## FANDOM

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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 numbers n such that the concatenation of n and the unit (1), i.e. 10n+1 (all squares of primes except 4 and 9 are of this form), is square, are all even numbers, and the half of these n are exactly the generalized pentagonal numbers, and such numbers are important to Euler's theory of partitions, as expressed in his pentagonal number theorem (the pentagonal number theorem, originally due to Euler, relates the product and series representations of the Euler function. It states that

$\prod_{n=1}^{\infty}\left(1-x^{n}\right)=\sum_{k=-\infty}^{\infty}\left(-1\right)^{k}x^{k\left(3k-1\right)/2}=1+\sum_{k=1}^\infty(-1)^k\left(x^{k(3k+1)/2}+x^{k(3k-1)/2}\right).$

In other words,

$(1-x)(1-x^2)(1-x^3)(1-x^4) \cdots = 1 - x - x^2 + x^5 + x^7 - x^{10} - x^{13} + x^{1\mathcal{X}} + x^{22} - \cdots.$

The exponents 1, 2, 5, 7, 10, 13, 1X, 22, ... on the right hand side are given by the formula Template:Math for k = 1, −1, 2, −2, 3, −3, 4, −4, ... and are called (generalized) pentagonal numbers. This holds as an identity of convergent power series for $|x|<1$, and also as an identity of formal power series.

A striking feature of this formula is the amount of cancellation in the expansion of the product), also, the identity implies a marvelous recurrence for calculating $p(n)$, the number of partitions of n (p(n)):

$p(n)=p(n-1)+p(n-2)-p(n-5)-p(n-7)+p(n-10)+p(n-13)-p(n-1\mathcal{X})-p(n-22)+\cdots$
$p(0)=1$

or more formally,

$p(n)=\sum_k (-1)^{k-1}p(n-g_k)$

Also the sum of divisors of n (σ(n)):

$\sigma(n)=\sigma(n-1)+\sigma(n-2)-\sigma(n-5)-\sigma(n-7)+\sigma(n-10)+\sigma(n-13)-\sigma(n-1\mathcal{X})-\sigma(n-22)+\cdots$

but if the last term is σ(0) (this situation appears if and only if n itself is generalized pentagonal number, i.e. the concatenation of 2n and 1 is square), then we change it to n.

where the summation is over all nonzero integers k (positive and negative) and $g_k$ is the kth generalized pentagonal number. Since $p(n)=0$ for all $n<0$, the series will eventually become zeroes, enabling discrete calculation, besides, generalized pentagonal numbers are closely related to centered hexagonal numbers (also called hex numbers, the hex numbers are end with 1, 7, 7, 1, 1, 7, 7, 1, ...). When the array corresponding to a centered hexagonal number is divided between its middle row and an adjacent row, it appears as the sum of two generalized pentagonal numbers, with the larger piece being a pentagonal number proper.

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

No repdigits with more than one digit are squares, in fact, a square cannot end with three same digits except 000.

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, 8EE8, 9119, X44X and E77E, but none of them are squares)

n n-digit palindromic squares square roots number of n-digit palindromic squares
1 1, 4, 91, 2, 33
2 nonenone0
3 121, 48411, 222
4 nonenone0
5 10201, 12321, 14641, 16661, 16E61, 40804, 41414, 44944101, 111, 121, 12E, 131, 202, 204, 2128
6 16006142E1
7 1002001, 102X201, 1093901, 1234321, 148X841, 4008004, 445X544, 49XXX941001, 1015, 1047, 1111, 1221, 2002, 2112, 22448
8 nonenone0
9 100020001, 102030201, 104060401, 1060E0601, 121242121, 123454321, 125686521, 1420E0241, 1444X4441, 1468E8641, 14X797X41, 1621E1261, 163151361, 1XX222XX1, 400080004, 404090404, 410212014, 4414X4144, 4456E6544, 496787694, 96384836910001, 10101, 10201, 10301, 11011, 11111, 11211, 11E21, 12021, 12121, 1229E, 1292E, 12977, 14685, 20002, 20102, 20304, 21012, 21112, 22344, 3105319
X 1642662461434X51
E 10000200001, 10221412201, 10444X44401, 12102420121, 12345654321, 141E1E1E141, 14404X40441, 16497679461, 40000800004, 40441X14404, 41496869414, 44104X40144, 49635653694100001, 101101, 102201, 110011, 111111, 11E13E, 120021, 12X391, 200002, 201102, 204204, 210012, 22334411
10 nonenone0

