# Duodecimal

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,[1] these digit forms are available as Unicode characters on computerized systems since June 2015[2] as ↊ (Code point 218A) and ↋ (Code point 218B), respectively.[3] 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.[4] 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 (​ 12, ​ 13, ​ 23, ​ 14 and ​ 34) 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[2]. 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;[5] the Chepang language of Nepal[6] 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.[7][8]

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.

Table of units from a base of 12
Relative
value
French unit
of length
English unit
of length
English
(Troy) unit
of weight
Roman unit
of weight
English unit
of mass
120 pied foot pound libra
12−1 pouce inch ounce uncia slinch
12−2 ligne line 2 scruples 2 scrupula slug
12−3 point point seed siliqua

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),[9][10] 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.[11][12]

## 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,[13] 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.[13] 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).[14]

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.[13] This notation was introduced by Sir Isaac Pitman.[13][15]

Until 2015, the Dozenal Society of America (DSA) used and , the symbols devised by William Addison Dwiggins.[13][16] After the Pitman digits (↊, U+218A and ↋, U+218B) were added to Unicode in 2015[2][17], the DSA took a vote and then began publishing content using the Pitman digits instead.[18][19] 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.[13] The symbols were chosen because they are available in typewriters and already present in telephone dials,[13] this notation was used in publications of the Dozenal Society of America in the period 1974–2008.[20][21] Many don't use any of the Hindu-Arabic symbols, under the principle of "separate identity."[13]

### 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:[19]

Common Base Abb. Letter Cardinal Decimal Duodecimal
binary bin b two 2 2
octal oct o eight 8 8
decimal dec d ten 10
dozenal (duodecimal) doz z twelve 12 10
hexadecimal hex x sixteen 16 14

This allows one to write "54z = 64d," "54twelve = 64ten" 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,[16][22] 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:[16]

Duodecimal Name Decimal Duodecimal fraction Name
1 one 1
10 do 12 0;1 edo
100 gro 144 0;01 egro
1,000 mo 1,728 0;001 emo
10,000 do-mo 20,736 0;000,1 edo-mo
100,000 gro-mo 248,832 0;000,01 egro-mo
1,000,000 bi-mo 2,985,984 0;000,001 ebi-mo
1,000,000,000 tri-mo 5,159,780,352 0;000,000,001 etri-mo

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.[22]

William James Sidis used 12 as the base for his constructed language Vendergood in 1906, noting it being the smallest number with four factors and the prevalence in commerce[23].

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.

A duodecimal clockface as in the logo of the Dozenal Society of America, here used to denote musical keys

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[24] 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)[25]

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)[26]

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.[citation needed]

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.[27]

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.[28]

### 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.[29] 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.[2][17]

Unicode points U+218C and U+218D seem to be reserved for the Dwiggins digits (stylized X and E).[30]

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.[31]

### Duodecimal metric systems

Systems of measurement proposed by dozenalists include:

• Tom Pendlebury's TGM system[32][33]
• Takashi Suga's Universal Unit System[34][33]

## Comparison to other numeral systems

A duodecimal multiplication table

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.

Duodecimal multiplication table
× 0 1 2 3 4 5 6 7 8 9 Ɛ 10 11 12 13 14 15 16 17 18 19 1ᘔ 20
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
1 0 1 2 3 4 5 6 7 8 9 Ɛ 10 11 12 13 14 15 16 17 18 19 1ᘔ 20
2 0 2 4 6 8 10 12 14 16 18 1ᘔ 20 22 24 26 28 2ᘔ 30 32 34 36 38 3ᘔ 40
3 0 3 6 9 10 13 16 19 20 23 26 29 30 33 36 39 40 43 46 49 50 53 56 59 60
4 0 4 8 10 14 18 20 24 28 30 34 38 40 44 48 50 54 58 60 64 68 70 74 78 80
5 0 5 13 18 21 26 34 39 42 47 50 55 5ᘔ 63 68 71 76 84 89 92 97 ᘔ0
6 0 6 10 16 20 26 30 36 40 46 50 56 60 66 70 76 80 86 90 96 ᘔ0 ᘔ6 Ɛ0 Ɛ6 100
7 0 7 12 19 24 36 41 48 53 5ᘔ 65 70 77 82 89 94 ᘔ6 Ɛ1 Ɛ8 103 10ᘔ 115 120
8 0 8 14 20 28 34 40 48 54 60 68 74 80 88 94 ᘔ0 ᘔ8 Ɛ4 100 108 114 120 128 134 140
9 0 9 16 23 30 39 46 53 60 69 76 83 90 99 ᘔ6 Ɛ3 100 109 116 123 130 139 146 153 160
0 18 26 34 42 50 5ᘔ 68 76 84 92 ᘔ0 ᘔᘔ Ɛ8 106 114 122 130 13ᘔ 148 156 164 172 180
Ɛ 0 Ɛ 1ᘔ 29 38 47 56 65 74 83 92 ᘔ1 Ɛ0 ƐƐ 10ᘔ 119 128 137 146 155 164 173 182 191 1ᘔ0
10 0 10 20 30 40 50 60 70 80 90 ᘔ0 Ɛ0 100 110 120 130 140 150 160 170 180 190 1ᘔ0 1Ɛ0 200
11 0 11 22 33 44 55 66 77 88 99 ᘔᘔ ƐƐ 110 121 132 143 154 165 176 187 198 1ᘔ9 1Ɛᘔ 20Ɛ 220
12 0 12 24 36 48 5ᘔ 70 82 94 ᘔ6 Ɛ8 10ᘔ 120 132 144 156 168 17ᘔ 190 1ᘔ2 1Ɛ4 206 218 22ᘔ 240
13 0 13 26 39 50 63 76 89 ᘔ0 Ɛ3 106 119 130 143 156 169 180 193 1ᘔ6 1Ɛ9 210 223 236 249 260
14 0 14 28 40 54 68 80 94 ᘔ8 100 114 128 140 154 168 180 194 1ᘔ8 200 214 228 240 254 268 280
15 0 15 2ᘔ 43 58 71 86 Ɛ4 109 122 137 150 165 17ᘔ 193 1ᘔ8 201 216 22Ɛ 244 259 272 287 2ᘔ0
16 0 16 30 46 60 76 90 ᘔ6 100 116 130 146 160 176 190 1ᘔ6 200 216 230 246 260 276 290 2ᘔ6 300
17 0 17 32 49 64 96 Ɛ1 108 123 13ᘔ 155 170 187 1ᘔ2 1Ɛ9 214 22Ɛ 246 261 278 293 2ᘔᘔ 305 320
18 0 18 34 50 68 84 ᘔ0 Ɛ8 114 130 148 164 180 198 1Ɛ4 210 228 244 260 278 294 2Ɛ0 308 324 340
19 0 19 36 53 70 89 ᘔ6 103 120 139 156 173 190 1ᘔ9 206 223 240 259 276 293 2Ɛ0 309 326 343 360
1ᘔ 0 1ᘔ 38 56 74 92 Ɛ0 10ᘔ 128 146 164 182 1ᘔ0 1Ɛᘔ 218 236 254 272 290 2ᘔᘔ 308 326 344 362 380
0 3ᘔ 59 78 97 Ɛ6 115 134 153 172 191 1Ɛ0 20Ɛ 22ᘔ 249 268 287 2ᘔ6 305 324 343 362 381 3ᘔ0
20 0 20 40 60 80 ᘔ0 100 120 140 160 180 1ᘔ0 200 220 240 260 280 2ᘔ0 300 320 340 360 380 3ᘔ0 400

