 0.1249724972497... recurring (rounded to 0.125)
• ⁠1/Ɛ⁠ = 0.111111111111... recurring (rounded to 0.111)
• ⁠1/11⁠ = 0.0Ɛ0Ɛ0Ɛ0Ɛ0Ɛ0Ɛ... recurring (rounded to 0.0Ɛ1)
• ⁠1/12⁠ = 0.0ᘔ35186ᘔ35186... recurring (rounded to 0.0ᘔ3)
• ⁠1/13⁠ = 0.0972497249724... recurring (rounded to 0.097)

As explained in recurring decimals, whenever an irreducible fraction is written in radix point notation in any base, the fraction can be expressed exactly (terminates) if and only if all the prime factors of its denominator are also prime factors of the base. Thus, in base-ten (= 2×5) system, fractions whose denominators are made up solely of multiples of 2 and 5 terminate: ⁠1/8⁠ = ⁠1/(2×2×2)⁠, ⁠1/20⁠ = ⁠1/(2×2×5)⁠ and ⁠1/500⁠ = ⁠1/(2×2×5×5×5)⁠ can be expressed exactly as 0.125, 0.05 and 0.002 respectively. ⁠1/3⁠ and ⁠1/7⁠, however, recur (0.333... and 0.142857142857...). In the duodecimal (= 2×2×3) system, ⁠1/8⁠ is exact; ⁠1/20⁠ and ⁠1/500⁠ recur because they include 5 as a factor; ⁠1/3⁠ is exact; and ⁠1/7⁠ recurs, just as it does in decimal.

The number of denominators which give terminating fractions within a given number of digits, say n, in a base b is the number of factors (divisors) of b, the nth power of the base b (although this includes the divisor 1, which does not produce fractions when used as the denominator). The number of factors of b is given using its prime factorization.

For decimal, 10 = 2 * 5. The number of divisors is found by adding one to each exponent of each prime and multiplying the resulting quantities together. Factors of 10 = (n+1)(n+1) = (n+1).

For example, the number 8 is a factor of 10 (1000), so 1/8 and other fractions with a denominator of 8 can not require more than 3 fractional decimal digits to terminate. 5/8 = 0.625_ten

For duodecimal, 12 = 2 * 3. This has (2n+1)(n+1) divisors. The sample denominator of 8 is a factor of a gross (12 = 144), so eighths can not need more than two duodecimal fractional places to terminate. 5/8 = 0.76_twelve

Because both ten and twelve have two unique prime factors, the number of divisors of b for b = 10 or 12 grows quadratically with the exponent n (in other words, of the order of n).

Recurring digits

The Dozenal Society of America argues that factors of 3 are more commonly encountered in real-life division problems than factors of 5. Thus, in practical applications, the nuisance of repeating decimals is encountered less often when duodecimal notation is used. Advocates of duodecimal systems argue that this is particularly true of financial calculations, in which the twelve months of the year often enter into calculations.

However, when recurring fractions do occur in duodecimal notation, they are less likely to have a very short period than in decimal notation, because 12 (twelve) is between two prime numbers, 11 (eleven) and 13 (thirteen), whereas ten is adjacent to the composite number 9. Nonetheless, having a shorter or longer period doesn't help the main inconvenience that one does not get a finite representation for such fractions in the given base (so rounding, which introduces inexactitude, is necessary to handle them in calculations), and overall one is more likely to have to deal with infinite recurring digits when fractions are expressed in decimal than in duodecimal, because one out of every three consecutive numbers contains the prime factor 3 in its factorization, whereas only one out of every five contains the prime factor 5. All other prime factors, except 2, are not shared by either ten or twelve, so they do not influence the relative likeliness of encountering recurring digits (any irreducible fraction that contains any of these other factors in its denominator will recur in either base). Also, the prime factor 2 appears twice in the factorization of twelve, whereas only once in the factorization of ten; which means that most fractions whose denominators are powers of two will have a shorter, more convenient terminating representation in duodecimal than in decimal representation (e.g. 1/(2) = 0.25 _ten = 0.3 _twelve; 1/(2) = 0.125 _ten = 0.16 _twelve; 1/(2) = 0.0625 _ten = 0.09 _twelve; 1/(2) = 0.03125 _ten = 0.046 _twelve; etc.).

Values in bold indicate that value is exact.