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 ≤ E, but not for n ≥ 10 (thus, for all odd number n ≤ 19, there 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.

The square numbers using no more than two distinct digits are 0, 1, 4, 9, 14, 21, 30, 41, 54, 69, 84, X1, 100, 121, 144, 344, 400, 441, 484, 554, 900, 3000, 4344, 9944, 10000, 11XX1, 16661, 40000, 41414, 44944, 47744, 66969, 90000, 111101, 114144, 300000, 444404, 454554, 999909, 1000000, 1141144, 3333030, 4000000, 4544554, 9000000, 11110100, 30000000, 41144144, 44440400, 99990900, XXXXXXX1, 100000000, 333303000, 400000000, 900000000, 1111010000, 3000000000, 4444040000, 9999090000, 10000000000, 33330300000, 40000000000, 90000000000, 111101000000, 300000000000, 444404000000, 999909000000, 1000000000000, ...

41414 is the largest undulating square (of the form ababab...), note that the only 2 distinct digits in it (1 and 4) and the only 2 distinct two-digit numbers formed by its digits (14 and 41) are also all squares.

A cube can end with all digits except 2, 6 and X (in fact, no perfect powers end with 2, 6 or X), 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 cube numbers using no more than two distinct digits are 0, 1, 8, 23, 54, X5, 1000, 1331, 8000, 1000000, 8000000, 1000000000, 8000000000, 1000000000000, ...

The digital root of a cube can be any number.

21 and 201 are both squares, and it is conjectured that no other numbers of the form 2000...0001 (i.e. of the form 2×10n+1) are squares.

10814 and 100814 are both squares, and 10854 and 100853 are both cubes.

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.

If x^2 + y^2 = z^2 (that is, {x, y, z} is a Pythagorean triple, then xy end with 0 (and thus xyz also end with 0).

Squares (and every powers) of 0, 1, 4, 9, 54, 69, 369, 854, 3854, 8369, E3854, 1E3854, X08369, ... 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 (dozadic numbers), these solutions are 0, 1, ...2E21E61E3854 and ...909X05X08369, since 10 is neither a prime nor 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)

The triangular numbers using no more than two distinct digits are 0, 1, 3, 6, X, 13, 19, 24, 30, 39, 47, 56, 66, 77, 89, X0, E4, 191, 303, 446, 550, 633, 66X, 6X6, 1117, 3X3X, 3EE3, 6060, 6161, 6366, 6999, 6EE6, 8989, 9779, 23223, 35553, 50050, 77677, 113113, 303333, 331331, 600600, X33X33, 3030330, 60006000, 333666333, 6000060000, 600000600000, ...

The pronic numbers using no more than two distinct digits are 0, 2, 6, 10, 18, 26, 36, 48, 60, 76, 92, E0, 110, 606, 656, 992, XX0, EE6, 1118, 2232, 7878, EE00, 10100, 33330, 46446, 6XXX6, X00X0, 118118, 226226, 606666, 662662, EEE000, 1001000, 6060660, EEEE0000, 100010000, EEEEE00000, 10000100000, EEEEEE000000, 1000001000000, ...

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 X.