## 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

Duod. Dec. Duod. Dec. Duod. Dec. Duod. Dec. Duod. Dec. Duod. Dec. Duod. Dec. Duod. Dec. Duod. Dec.
1,000,000 2,985,984 100,000 248,832 10,000 20,736 1,000 1,728 100 144 10 12 1 1 0.1 0.083 0.01 0.00694
2,000,000 5,971,968 200,000 497,664 20,000 41,472 2,000 3,456 200 288 20 24 2 2 0.2 0.16 0.02 0.0138
3,000,000 8,957,952 300,000 746,496 30,000 62,208 3,000 5,184 300 432 30 36 3 3 0.3 0.25 0.03 0.02083
4,000,000 11,943,936 400,000 995,328 40,000 82,944 4,000 6,912 400 576 40 48 4 4 0.4 0.3 0.04 0.027
5,000,000 14,929,920 500,000 1,244,160 50,000 103,680 5,000 8,640 500 720 50 60 5 5 0.5 0.416 0.05 0.03472
6,000,000 17,915,904 600,000 1,492,992 60,000 124,416 6,000 10,368 600 864 60 72 6 6 0.6 0.5 0.06 0.0416
7,000,000 20,901,888 700,000 1,741,824 70,000 145,152 7,000 12,096 700 1,008 70 84 7 7 0.7 0.583 0.07 0.04861
8,000,000 23,887,872 800,000 1,990,656 80,000 165,888 8,000 13,824 800 1,152 80 96 8 8 0.8 0.6 0.08 0.05
9,000,000 26,873,856 900,000 2,239,488 90,000 186,624 9,000 15,552 900 1,296 90 108 9 9 0.9 0.75 0.09 0.0625
ᘔ,000,000 29,859,840 ᘔ00,000 2,488,320 ᘔ0,000 207,360 ᘔ,000 17,280 ᘔ00 1,440 ᘔ0 120 10 0.ᘔ 0.83 0.0ᘔ 0.0694
Ɛ,000,000 32,845,824 Ɛ00,000 2,737,152 Ɛ0,000 228,096 Ɛ,000 19,008 Ɛ00 1,584 Ɛ0 132 Ɛ 11 0.Ɛ 0.916 0.0Ɛ 0.07638

### Decimal to duodecimal digit conversion

Dec. Duod. Dec. Duod. Dec. Duod. Dec. Duod. Dec. Duod. Dec. Duod. Dec. Duod. Dec. Duod.
100,000 49,ᘔ54 10,000 5,954 1,000 6Ɛ4 100 84 10 1 1 0.1 0.12497 0.01 0.015343ᘔ0Ɛ62ᘔ68781Ɛ059
200,000 97,8ᘔ8 20,000 Ɛ,6ᘔ8 2,000 1,1ᘔ8 200 148 20 18 2 2 0.2 0.2497 0.02 0.02ᘔ68781Ɛ05915343ᘔ0Ɛ6
300,000 125,740 30,000 15,440 3,000 1,8ᘔ0 300 210 30 26 3 3 0.3 0.37249 0.03 0.043ᘔ0Ɛ62ᘔ68781Ɛ059153
400,000 173,594 40,000 1Ɛ,194 4,000 2,394 400 294 40 34 4 4 0.4 0.4972 0.04 0.05915343ᘔ0Ɛ62ᘔ68781Ɛ
500,000 201,428 50,000 24,Ɛ28 5,000 2,ᘔ88 500 358 50 42 5 5 0.5 0.6 0.05 0.07249
600,000 24Ɛ,280 60,000 2ᘔ,880 6,000 3,580 600 420 60 50 6 6 0.6 0.7249 0.06 0.08781Ɛ05915343ᘔ0Ɛ62ᘔ6
700,000 299,114 70,000 34,614 7,000 4,074 700 4ᘔ4 70 5ᘔ 7 7 0.7 0.84972 0.07 0.0ᘔ0Ɛ62ᘔ68781Ɛ05915343
800,000 326,Ɛ68 80,000 3ᘔ,368 8,000 4,768 800 568 80 68 8 8 0.8 0.9724 0.08 0.0Ɛ62ᘔ68781Ɛ05915343ᘔ
900,000 374,ᘔ00 90,000 44,100 9,000 5,260 900 630 90 76 9 9 0.9 0.ᘔ9724 0.09 0.10Ɛ62ᘔ68781Ɛ05915343ᘔ

### Conversion of powers

Exponent b=2 b=3 b=4 b=5 b=6 b=7
Dec. Duod. Dec. Duod. Dec. Duod. Dec. Duod. Dec. Duod. Dec. Duod.
b6 64 54 729 509 4,096 2,454 15,625 9,061 46,656 23,000 117,649 58,101
b5 32 28 243 183 1,024 714 3,125 1,985 7,776 4,600 16,807 9,887
b4 16 14 81 69 256 194 625 441 1,296 900 2,401 1,481
b3 8 8 27 23 64 54 125 ᘔ5 216 160 343 247
b2 4 4 9 9 16 14 25 21 36 30 49 41
b1 2 2 3 3 4 4 5 5 6 6 7 7
b−1 0.5 0.6 0.3 0.4 0.25 0.3 0.2 0.2497 0.16 0.2 0.142857 0.186ᘔ35
b−2 0.25 0.3 0.1 0.14 0.0625 0.09 0.04 0.05915343ᘔ0
Ɛ62ᘔ68781Ɛ
0.027 0.04 0.0204081632653
06122448979591
836734693877551
0.02Ɛ322547ᘔ05ᘔ
644ᘔ9380Ɛ908996
741Ɛ615771283Ɛ
Exponent b=8 b=9 b=10 b=11 b=12
Dec. Duod. Dec. Duod. Dec. Duod. Dec. Duod. Dec. Duod.
b6 262,144 107,854 531,441 217,669 1,000,000 402,854 1,771,561 715,261 2,985,984 1,000,000
b5 32,768 16,Ɛ68 59,049 2ᘔ,209 100,000 49,ᘔ54 161,051 79,24Ɛ 248,832 100,000
b4 4,096 2,454 6,561 3,969 10,000 5,954 14,641 8,581 20,736 10,000
b3 512 368 729 509 1,000 6Ɛ4 1,331 92Ɛ 1,728 1,000
b2 64 54 81 69 100 84 121 ᘔ1 144 100
b1 8 8 9 9 10 11 Ɛ 12 10
b−1 0.125 0.16 0.1 0.14 0.1 0.12497 0.09 0.1 0.083 0.1
b−2 0.015625 0.023 0.012345679 0.0194 0.01 0.015343ᘔ0Ɛ6
2ᘔ68781Ɛ059
0.00826446280
99173553719
0.0123456789Ɛ 0.00694 0.01

## 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 n2 ends with 0, if n is not divisible by 2 or 3, then n2 ends with 1, if n is divisible by 2 but not by 3, then n2 ends with 4, if n is not divisible by 2 but by 3, then n2 ends with 9. If the unit digit of n2 is 0, then the dozens digit of n2 is either 0 or 3, if the unit digit of n2 is 1, then the dozens digit of n2 is even, if the unit digit of n2 is 4, then the dozen digit of n2 is 0, 1, 4, 5, 8 or 9, if the unit digit of n2 is 9, then the dozen digit of n2 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 322 = ᘔ04 = 1444)

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 112 = 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)

n n-digit palindromic squares square roots number of n-digit palindromic squares
1 1, 4, 9 1, 2, 3 3
2 none none 0
3 121, 484 11, 22 2
4 none none 0
5 10201, 12321, 14641, 16661, 16Ɛ61, 40804, 41414, 44944 101, 111, 121, 12Ɛ, 131, 202, 204, 212 8
6 160061 42Ɛ 1
7 1002001, 102ᘔ201, 1093901, 1234321, 148ᘔ841, 4008004, 445ᘔ544, 49ᘔᘔᘔ94 1001, 1015, 1047, 1111, 1221, 2002, 2112, 2244 8
8 none none 0
9 100020001, 102030201, 104060401, 1060Ɛ0601, 121242121, 123454321, 125686521, 1420Ɛ0241, 1444ᘔ4441, 1468Ɛ8641, 14ᘔ797ᘔ41, 1621Ɛ1261, 163151361, 1ᘔᘔ222ᘔᘔ1, 400080004, 404090404, 410212014, 4414ᘔ4144, 4456Ɛ6544, 496787694, 963848369 10001, 10101, 10201, 10301, 11011, 11111, 11211, 11Ɛ21, 12021, 12121, 1229Ɛ, 1292Ɛ, 12977, 14685, 20002, 20102, 20304, 21012, 21112, 22344, 31053 19
1642662461 434ᘔ5 1
Ɛ 10000200001, 10221412201, 10444ᘔ44401, 12102420121, 12345654321, 141Ɛ1Ɛ1Ɛ141, 14404ᘔ40441, 16497679461, 40000800004, 40441ᘔ14404, 41496869414, 44104ᘔ40144, 49635653694 100001, 101101, 102201, 110011, 111111, 11Ɛ13Ɛ, 120021, 12ᘔ391, 200002, 201102, 204204, 210012, 223344 11
10 none none 0

It is conjectured that if n is divisible by 4, then there are no n-digit palindromic squares.

Rn2 (where Rn is the repunit with length n) is a palindromic number for n ≤ Ɛ, but not for n ≥ 10 (thus, for all odd number n ≤ 19, ther is n-digit palindromic square 123...321), besides, 11n (also 1{0}1n, i.e. 101n, 1001n, 10001n, 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 n3 ends with the same digit as n; if n is congruent to 2 mod 4, then n3 ends with the digit (the last digit of n +− 6).