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
0 1000000000000000000000000 1
1 1111111111111111111111111 1
2 1248484848484848484848484 2
3 1393939393939393939393939 2
4 1444444444444444444444444 1
5 1515151515151515151515151 2
6 1600000000000000000000000 1
7 1717171717171717171717171 2
8 1848484848484848484848484 2
9 1999999999999999999999999 1
X 1X44444444444444444444444 1
E 1E1E1E1E1E1E1E1E1E1E1E1E1 2
10 1000000000000000000000000 1
11 1111111111111111111111111 1
12 1248484848484848484848484 2
13 1393939393939393939393939 2
14 1444444444444444444444444 1
15 1515151515151515151515151 2
16 1600000000000000000000000 1
17 1717171717171717171717171 2
18 1848484848484848484848484 2
19 1999999999999999999999999 1
1X 1X44444444444444444444444 1
1E 1E1E1E1E1E1E1E1E1E1E1E1E1 2
20 1000000000000000000000000 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, E (all digits != 2 mod 4)
final two digits of nk
0 01000000000000000000000000000000000000000000000000 1
1 01010101010101010101010101010101010101010101010101 1
2 01020408142854X89468142854X89468142854X89468142854 6
3 01030923698309236983092369830923698309236983092369 4
4 01041454941454941454941454941454941454941454941454 3
5 010521X5418561658145X125010521X5418561658145X12501 10
6 01063060000000000000000000000000000000000000000000 1
7 01074147818701074147818701074147818701074147818701 6
8 01085468546854685468546854685468546854685468546854 2
9 01096909690969096909690969096909690969096909690969 2
X 010X84E4545454545454545454545454545454545454545454 1
E 010EX12E814E616E418E21XE010EX12E814E616E418E21XE01 10
10 01100000000000000000000000000000000000000000000000 1
11 01112131415161718191X1E101112131415161718191X1E101 10
12 0112440894X85428146894X85428146894X85428146894X854 6
13 01136953695369536953695369536953695369536953695369 2
14 01149454149454149454149454149454149454149454149454 3
15 01150115011501150115011501150115011501150115011501 2
16 01163060000000000000000000000000000000000000000000 1
17 01176177011761770117617701176177011761770117617701 4
18 01189468142854X89468142854X89468142854X89468142854 6
19 01190939699909396999093969990939699909396999093969 4
1X 011X44E4941454941454941454941454941454941454941454 3
1E 011E815E419E011E815E419E011E815E419E011E815E419E01 6
20 01200000000000000000000000000000000000000000000000 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
0 1000000000000000000000000 1
1 1111111111111111111111111 1
2 12485X973612485X973612485 X
3 1395413954139541395413954 5
4 1459314593145931459314593 5
5 1534915349153491534915349 5
6 16379X584216379X584216379 X
7 17523X469817523X469817523 X
8 18964X325718964X325718964 X
9 1943519435194351943519435 5
X 1X1X1X1X1X1X1X1X1X1X1X1X1 2
E 1EEEEEEEEEEEEEEEEEEEEEEEE 1
10 1111111111111111111111111 1
11 12485X973612485X973612485 X
12 1395413954139541395413954 5
13 1459314593145931459314593 5
14 1534915349153491534915349 5
15 16379X584216379X584216379 X
16 17523X469817523X469817523 X
17 18964X325718964X325718964 X
18 1943519435194351943519435 5
19 1X1X1X1X1X1X1X1X1X1X1X1X1 2
1X 1EEEEEEEEEEEEEEEEEEEEEEEE 1
1E 1111111111111111111111111 1
20 12485X973612485X973612485 X

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

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

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.

nFndigit root of FnLndigit root of LnnFndigit root of FnLndigit root of Ln
111112137501581101E
21133225X30181111037
32244239580221922047
4337724133E03X2X33073
555EE25209705147550EX
68816726341608E7588162
71122572754E111110121251
819X3E3288907191176X93E3
92X164X29121E82X22780X644
X47EX322X1XE03473432E7X37
E75114712E310EE7556XE0647E
10100122X33050003008E22022X7
11175237543181102752161108757
1227535X373211110575X25330XX33
1342X5958E331922082X13E441758X
146X38133E7342X3311X3E6477263E2
15E112209773547551X111X3EE41971
1615E4X341633675882EE411487668163
172505154E1X37101214X05223075X9E14
183XE9E8907238176X979E933793056077
1964021121E81392780E080255X9X643E8E
1XX2EE11XE0333X432E885EE8967169X037
1E147012310EE43E6XE079201213550121EE7
2022X00350002740E22045800X21001800023

(Note that F2X begins with L1X, and F2E begins with L1E)

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

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 Template:Oeis).