The digital root of a cube can be any number.

If k≥2, then nk+2 ends with the same digit as nk, thus, if i≥2, j≥2 and i and j have the same parity, then ni and nj 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 x2x=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.

unit digit of nk
k
n
0 1 2 3 4 5 6 7 8 9 Ɛ 10 11 12 13 14 15 16 17 18 19 1ᘔ 20 Period
0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
2 1 2 4 8 4 8 4 8 4 8 4 8 4 8 4 8 4 8 4 8 4 8 4 8 4 2
3 1 3 9 3 9 3 9 3 9 3 9 3 9 3 9 3 9 3 9 3 9 3 9 3 9 2
4 1 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 1
5 1 5 1 5 1 5 1 5 1 5 1 5 1 5 1 5 1 5 1 5 1 5 1 5 1 2
6 1 6 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1
7 1 7 1 7 1 7 1 7 1 7 1 7 1 7 1 7 1 7 1 7 1 7 1 7 1 2
8 1 8 4 8 4 8 4 8 4 8 4 8 4 8 4 8 4 8 4 8 4 8 4 8 4 2
9 1 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 1
1 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 1
Ɛ 1 Ɛ 1 Ɛ 1 Ɛ 1 Ɛ 1 Ɛ 1 Ɛ 1 Ɛ 1 Ɛ 1 Ɛ 1 Ɛ 1 Ɛ 1 Ɛ 1 2
10 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1
11 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
12 1 2 4 8 4 8 4 8 4 8 4 8 4 8 4 8 4 8 4 8 4 8 4 8 4 2
13 1 3 9 3 9 3 9 3 9 3 9 3 9 3 9 3 9 3 9 3 9 3 9 3 9 2
14 1 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 1
15 1 5 1 5 1 5 1 5 1 5 1 5 1 5 1 5 1 5 1 5 1 5 1 5 1 2
16 1 6 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1
17 1 7 1 7 1 7 1 7 1 7 1 7 1 7 1 7 1 7 1 7 1 7 1 7 1 2
18 1 8 4 8 4 8 4 8 4 8 4 8 4 8 4 8 4 8 4 8 4 8 4 8 4 2
19 1 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 1
1ᘔ 1 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 1
1 Ɛ 1 Ɛ 1 Ɛ 1 Ɛ 1 Ɛ 1 Ɛ 1 Ɛ 1 Ɛ 1 Ɛ 1 Ɛ 1 Ɛ 1 Ɛ 1 2
20 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1

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).

n possible unit digit of an nth power
0 1
1 any number
even number ≥ 2 0, 1, 4, 9 (the square digits)
odd number ≥ 3 0, 1, 3, 4, 5, 7, 8, 9, Ɛ (all digits != 2 mod 4)
final two digits of nk
k
n
0 1 2 3 4 5 6 7 8 9 Ɛ 10 11 12 13 14 15 16 17 18 19 1ᘔ 20 Period
0 01 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 1
1 01 01 01 01 01 01 01 01 01 01 01 01 01 01 01 01 01 01 01 01 01 01 01 01 01 1
2 01 02 04 08 14 28 54 ᘔ8 94 68 14 28 54 ᘔ8 94 68 14 28 54 ᘔ8 94 68 14 28 54 6
3 01 03 09 23 69 83 09 23 69 83 09 23 69 83 09 23 69 83 09 23 69 83 09 23 69 4
4 01 04 14 54 94 14 54 94 14 54 94 14 54 94 14 54 94 14 54 94 14 54 94 14 54 3
5 01 05 21 ᘔ5 41 85 61 65 81 45 ᘔ1 25 01 05 21 ᘔ5 41 85 61 65 81 45 ᘔ1 25 01 10
6 01 06 30 60 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 1
7 01 07 41 47 81 87 01 07 41 47 81 87 01 07 41 47 81 87 01 07 41 47 81 87 01 6
8 01 08 54 68 54 68 54 68 54 68 54 68 54 68 54 68 54 68 54 68 54 68 54 68 54 2
9 01 09 69 09 69 09 69 09 69 09 69 09 69 09 69 09 69 09 69 09 69 09 69 09 69 2
01 0ᘔ 84 Ɛ4 54 54 54 54 54 54 54 54 54 54 54 54 54 54 54 54 54 54 54 54 54 1
Ɛ 01 ᘔ1 81 61 41 21 ᘔƐ 01 ᘔ1 81 61 41 21 ᘔƐ 01 10
10 01 10 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 1
11 01 11 21 31 41 51 61 71 81 91 ᘔ1 Ɛ1 01 11 21 31 41 51 61 71 81 91 ᘔ1 Ɛ1 01 10
12 01 12 44 08 94 ᘔ8 54 28 14 68 94 ᘔ8 54 28 14 68 94 ᘔ8 54 28 14 68 94 ᘔ8 54 6
13 01 13 69 53 69 53 69 53 69 53 69 53 69 53 69 53 69 53 69 53 69 53 69 53 69 2
14 01 14 94 54 14 94 54 14 94 54 14 94 54 14 94 54 14 94 54 14 94 54 14 94 54 3
15 01 15 01 15 01 15 01 15 01 15 01 15 01 15 01 15 01 15 01 15 01 15 01 15 01 2
16 01 16 30 60 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 1
17 01 17 61 77 01 17 61 77 01 17 61 77 01 17 61 77 01 17 61 77 01 17 61 77 01 4
18 01 18 94 68 14 28 54 ᘔ8 94 68 14 28 54 ᘔ8 94 68 14 28 54 ᘔ8 94 68 14 28 54 6
19 01 19 09 39 69 99 09 39 69 99 09 39 69 99 09 39 69 99 09 39 69 99 09 39 69 4
1ᘔ 01 1ᘔ 44 Ɛ4 94 14 54 94 14 54 94 14 54 94 14 54 94 14 54 94 14 54 94 14 54 3
01 81 41 01 81 41 01 81 41 01 81 41 01 6
20 01 20 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 1

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 10n−1 (= λ(10n)).

digital root of nk
k
n
0 1 2 3 4 5 6 7 8 9 Ɛ 10 11 12 13 14 15 16 17 18 19 1ᘔ 20 Period
0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
2 1 2 4 8 5 9 7 3 6 1 2 4 8 5 9 7 3 6 1 2 4 8 5
3 1 3 9 5 4 1 3 9 5 4 1 3 9 5 4 1 3 9 5 4 1 3 9 5 4 5
4 1 4 5 9 3 1 4 5 9 3 1 4 5 9 3 1 4 5 9 3 1 4 5 9 3 5
5 1 5 3 4 9 1 5 3 4 9 1 5 3 4 9 1 5 3 4 9 1 5 3 4 9 5
6 1 6 3 7 9 5 8 4 2 1 6 3 7 9 5 8 4 2 1 6 3 7 9
7 1 7 5 2 3 4 6 9 8 1 7 5 2 3 4 6 9 8 1 7 5 2 3
8 1 8 9 6 4 3 2 5 7 1 8 9 6 4 3 2 5 7 1 8 9 6 4
9 1 9 4 3 5 1 9 4 3 5 1 9 4 3 5 1 9 4 3 5 1 9 4 3 5 5
1 1 1 1 1 1 1 1 1 1 1 1 1 2
Ɛ 1 Ɛ Ɛ Ɛ Ɛ Ɛ Ɛ Ɛ Ɛ Ɛ Ɛ Ɛ Ɛ Ɛ Ɛ Ɛ Ɛ Ɛ Ɛ Ɛ Ɛ Ɛ Ɛ Ɛ Ɛ 1
10 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
11 1 2 4 8 5 9 7 3 6 1 2 4 8 5 9 7 3 6 1 2 4 8 5
12 1 3 9 5 4 1 3 9 5 4 1 3 9 5 4 1 3 9 5 4 1 3 9 5 4 5
13 1 4 5 9 3 1 4 5 9 3 1 4 5 9 3 1 4 5 9 3 1 4 5 9 3 5
14 1 5 3 4 9 1 5 3 4 9 1 5 3 4 9 1 5 3 4 9 1 5 3 4 9 5
15 1 6 3 7 9 5 8 4 2 1 6 3 7 9 5 8 4 2 1 6 3 7 9
16 1 7 5 2 3 4 6 9 8 1 7 5 2 3 4 6 9 8 1 7 5 2 3
17 1 8 9 6 4 3 2 5 7 1 8 9 6 4 3 2 5 7 1 8 9 6 4
18 1 9 4 3 5 1 9 4 3 5 1 9 4 3 5 1 9 4 3 5 1 9 4 3 5 5
19 1 1 1 1 1 1 1 1 1 1 1 1 1 2
1ᘔ 1 Ɛ Ɛ Ɛ Ɛ Ɛ Ɛ Ɛ Ɛ Ɛ Ɛ Ɛ Ɛ Ɛ Ɛ Ɛ Ɛ Ɛ Ɛ Ɛ Ɛ Ɛ Ɛ Ɛ Ɛ 1
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
20 1 2 4 8 5 9 7 3 6 1 2 4 8 5 9 7 3 6 1 2 4 8 5

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

n possible digital root of an nth power
0 1
= 1, 3, 7, 9 (mod ᘔ) any number
= 2, 4, 6, 8 (mod ᘔ) 1, 3, 4, 5, 9, Ɛ
= 5 (mod ᘔ) 1, ᘔ, Ɛ
> 0 and divisible by ᘔ 1, 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 104.

In the following table, Fn is the n-th Fibonacci number, and Ln is the n-th Lucas number.

n Fn digit root of Fn Ln digit root of Ln n Fn digit root of Fn Ln digit root of Ln
1 1 1 1 1 21 37501 5 81101 Ɛ
2 1 1 3 3 22 5ᘔ301 8 111103 7
3 2 2 4 4 23 95802 2 192204 7
4 3 3 7 7 24 133Ɛ03 2ᘔ3307 3
5 5 5 Ɛ Ɛ 25 209705 1 47550Ɛ
6 8 8 16 7 26 341608 Ɛ 758816 2
7 11 2 25 7 27 54Ɛ111 1 1012125 1
8 19 3 28 890719 1 176ᘔ93Ɛ 3
9 2ᘔ 1 64 29 121Ɛ82ᘔ 2 2780ᘔ64 4
47 Ɛ ᘔ3 2 2ᘔ 1ᘔƐ0347 3 432Ɛ7ᘔ3 7
Ɛ 75 1 147 1 310ƐƐ75 5 6ᘔƐ0647 Ɛ
10 100 1 22ᘔ 3 30 5000300 8 Ɛ22022ᘔ 7
11 175 2 375 4 31 8110275 2 16110875 7
12 275 3 5ᘔ3 7 32 11110575 25330ᘔᘔ3 3
13 42ᘔ 5 958 Ɛ 33 1922082ᘔ 1 3Ɛ441758
14 6ᘔ3 8 133Ɛ 7 34 2ᘔ3311ᘔ3 Ɛ 6477263Ɛ 2
15 Ɛ11 2 2097 7 35 47551ᘔ11 1 ᘔ3ƐƐ4197 1
16 15Ɛ4 3416 3 36 75882ƐƐ4 1 148766816 3
17 2505 1 54Ɛ1 37 101214ᘔ05 2 23075ᘔ9Ɛ1 4
18 3ᘔƐ9 Ɛ 8907 2 38 176ᘔ979Ɛ9 3 379305607 7
19 6402 1 121Ɛ8 1 39 2780Ɛ0802 5 5ᘔ9ᘔ643Ɛ8 Ɛ
1ᘔ ᘔ2ƐƐ 1 1ᘔƐ03 3 3ᘔ 432Ɛ885ƐƐ 8 967169ᘔ03 7
14701 2 310ƐƐ 4 6ᘔƐ079201 2 13550121ƐƐ 7
20 22ᘔ00 3 50002 7 40 Ɛ22045800 2100180002 3

(Note that F2ᘔ begins with L1ᘔ, and F begins with L)

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 2×10n−1 (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 ).

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.

n 2n n 2n n 2n n 2n n 2n n 2n
1 2 21 Ɛ2ᘔ20ᘔ8 41 5317Ɛ5804588ᘔ8 61 256906ᘔ1ᘔ93096Ɛ8934ᘔ8 81 11ᘔ12ᘔ743504482569888538ᘔ0ᘔ8 ᘔ1 65933Ɛ8691303ᘔ448Ɛ712227ᘔ7Ɛ11448ᘔ8
2 4 22 1ᘔ584194 42 ᘔ633ᘔƐ408Ɛ5594 62 4Ɛ161183966171Ɛ566994 82 238259286ᘔ08944Ɛ17554ᘔ758194 ᘔ2 10Ɛ667Ɛ51626078895Ɛ22445393ᘔ2289594
3 8 23 38Ɛ48368 43 190679ᘔ815ᘔᘔƐ68 63 9ᘔ302347710323ᘔƐ11768 83 4744Ɛ6551815689ᘔ32ᘔᘔ992Ɛ4368 ᘔ3 21Ɛ113ᘔᘔ305013556Ɛᘔ4488ᘔ76784556Ɛ68
4 14 24 75ᘔ94714 44 361137942Ɛ99Ɛ14 64 1786046932206479ᘔ23314 84 9289Ɛ0ᘔᘔ342Ɛ15786599765ᘔ8714 ᘔ4 43ᘔ2279860ᘔ026ᘔƐ1Ɛ88955931348ᘔƐ1Ɛ14
5 28 25 12Ɛ969228 45 702273685Ɛ77ᘔ28 65 3350091664410937846628 85 16557ᘔ198685ᘔ2Ɛ350Ɛ7730Ɛ95228 ᘔ5 878453750180519ᘔ3Ɛ556ᘔƐ6626959ᘔ3ᘔ28
6 54 26 25Ɛ716454 46 120452714ƐƐ33854 66 66ᘔ0163108821673491054 86 30ᘔƐ3837514Ɛ85ᘔ6ᘔ1Ɛ3261Ɛ6ᘔ454 ᘔ6 15348ᘔ72ᘔ0340ᘔ3787ᘔᘔƐ19Ɛ10516Ɛ787854
7 ᘔ8 27 4ƐƐ2308ᘔ8 47 2408ᘔ5229Ɛᘔ674ᘔ8 67 111803062154431269620ᘔ8 87 619ᘔ7472ᘔ29Ɛ4Ɛ9183ᘔ6503Ɛ188ᘔ8 ᘔ7 2ᘔ695925806818735399ᘔ37ᘔ20ᘔ31Ɛ3534ᘔ8
8 194 28 9Ɛᘔ461594 48 48158ᘔ457Ɛ912994 68 2234061042ᘔ886251704194 88 103792925857ᘔ9Ɛ634790ᘔ07ᘔ35594 ᘔ8 5916Ɛ64Ɛ41143526ᘔ777873841863ᘔ6ᘔ6994
9 368 29 17Ɛ8902Ɛ68 49 942Ɛ588Ɛ3Ɛ625768 69 446810208595504ᘔ3208368 89 20736564Ɛ4Ɛ397Ɛ06936181386ᘔƐ68 ᘔ9 Ɛ631Ɛ09ᘔ82286ᘔ5193335274835079191768
714 2ᘔ 33Ɛ5605Ɛ14 4ᘔ 1685ᘔƐ55ᘔ7Ɛ04Ɛ314 6ᘔ 891420414Ɛ6ᘔᘔ0986414714 8ᘔ 41270Ɛ09ᘔ9ᘔ773ᘔ116703427519Ɛ14 ᘔᘔ 1Ɛ063ᘔ179445518ᘔ36666ᘔ52946ᘔ136363314
Ɛ 1228 67ᘔƐ00Ɛᘔ28 314Ɛ9ᘔᘔƐ93ᘔ09ᘔ628 1562840829Ɛ1981750829228 82521ᘔ179793278231206852ᘔ37ᘔ28 ᘔƐ 3ᘔ107833688ᘔᘔ358711118ᘔ56918270706628
10 2454 30 1139ᘔ01Ɛ854 50 629Ɛ799Ɛ678179054 70 2Ɛ05481457ᘔ37432ᘔ1456454 90 144ᘔ4383373665344624114ᘔ5873854 Ɛ0 78213467155986Ɛ52222358Ɛ1634521211054
11 48ᘔ8 31 2277803Ɛ4ᘔ8 51 1057Ɛ377Ɛ1343360ᘔ8 71 5ᘔ0ᘔ9428Ɛ3872865828Ɛ08ᘔ8 91 2898874672710ᘔ6890482298Ɛ5274ᘔ8 Ɛ1 1344269122ᘔƐ751ᘔᘔ44446Ɛ5ᘔ3068ᘔ424220ᘔ8
12 9594 32 4533407ᘔ994 52 20Ɛ3ᘔ733ᘔ268670194 72 Ɛ8196855ᘔ752550Ɛ455ᘔ1594 92 557552912522191560944575ᘔᘔ52994 Ɛ2 26885162459Ɛ2ᘔ39888891ᘔƐ86115884844194
13 16Ɛ68 33 8ᘔ668139768 53 41ᘔ792678515120368 73 1Ɛ43714ᘔƐ92ᘔ4ᘔᘔ1ᘔ8ᘔƐ82Ɛ68 93 ᘔƐ2ᘔᘔ5624ᘔ44362Ɛ01688Ɛ2Ɛ98ᘔ5768 Ɛ3 5154ᘔ3048Ɛ7ᘔ58775555639Ɛ5022Ɛ549488368
14 31Ɛ14 34 159114277314 54 839365134ᘔ2ᘔ240714 74 3ᘔ872299Ɛ6589983959Ɛ45Ɛ14 94 19ᘔ598Ɛ049888705ᘔ03155ᘔ5Ɛ758Ɛ314 Ɛ4 ᘔ2ᘔ986095Ɛ38Ɛ532ᘔᘔᘔƐ077ᘔᘔ045ᘔᘔ96954714
15 63ᘔ28 35 2Ɛ6228532628 55 147670ᘔ269858481228 75 79524577Ɛ0Ɛ577476Ɛ7ᘔ8Ɛᘔ28 95 378Ɛ75ᘔ09755520Ɛ8062ᘔƐ8ƐƐ2Ɛ5ᘔ628 Ɛ5 185975016Ɛᘔ75ᘔᘔ65999ᘔ1339808Ɛ99716ᘔ9228
16 107854 36 5Ɛ0454ᘔ65054 56 29312185174Ɛ4942454 76 136ᘔ48Ɛ33ᘔ1ᘔƐ32931Ɛ395Ɛ854 96 735Ɛ2Ɛ8172ᘔᘔᘔ41Ɛ41059Ɛ5Ɛᘔ5ᘔƐ9054 Ɛ6 34Ɛ72ᘔ031Ɛ92Ɛ990Ɛ77782677415Ɛ7723196454
17 2134ᘔ8 37 Ɛᘔ08ᘔ990ᘔ0ᘔ8 57 5662434ᘔ329ᘔ96848ᘔ8 77 271895ᘔ67839ᘔ65663ᘔ76ƐƐ4ᘔ8 97 126Ɛᘔ5Ɛ432599883ᘔ820Ɛ7ᘔƐƐ8Ɛ9Ɛ60ᘔ8 Ɛ7 69Ɛ258063Ɛ65Ɛ761Ɛ3334513282ƐƐ32463708ᘔ8
18 426994 38 1Ɛ81597618194 58 Ɛ104869865797149594 78 52356Ɛ91347790Ɛ107931Ɛᘔ994 98 251Ɛ8Ɛᘔ864Ɛ775479441Ɛ39ƐƐ5Ɛ7Ɛ0194 Ɛ8 117ᘔ4Ɛ4107Ɛ0ƐƐ303ᘔ6668ᘔ26545Ɛᘔ6490721594
19 851768 39 3Ɛ42Ɛ73034368 59 1ᘔ20951750Ɛ372296Ɛ68 79 ᘔ46Ɛ1Ɛ62693361ᘔ213663Ɛ9768 99 4ᘔ3Ɛ5Ɛ9509Ɛ32ᘔ936883ᘔ77ƐᘔƐƐ3ᘔ0368 Ɛ9 23389ᘔ8213ᘔ1Ɛᘔ60791115850ᘔ8ƐƐ90961242Ɛ68
1ᘔ 14ᘔ3314 3ᘔ 7ᘔ85Ɛ26068714 5ᘔ 38416ᘔ32ᘔ1ᘔ724571Ɛ14 7ᘔ 1891ᘔ3Ɛ051667038427107Ɛ7314 9ᘔ 987ᘔƐƐ6ᘔ17ᘔ659671547933Ɛ9Ɛᘔ780714 Ɛᘔ 467579442783Ɛ90136222Ɛ4ᘔ195ƐƐ61702485Ɛ14
2986628 1394Ɛᘔ50115228 74831865839248Ɛ23ᘔ28 356387ᘔ0ᘔ3112074852213Ɛ2628 17539ƐƐ183390Ɛ7122ᘔ93667Ɛ7Ɛ9341228 ƐƐ 912Ɛ36885347Ɛ60270445ᘔ9836ƐƐƐ0320494Ɛᘔ28
20 5751054 40 2769Ɛ8ᘔ022ᘔ454 60 12946350Ɛ476495ᘔ47854 80 6Ɛ075381862241294ᘔ4427ᘔ5054 ᘔ0 32ᘔ77Ɛᘔ346761Ɛ2245967113Ɛ3Ɛ6682454 100 1625ᘔ7154ᘔ693Ɛ0052088Ɛ97471ƐƐᘔ0640969Ɛ854