Except 0 = F0 and 1 = F1 = F2, the only square Fibonacci number is 100 = F10 (100 is exactly 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.

n2nn2nn2nn2nn2nn2n
1221E2X20X8415317E5804588X861256906X1X93096E8934X88111X12X743504482569888538X0X8X165933E8691303X448E712227X7E11448X8
24221X58419442X633XE408E5594624E161183966171E56699482238259286X08944E17554X758194X210E667E51626078895E22445393X2289594
382338E4836843190679X815XXE68639X302347710323XE11768834744E6551815689X32XX992E4368X321E113XX305013556EX4488X76784556E68
4142475X9471444361137942E99E14641786046932206479X23314849289E0XX342E15786599765X8714X443X2279860X026XE1E88955931348XE1E14
5282512E96922845702273685E77X286533500916644109378466288516557X198685X2E350E7730E95228X5878453750180519X3E556XE6626959X3X28
6542625E71645446120452714EE338546666X01631088216734910548630XE3837514E85X6X1E3261E6X454X615348X72X0340X3787XXE19E10516E787854
7X8274EE2308X8472408X5229EX674X867111803062154431269620X887619X7472X29E4E9183X6503E188X8X72X695925806818735399X37X20X31E3534X8
8194289EX4615944848158X457E912994682234061042X88625170419488103792925857X9E634790X07X35594X85916E64E41143526X777873841863X6X6994
93682917E8902E6849942E588E3E62576869446810208595504X32083688920736564E4E397E06936181386XE68X9E631E09X82286X5193335274835079191768
X7142X33E5605E144X1685XE55X7E04E3146X891420414E6XX09864147148X41270E09X9X773X116703427519E14XX1E063X179445518X36666X52946X136363314
E12282E67XE00EX284E314E9XXE93X09X6286E1562840829E19817508292288E82521X179793278231206852X37X28XE3X107833688XX358711118X56918270706628
102454301139X01E85450629E799E678179054702E05481457X37432X145645490144X4383373665344624114X5873854E078213467155986E52222358E1634521211054
1148X8312277803E4X8511057E377E1343360X8715X0X9428E3872865828E08X8912898874672710X6890482298E5274X8E11344269122XE751XX44446E5X3068X424220X8
129594324533407X9945220E3X733X26867019472E8196855X752550E455X159492557552912522191560944575XX52994E226885162459E2X39888891XE86115884844194
1316E68338X6681397685341X792678515120368731E43714XE92X4XX1X8XE82E6893XE2XX5624X44362E01688E2E98X5768E35154X3048E7X58775555639E5022E549488368
1431E143415911427731454839365134X2X240714743X872299E6589983959E45E149419X598E049888705X03155X5E758E314E4X2X986095E38E532XXXE077XX045XX96954714
1563X28352E622853262855147670X2698584812287579524577E0E577476E7X8EX2895378E75X09755520E8062XE8EE2E5X628E5185975016EX75XX65999X1339808E99716X9228
16107854365E0454X650545629312185174E494245476136X48E33X1XE32931E395E85496735E2E8172XXX41E41059E5EX5XE9054E634E72X031E92E990E77782677415E7723196454
172134X837EX08X990X0X8575662434X329X96848X877271895X67839X65663X76EE4X897126EX5E432599883X820E7XEE8E9E60X8E769E258063E65E761E3334513282EE32463708X8
18426994381E8159761819458E1048698657971495947852356E91347790E107931EX99498251E8EX864E775479441E39EE5E7E0194E8117X4E4107E0EE303X6668X26545EX6490721594
19851768393E42E73034368591X20951750E372296E6879X46E1E62693361X213663E9768994X3E5E9509E32X936883X77EXEE3X0368E923389X8213X1EX60791115850X8EE90961242E68
1X14X33143X7X85E260687145X38416X32X1X724571E147X1891X3E051667038427107E73149X987XEE6X17X659671547933E9EX780714EX467579442783E90136222E4X195EE61702485E14
1E29866283E1394EX501152285E74831865839248E23X287E356387X0X3112074852213E26289E17539EE183390E7122X93667E7E9341228EE912E36885347E60270445X9836EEE0320494EX28
205751054402769E8X022X4546012946350E476495X47854806E075381862241294X4427X5054X032X77EX346761E2245967113E3E66824541001625X7154X693E0052088E97471EEX0640969E854