The smallest power of 2 starts with the digit Ɛ is 221 = Ɛ2ᘔ20ᘔ8, for all digits 1≤d≤Ɛ, there exists 0≤n≤21 such that 2n starts with the digit d.

21ᘔƐ = 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 272 = Ɛ8196855ᘔ752550Ɛ455ᘔ1594 = 77371252455336267181195264, it has only 22 digits in base ᘔ)

The number 229 = 2368 (see power of 2#Powers of two whose exponents are powers of two) is very close to googol (10100), since it has ƐƐ digits. (thus, the Fermat number F9 (=229+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 (=13 + 103) = 509 + 6Ɛ4 (=93 + ᘔ3) (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 Nn is nth NSW number, and Pn is nth Pell number, Nn/Pn 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 Ln is nth Lucas number, and Fn is nth Fibonacci number, Ln/Fn 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 m2 = 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 ≤ kn.

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 ${\displaystyle Z[i]}$, 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 ${\displaystyle Z[\omega ]}$, 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 x2 + y2 if and only if p ends with 1 or 5 (or p=2).

A prime p can be written as x2 + 3y2 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 ai + ai+n/2 = Ɛ for every 1 ≤ in/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 1060 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 = 51, 131 = ᘔƐ, 181 = 131, 461 = 325, 491 = 34Ɛ, see ) (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 = 7, 23 = 1Ɛ, 47 = 3Ɛ, 59 = 4Ɛ, 167 = 11Ɛ, see ) (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 k2) 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 3k2, only 5 and 7 can be such primes, the proof for these bases is completely the same as that for base 10)

p period length of 1/p p period length of 1/p p period length of 1/p p period length of 1/p p period length of 1/p p period length of 1/p p period length of 1/p p period length of 1/p p period length of 1/p p period length of 1/p p period length of 1/p p period length of 1/p
2 0 111 3 267 266 41Ɛ 20Ɛ 591 3ᘔ 767 766 927 926 Ɛ1Ɛ 56Ɛ 1107 1106 12Ɛ5 12Ɛ4 14Ɛ1 14Ɛ 16ᘔ7 16ᘔ6
3 0 117 116 271 27 421 63 59Ɛ 2ᘔƐ 76Ɛ 395 955 954 Ɛ21 570 1115 1114 1301 760 14Ɛ5 14Ɛ4 16Ɛ5 188
5 4 11Ɛ 277 276 427 426 5Ɛ1 159 771 132 95Ɛ 48Ɛ Ɛ25 Ɛ24 1125 1124 1317 506 14ƐƐ 85Ɛ 16Ɛ7 16Ɛ6
7 6 125 124 27Ɛ 13Ɛ 431 109 5Ɛ5 5Ɛ4 775 774 965 964 Ɛ2Ɛ 575 112Ɛ 675 1337 512 150Ɛ 865 1705 398
Ɛ 1 12Ɛ 75 285 284 435 434 5Ɛ7 66 77Ɛ 39Ɛ 971 172 Ɛ31 116 1135 1134 133Ɛ 77Ɛ 1517 1ᘔᘔ 1711 493
11 2 131 76 291 83 437 436 5ƐƐ 2ƐƐ 785 784 987 32ᘔ Ɛ37 Ɛ36 114Ɛ 685 1345 1344 1521 436 1715 1714
15 14 13Ɛ 295 294 447 446 611 163 791 3ᘔ6 995 994 Ɛ45 Ɛ44 1151 115 1351 166 1525 1524 1727 1726
17 6 141 20 2ᘔ1 150 455 454 615 614 797 796 9ᘔ7 9ᘔ6 Ɛ61 16 1165 1164 1365 1364 1547 1Ɛ2 1735 1734
Ɛ 145 144 2ᘔƐ 155 457 15ᘔ 617 206 7ᘔ1 138 9ᘔƐ 4Ɛ5 Ɛ67 3ᘔ2 1167 42 1367 1366 1561 89 1745 1744
25 4 147 56 2Ɛ1 26 45Ɛ 22Ɛ 61Ɛ 30Ɛ 7ƐƐ 3ƐƐ 9Ɛ1 Ɛ6Ɛ 595 1185 1184 136Ɛ 795 156Ɛ 97 1747 1746
27 26 157 12 2ƐƐ 37 465 464 637 212 801 140 9Ɛ5 9Ɛ4 Ɛ71 596 118Ɛ 6ᘔ5 1377 106 1577 1576 1751 32ᘔ
31 9 167 166 301 90 46Ɛ 7 63Ɛ 31Ɛ 80Ɛ 405 9ƐƐ 4ƐƐ Ɛ91 2Ɛ3 1197 472 138Ɛ 7ᘔ5 157Ɛ 89Ɛ 1755 1754
35 34 16Ɛ 95 307 102 471 13 647 216 817 816 ᘔ07 ᘔ06 Ɛ95 Ɛ94 11ᘔ1 1391 3Ɛ3 1585 1584 1757 1756
37 36 171 96 30Ɛ 165 481 24 655 654 825 824 ᘔ0Ɛ 505 Ɛ97 Ɛ96 11ᘔ5 11ᘔ4 1395 1394 1587 2ᘔ 176Ɛ 995
175 8 315 314 485 44 661 176 82Ɛ 415 ᘔ11 56 Ɛᘔ5 Ɛᘔ4 11ᘔ7 11ᘔ6 13ᘔ1 7Ɛ0 1591 2Ɛ6 1781 4Ɛ0
45 44 17Ɛ 321 170 48Ɛ 245 665 138 835 834 ᘔ17 ᘔ16 ƐƐ5 ƐƐ4 11ᘔƐ 6Ɛ5 13ᘔ7 536 15ᘔƐ 8Ɛ5 1785 1784
25 181 ᘔ0 325 324 497 496 66Ɛ 335 841 84 ᘔ27 156 ƐƐ7 ƐƐ6 11Ɛ7 11Ɛ6 13Ɛ1 13Ɛ 15ƐƐ 8ƐƐ 178Ɛ 9ᘔ5
51 13 18Ɛ ᘔ5 327 10ᘔ 4ᘔ5 4ᘔ4 675 674 851 14ᘔ ᘔ35 ᘔ34 1005 1004 1201 700 13Ɛ5 13Ɛ4 1601 160 1797 1796
57 56 195 194 32Ɛ 175 4Ɛ1 9ᘔ 687 686 855 854 ᘔ37 ᘔ36 1011 73 120Ɛ 705 1405 1404 1615 1614 17ᘔ1 9Ɛ0
19Ɛ ᘔƐ 33Ɛ 17Ɛ 4ƐƐ 25Ɛ 68Ɛ 345 85Ɛ 42Ɛ ᘔ3Ɛ 51Ɛ 1017 1016 1211 706 1407 326 1621 910 17ᘔ5 17ᘔ4
61 30 1ᘔ5 1ᘔ4 347 46 507 182 695 694 865 864 ᘔ41 188 1021 610 121Ɛ 70Ɛ 1425 1424 1625 1624 17ƐƐ 9ƐƐ
67 22 1ᘔ7 46 34Ɛ 511 266 69Ɛ 34Ɛ 867 2ᘔ2 ᘔ45 ᘔ44 1027 1026 1231 123 142Ɛ 815 1635 274 1807 682
35 1Ɛ1 Ɛ6 357 11ᘔ 517 516 6ᘔ7 6ᘔ6 871 152 ᘔ4Ɛ 525 1041 620 123Ɛ 71Ɛ 1431 286 1647 276 1815 1814
75 8 1Ɛ5 1Ɛ4 35Ɛ 18Ɛ 51Ɛ 45 6Ɛ1 881 440 ᘔ5Ɛ 52Ɛ 1047 1046 1245 1244 1437 1436 1655 1654 181Ɛ ᘔ0Ɛ
81 14 1Ɛ7 1Ɛ6 365 364 527 526 701 360 88Ɛ 445 ᘔ6Ɛ 535 104Ɛ 625 1255 114 143Ɛ 81Ɛ 1657 61ᘔ 1825 1824
85 84 205 204 375 34 531 276 705 704 8ᘔ5 98 ᘔ77 ᘔ76 1051 313 1257 49ᘔ 1445 1444 165Ɛ 92Ɛ 1831 509
87 86 217 86 377 376 535 534 70Ɛ 365 8ᘔ7 2Ɛ6 ᘔ87 ᘔ86 1061 16 125Ɛ 72Ɛ 1457 1456 1667 622 183Ɛ ᘔ1Ɛ
45 21Ɛ 10Ɛ 391 1ᘔ6 541 54 711 71 8ᘔƐ 455 ᘔ91 283 106Ɛ 635 1261 730 1461 38 1671 936 184Ɛ ᘔ25
91 46 221 66 397 396 545 544 71Ɛ 36Ɛ 8Ɛ5 8Ɛ4 ᘔ95 ᘔ94 107Ɛ 63Ɛ 126Ɛ 735 1465 1464 1677 20ᘔ 1861 269
95 94 225 224 3ᘔ5 3ᘔ4 557 556 721 370 8Ɛ7 8Ɛ6 ᘔ9Ɛ 54Ɛ 1087 1086 127Ɛ 73Ɛ 1467 562 167Ɛ 93Ɛ 1865 1864
ᘔ7 ᘔ6 237 92 3ᘔƐ 1Ɛ5 565 564 727 24ᘔ 901 230 ᘔᘔ7 376 109Ɛ 64Ɛ 1281 740 1471 419 1681 140 186Ɛ ᘔ35
ᘔƐ 55 241 120 3Ɛ5 3Ɛ4 575 574 735 734 905 198 ᘔᘔƐ 555 10Ɛ1 329 1295 94 1475 1474 1685 1684 1875 1874
Ɛ5 Ɛ4 24Ɛ 125 3Ɛ7 3Ɛ6 577 116 737 736 907 906 ᘔƐ7 ᘔƐ6 10Ɛ7 10Ɛ6 1297 1296 147Ɛ 83Ɛ 168Ɛ 945 1877 146
Ɛ7 Ɛ6 251 73 401 100 585 584 745 744 90Ɛ 465 ᘔƐƐ 55Ɛ 10ƐƐ 65Ɛ 12ᘔ1 75 148Ɛ 845 1697 1696 189Ɛ ᘔ4Ɛ
105 104 255 254 40Ɛ 205 587 1ᘔᘔ 747 9ᘔ 91Ɛ 46Ɛ Ɛ11 1ᘔ2 1101 220 12ᘔ5 2Ɛ8 1495 1494 169Ɛ 94Ɛ 18ᘔ1 210
107 106 25Ɛ 12Ɛ 415 414 58Ɛ 2ᘔ5 751 1ᘔ3 921 236 Ɛ15 228 1105 1104 12ᘔ7 12ᘔ6 149Ɛ 84Ɛ 16ᘔ1 950 18ᘔƐ ᘔ55
period length primes period length primes
1 Ɛ 11 1Ɛ0411, 69ᘔ3901
2 11 12 157, 7687
3 111 13 51, 471, 57Ɛ1
4 5, 25 14 15, 81, 106ᘔ95
5 11111 15 ᘔ9ᘔ9ᘔƐ, 126180ƐƐ0ƐƐ
6 7, 17 16 Ɛ61, 1061
7 46Ɛ, 2ᘔ3Ɛ 17 1111111111111111111
8 75, 175 18 24727225
9 31, 3ᘔ891 19 Ɛ00Ɛ00ƐƐ0ƐƐ1
Ɛ0Ɛ1 1ᘔ Ɛ0Ɛ0Ɛ0Ɛ0Ɛ1
Ɛ 1Ɛ, 754Ɛ2Ɛ41 3Ɛ, 78935Ɛᘔ441, 523074ᘔ3ᘔᘔƐ
10 ƐƐ01 20 141, 8Ɛ5281

The period level of a prime p ≥ 5 is (p−1)/(period length of 1/p), e.g., ${\displaystyle {\frac {1}{17}}}$ 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 R4ᘔ5 (Rn is the repunit with length n). If p is a Sophie Germain prime other than 2, 3 and 5, then Rp is divisible by 2p+1, thus Rp 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 10c-1 = 1 mod c) which are not divisible by Ɛ, they are called "deceptive primes", if n is deceptive prime, then Rn 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.