For all digits 1 ≤ d ≤ X (i.e. all digits other than the largest digit (E)), there exists 0 ≤ n ≤ 20 such that 2n starts with the digit d. (This is not true for the digit E, the smallest power of 2 starts with the digit E is indeed 221 = E2X20X8)

21XE = 59E18922E81631X39875663E89X853X91E595336X6114815X5X6929933X288E774E479575X628 may be the largest power of 2 not contain the digit 0, it has 65 digits.

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 EE digits. (thus, the Fermat number F9 (=229+1) is very close to googol) (it is a mathematical coincidence: 229 ≈ 10102)

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 + 6E4 (=93 + X3) (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 3X9).

All values of n > 45 (45 is the largest n such that this is not true, note that 45 is also the smallest prime that produces prime reciprocal magic square) for incrementally largest values of minimal x > 1 (or minimal y > 0) satisfying Pell's equation $x^2-ny^2=1$ end with 1, and the dozens digit of all such values n > 2X1 are odd. (these values n are 2, 5, X, 11, 25, 3X, 45, 51, 91, 131, 1E1, 291, 2X1, 2E1, 391, 471, 711, 751, 971, X91, E31, ...)

The denominator of every nonzero Bernoulli number (except $B_0=1$ and $B_1=-\frac{1}{2}$) ends with 6.

If n ends with 2 and n/2 is prime (or 1), then the denominator of the Bernoulli number $B_n$ is 6 (this is also true for some (but not all) n ends with X and n/2 is prime). (if the denominator of the Bernoulli number $B_n$ is 6, then n ends with 2 or X, but n/2 needs not to be prime or 1, the first counterexample is n = 82, the denominator of the Bernoulli number $B_{82}$ is 6, but 82/2 = 41 = 72 is neither prime nor 1)

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

The recurring dozenal of the reciprocal of n terminates if and only if n is 3-smooth number (or harmonic number[1]) (i.e. n is regular to 10 if and only if n is 3-smooth number (or harmonic number)), since the 3-smooth numbers (or the harmonic numbers) are the numbers that evenly divide powers of 10. 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, X8, 100, 116, 140, 160, 183, 194, 200, 230, 280, 300, 346, 368, 400, 460, 509, 540, 600, 690, 714, 800, 900, X16, X80, 1000. They are exactly the numbers k such that $\phi(6k)=2k$, where $\phi$ is the Euler's totient function. The sum of the reciprocals of the 3-smooth numbers is equal to 3, i.e. 1/1 + 1/2 + 1/3 + 1/4 + 1/6 + 1/8 + 1/9 + 1/10 + ... = 1 + 0.6 + 0.4 + 0.3 + 0.2 + 0.16 + 0.14 + 0.1 + ... = 3. Brief proof: 1/1 + 1/2 + 1/3 + 1/4 + 1/6 + 1/8 + 1/9 + 1/10 + ... = (Sum_{m>=0} 1/(2^m)) * (Sum_{n>=0} 1/(3^n)) = (1/(1-1/2)) * (1/(1-1/3)) = (2/(2-1)) * (3/(3-1)) = 3.