smallest multiple of n uses only 0's and 1's
n +1 +2 +3 +4 +5 +6 +7 +8 +9 +ᘔ +10
0+ 1 10 10 10 101 10 1001 100 100 1010 11111111111 10
10+ 11 10010 1010 100 10111 100 1001 1010 10010 111111111110 11101 100
20+ 110111 110 1000 10010 101 1010 101011 1000 111111111110 101110 101101 100
30+ 1001001 10010 110 10100 111001 10010 101001 111111111110 10100 111010 10001111 100
40+ 10111101 1101110 101110 110 1100101 1000 1011101111111 100100 10010 1010 101011 1010
50+ 10010101 1010110 100100 1000 1111 111111111110 1100101 101110 111010 1011010 1100111 100
60+ 10101101 10010010 1101110 10010 1011101111111 110 10101011 10100 10000 1110010 100111001 10010
70+ 1101001 1010010 1010 1111111111100 10001 10100 1001 111010 1010110 100011110 101101 1000
80+ 111011 101111010 1111111111100 1101110 110001 101110 10100111 1100 1011010 11001010 100111 1000
90+ 1010111111 10111011111110 10010010 100100 10101001 10010 110101001 1010 1100 1010110 101100011 10100
ᘔ0+ 111111111101 100101010 1110010 1010110 11100001 100100 1100001 10000 1010010 11110 1111011111 111111111110
Ɛ0+ 1001 11001010 101000 1011100 101011 111010 11010111 1011010 100011110 11001110 1111111111111111111111 100
 n 1 5 7 Ɛ 11 15 17 1Ɛ 21 25 27 2Ɛ smallest k such that k×n is a repunit 1 275 1ᘔ537 123456789Ɛ 1 92ᘔ79Ɛ43715865 8327 69Ɛ63848Ɛ 634ᘔ159788253ᘔ72Ɛ1 55 509867481Ɛ793ᘔᘔ5ᘔ1243628Ɛ317 45ᘔ3976ᘔ7Ɛ the length of the repunit k×n 1 4 6 Ɛ 2 14 6 Ɛ 18 4 26 10

(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 = RᘔƐ, 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 Rn 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 10c = 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{1n} = 34×10n+(10n−1)/Ɛ = (309×10n−1)/Ɛ and it can be factored to ((19×10n/2−1)/Ɛ) × (19×10n/2+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{1Ɛ78}, 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{1n} 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 10×582781Ɛ5+1, it has 459655 digits.

All numbers of the form 1{5}1 are composite (proof: 1{5n}1 = (14×10n+1−41)/Ɛ and it can be factored to (4×10(n+1)/2−7) × ((4×10(n+1)/2−7)/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 1010100 are 15, 57, 5Ɛ, 117, 11Ɛ, 5ƐƐƐ (the smallest representative prime of the permutation set).

The non-repdigit circular primes below 1010100 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 below 1010 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 (= 23) and 14 (= 24), 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 )

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 (=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 (=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 = 93 + ᘔ3, 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, ...)

n fixed points and cycled for the sequence for sum of n-th powers of the digits length of these cycles
1 {1}, {2}, {3}, {4}, {5}, {6}, {7}, {8}, {9}, {ᘔ}, {Ɛ} 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1
2 {1}, {5, 21}, {8, 54, 35, 2ᘔ, 88, ᘔ8, 118, 56, 51, 22}, {18, 55, 42}, {25}, {68, 84}, {ᘔ5} 1, 2, ᘔ, 3, 1, 2, 1
3 {1}, {8, 368, 52Ɛ, ᘔ20, 700, 247, 2ᘔ7, 947, 7ᘔ8, 10ᘔ7, 940, 561, 246, 200}, {577}, {668}, {6Ɛ5, Ɛ74, 100ᘔ}, {ᘔ83}, {11ᘔᘔ} 1, 12, 1, 1, 3, 1, 1
4 {1}, {ᘔ6ᘔ, 103ᘔ8, 8256, 35ᘔ9, 9ƐᘔƐ, 22643, Ɛ69, 1102ᘔ, 596ᘔ, ᘔ842, 8394, 6442, 1080, 2455}, {206ᘔ, 6668, 4754}, {3ᘔ2Ɛ, 12396, 472Ɛ, ᘔ02ᘔ, Ɛ700, 9ᘔ42, 98ᘔ9, 13902} 1, 12, 3, 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 two-digit number which is not repdigit converges to the cycle {0Ɛ, ᘔ1, 83, 47, 29, 65}.

n cycles for Kaprekar's routine for n-digit numbers length of these cycles
1 {0} 1
2 {00}, {0Ɛ, ᘔ1, 83, 47, 29, 65} 1, 6
3 {000}, {5Ɛ6} 1, 1
4 {0000}, {3ƐƐ8, 8284, 6376}, {4198, 8374, 5287, 6196, 7ƐƐ4, 7375} 1, 3, 6
5 {00000}, {64Ɛ66, 6ƐƐƐ5}, {83Ɛ74} 1, 2, 1
6 {000000}, {420ᘔ98, ᘔ73742, 842874, 642876, 62ƐƐ86, 951963, 860ᘔ54, ᘔ40ᘔ72, ᘔ82832, 864654}, {65ƐƐ56} 1, ᘔ, 1

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=112, 127=7×21, 135=5×31, 144=4×41, 163=3×61, 368=86−3, 376=6×73, 441=(4+1)4, 445=54+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(Ɛ2) and φ(182) = φ(2×Ɛ2), next counterexample is 362, which equals φ(381) = φ(1Ɛ2) and φ(742) = φ(2×1Ɛ2), 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(p2) = 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 (xM7) is {x, x+4, x+7, x+Ɛ}, and the x minor 7th chord (xm7) is {x, x+3, x+7, x+ᘔ}, and the x dominant 7th chord (x7) is {x, x+4, x+7, x+ᘔ}, and the x diminished 7th triad (xdim7) 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 1.610 = ᘔ9.8Ɛ5809 is very close to 27 = ᘔ8, since 217 = 2134ᘔ8 is very close to 310 = 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 ${\displaystyle {\sqrt[{10}]{2}}}$ (=20.1) 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 ${\displaystyle {\sqrt[{10}]{2}}}$), and thus we have that f(5)/f(0) = (9/8) × (ᘔ/9) × (14/13) = 4/3. Also, we can found that 20.5 is very close to 1.4, and 20.7 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 = 22 × 3 × 5, next possible such order is 70 = 22 × 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 , 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 , 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 dp 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 10n (i.e. with at most n digits), and ln(10) = 2.599Ɛ035Ɛ8169...

N total numbers of primes ≤ N numbers of primes end with 1 ≤ N numbers of primes end with 5 ≤ N numbers of primes end with 7 ≤ N numbers of primes end with Ɛ ≤ N
10 5 0 1 1 1
40 13 2 4 4 3
100 2ᘔ 6 9 9 8
400 89 1ᘔ 23 23 23
1000 1ᘔ5 51 59 59 58
4000 621 157 16ᘔ 170 166
10000 1426 3Ɛ8 410 412 406
40000 4833 11ᘔ4 121Ɛ 1219 1211
100000 10852 31ᘔ4 3225 3225 321ᘔ
400000 3928Ɛ Ɛ333 Ɛ377 Ɛ3Ɛ9 Ɛ3ᘔ2
1000000 ᘔ4Ɛ20 27204 27295 2730ᘔ 27333

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

 1 2 3 4 5 6 7 8 9 ᘔ Ɛ 10 11 12 13 14 15 16 17 18 19 1ᘔ 1Ɛ 20 21 22 23 24 25 26 27 28 29 2ᘔ 2Ɛ 30 31 32 33 34 35 36 37 38 39 3ᘔ 3Ɛ 40 41 42 43 44 45 46 47 48 49 4ᘔ 4Ɛ 50 51 52 53 54 55 56 57 58 59 5ᘔ 5Ɛ 60 61 62 63 64 65 66 67 68 69 6ᘔ 6Ɛ 70 71 72 73 74 75 76 77 78 79 7ᘔ 7Ɛ 80 81 82 83 84 85 86 87 88 89 8ᘔ 8Ɛ 90 91 92 93 94 95 96 97 98 99 9ᘔ 9Ɛ ᘔ0 ᘔ1 ᘔ2 ᘔ3 ᘔ4 ᘔ5 ᘔ6 ᘔ7 ᘔ8 ᘔ9 ᘔᘔ ᘔƐ Ɛ0 Ɛ1 Ɛ2 Ɛ3 Ɛ4 Ɛ5 Ɛ6 Ɛ7 Ɛ8 Ɛ9 Ɛᘔ ƐƐ 100

## 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.


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

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/ 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/ = 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)
Examples in duodecimal Decimal equivalent
1 × (5/8) = 0.76 1 × (5/8) = 0.625
100 × (5/8) = 76 144 × (5/8) = 90
576/9 = 76 810/9 = 90
400/9 = 54 576/9 = 64
1ᘔ.6 + 7.6 = 26 22.5 + 7.5 = 30

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 bn, 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 bn is given using its prime factorization.

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

For example, the number 8 is a factor of 103 (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 = 0.625ten

For duodecimal, 12n = 22n * 3n. This has (2n+1)(n+1) divisors, the sample denominator of 8 is a factor of a gross (122 = 144), so eighths can not need more than two duodecimal fractional places to terminate. 5/8 = 0.76twelve

Because both ten and twelve have two unique prime factors, the number of divisors of bn for b = 10 or 12 grows quadratically with the exponent n (in other words, of the order of n2).

### 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.[35] 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/(22) = 0.25 ten = 0.3 twelve; 1/(23) = 0.125 ten = 0.16 twelve; 1/(24) = 0.0625 ten = 0.09 twelve; 1/(25) = 0.03125 ten = 0.046 twelve; etc.).

Values in bold indicate that value is exact.

Fraction Prime factorsof the denominator Positional representation Positional representation Prime factorsof the denominator Fraction Decimal basePrime factors of the base: 2, 5Prime factors of one below the base: 3Prime factors of one above the base: 11All other primes: 7, 13, 17, 19, 23, 29, 31 Duodecimal basePrime factors of the base: 2, 3Prime factors of one below the base: ƐPrime factors of one above the base: 11All other primes: 5, 7, 15, 17, 1Ɛ, 25, 27 1/2 2 0.5 0.6 2 1/2 1/3 3 0.3 0.4 3 1/3 1/4 2 0.25 0.3 2 1/4 1/5 5 0.2 0.2497 5 1/5 1/6 2, 3 0.16 0.2 2, 3 1/6 1/7 7 0.142857 0.186ᘔ35 7 1/7 1/8 2 0.125 0.16 2 1/8 1/9 3 0.1 0.14 3 1/9 1/10 2, 5 0.1 0.12497 2, 5 1/ᘔ 1/11 11 0.09 0.1 Ɛ 1/Ɛ 1/12 2, 3 0.083 0.1 2, 3 1/10 1/13 13 0.076923 0.0Ɛ 11 1/11 1/14 2, 7 0.0714285 0.0ᘔ35186 2, 7 1/12 1/15 3, 5 0.06 0.09724 3, 5 1/13 1/16 2 0.0625 0.09 2 1/14 1/17 17 0.0588235294117647 0.08579214Ɛ36429ᘔ7 15 1/15 1/18 2, 3 0.05 0.08 2, 3 1/16 1/19 19 0.052631578947368421 0.076Ɛ45 17 1/17 1/20 2, 5 0.05 0.07249 2, 5 1/18 1/21 3, 7 0.047619 0.06ᘔ3518 3, 7 1/19 1/22 2, 11 0.045 0.06 2, Ɛ 1/1ᘔ 1/23 23 0.0434782608695652173913 0.06316948421 1Ɛ 1/1Ɛ 1/24 2, 3 0.0416 0.06 2, 3 1/20 1/25 5 0.04 0.05915343ᘔ0Ɛ62ᘔ68781Ɛ 5 1/21 1/26 2, 13 0.0384615 0.056 2, 11 1/22 1/27 3 0.037 0.054 3 1/23 1/28 2, 7 0.03571428 0.05186ᘔ3 2, 7 1/24 1/29 29 0.0344827586206896551724137931 0.04Ɛ7 25 1/25 1/30 2, 3, 5 0.03 0.04972 2, 3, 5 1/26 1/31 31 0.032258064516129 0.0478ᘔᘔ093598166Ɛ74311Ɛ28623ᘔ55 27 1/27 1/32 2 0.03125 0.046 2 1/28 1/33 3, 11 0.03 0.04 3, Ɛ 1/29 1/34 2, 17 0.02941176470588235 0.0429ᘔ708579214Ɛ36 2, 15 1/2ᘔ 1/35 5, 7 0.0285714 0.0414559Ɛ3931 5, 7 1/2Ɛ 1/36 2, 3 0.027 0.04 2, 3 1/30

The duodecimal period length of 1/n are

0, 0, 0, 0, 4, 0, 6, 0, 0, 4, 1, 0, 2, 6, 4, 0, 16, 0, 6, 4, 6, 1, 11, 0, 20, 2, 0, 6, 4, 4, 30, 0, 1, 16, 12, 0, 9, 6, 2, 4, 40, 6, 42, 1, 4, 11, 23, 0, 42, 20, 16, 2, 52, 0, 4, 6, 6, 4, 29, 4, 15, 30, 6, 0, 4, 1, 66, 16, 11, 12, 35, 0, ... (sequence A246004 in the OEIS)

The duodecimal period length of 1/(nth prime) are

0, 0, 4, 6, 1, 2, 16, 6, 11, 4, 30, 9, 40, 42, 23, 52, 29, 15, 66, 35, 36, 26, 41, 8, 16, 100, 102, 53, 54, 112, 126, 65, 136, 138, 148, 150, 3, 162, 83, 172, 89, 90, 95, 24, 196, 66, 14, 222, 113, 114, 8, 119, 120, 125, 256, 131, 268, 54, 138, 280, ... (sequence A246489 in the OEIS)

Smallest prime with duodecimal period n are

11, 13, 157, 5, 22621, 7, 659, 89, 37, 19141, 23, 20593, 477517, 211, 61, 17, 2693651, 1657, 29043636306420266077, 85403261, 8177824843189, 57154490053, 47, 193, 303551, 79, 306829, 673, 59, 31, 373, 153953, 886381, 2551, 71, 73, ... (sequence A252170 in the OEIS)

### Irrational numbers

As for irrational numbers, none of them have a finite representation in any of the rational-based positional number systems (such as the decimal and duodecimal ones); this is because a rational-based positional number system is essentially nothing but a way of expressing quantities as a sum of fractions whose denominators are powers of the base, and by definition no finite sum of rational numbers can ever result in an irrational number. For example, 123.456 = 1 × 102 + 2 × 101 + 3 × 100 + 4 × 1/101 + 5 × 1/102 + 6 × 1/103 (this is also the reason why fractions that contain prime factors in their denominator not in common with those of the base do not have a terminating representation in that base). Moreover, the infinite series of digits of an irrational number does not exhibit a strictly repeating pattern; instead, the different digits often succeed in a seemingly random fashion. The following chart compares the first few digits of the decimal and duodecimal representation of several of the most important algebraic and transcendental irrational numbers, some of these numbers may be perceived as having fortuitous patterns, making them easier to memorize, when represented in one base or the other.

Algebraic irrational number In decimal In duodecimal
2 (the length of the diagonal of a unit square) 1.414213562373... (≈ 1.414) 1.4Ɛ79170ᘔ07Ɛ8... (≈ 1.4Ɛ7)
3 (the length of the diagonal of a unit cube, or twice the height of an equilateral triangle of unit side) 1.732050807568... (≈ 1.732) 1.894Ɛ97ƐƐ9687... (≈ 1.895)
5 (the length of the diagonal of a 1×2 rectangle) 2.236067977499... (≈ 2.236) 2.29ƐƐ13254058... (≈ 2.2ᘔ)
φ (phi, the golden ratio = ${\displaystyle \scriptstyle {\frac {1+{\sqrt {5}}}{2}}}$) 1.618033988749... (≈ 1.618) 1.74ƐƐ6772802ᘔ... (≈ 1.75)
Transcendental irrational number In decimal In duodecimal
π (pi, the ratio of circumference to diameter) 3.141592653589793238462643... (≈ 3.142) 3.184809493Ɛ918664573ᘔ6211... (≈ 3.185)
e (the base of the natural logarithm) 2.718281828459... (≈ 2.718) 2.875236069821... (≈ 2.875)

The first few digits of the decimal and duodecimal representation of another important number, the Euler–Mascheroni constant (the status of which as a rational or irrational number is not yet known), are:

Number In decimal In duodecimal
γ (the limiting difference between the harmonic series and the natural logarithm) 0.577215664901... (≈ 0.577) 0.6Ɛ15188ᘔ6760... (≈ 0.6Ɛ1)

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