The 3-smooth numbers (or the numbers n such that the reciprocal of n terminates) ≤ 10 are 1, 2, 3, 4, 6, 8, 9, and 10, all of these numbers except 8 and 9 are divisors of 10 (8 is because it has more prime factors 2 than 10, and 9 is because it has more prime factors 3 than 10) (thus, the numbers of digits of the reciprocal of all these n except 8 and 9 are all 1, while the numbers of digits of the reciprocal of 8 and 9 are 2), and the numbers 8 and 9 are in the Catalan's conjecture (i.e. 8 and 9 are the only case of two consecutive perfect powers), besides, (8, 9) is the largest pair of regular numbers which differ by 1, besides, the product of 8 and 9 is 60, which is the smallest Achilles number, besides, the concatenation of 8 and 9 is 89, which is the smallest Ziesel number and the smallest integer such that the factorization of $x^n-1$ over Q includes coefficients other than $\pm 1$ (i.e. the 89th cyclotomic polynomial, $\Phi_{89}$, is the first with coefficients other than $\pm 1$), besides, the repunit with length k (Rk) (where k = the concatenation of n and the unit (1), i.e. k = 10n+1) is prime for both n = 8 and n = 9, and not for any other n ≤ 1000, besides, the squares of 8 and 9 are the only two 2-digit automorphic numbers, besides, 8 and 9 are the only two natural numbers n such that centered n-gonal numbers (the kth centered n-gonal number is n×Tk+1, where Tk is the kth triangular number) cannot be primes (8 is because all centered 8-gonal numbers are square numbers (4-gonal numbers), 9 is because all centered 9-gonal numbers are triangular numbers (3-gonal numbers) not equal to 3, but all square numbers and all triangular numbers not equal to 3 are not primes, in fact, all polygonal numbers with rank > 2 are not primes, i.e. all primes p cannot be a polygonal number (except the trivial case, i.e. each p is the second p-gonal number)), assuming the Bunyakovsky conjecture is true. (i.e. 8 and 9 are the only two natural number n such that $\frac{n}{2}x^2+\frac{n}{2}x+1$ is not irreducible) (Note that for n = 10, the centered 10-gonal numbers are exactly the star numbers)

The smallest n≥1 such that 4×60n−1 is prime (where 4 is the smallest composite number, and 60 is the smallest Achilles number, note that 4×60n is 3-smooth for all n≥1, thus 1/(4×60n) terminates in dozenal, and note that 4×60n is exactly 10000 (104) when n=2) is 460089, which contains the only four composite 3-smooth digits (4, 6, 8, 9) exactly once and from small to large in order (4 —> 6 —> 8 —> 9), and with two 0’s (zero digits) inside the middle, this prime proves the generalized Riesel problem base 60 (proves that 205 is the smallest generalized Riesel number in base 60, i.e. 205 is the smallest k such that gcd(k−1, 60−1) = 1 and k×60n−1 is composite for all n≥1, note that 205 = 4×60+4+1 = 4×(60+1)+1).

The 3-smooth numbers (or the numbers n such that the reciprocal of n terminates) ≤ 20 are 1, 2, 3, 4, 6, 8, 9, 10, 14, 16, and 20, all of these numbers are divisors of 100, note that the next two 3-smooth numbers (23 and 28) are not divisors of 100 (23 is because it has more prime factors 3 than 100, and 28 is because it has more prime factors 2 than 100) (thus, the numbers of digits of the reciprocal of all these n are all ≤2, while the numbers of digits of the reciprocal of 23 and 28 are 3).

Regular n-gon is constructible using neusis, or an angle trisector if and only if the reciprocal of $\varphi(n)$ is terminating number (where $\varphi$ is Euler's totient function) (i.e. $\varphi(n)$ is 3-smooth, or $\varphi(n)$ is regular to 10), thus the n ≤ 1000 such that regular n-gon is constructible using neusis, or an angle trisector are 3, 4, 5, 6, 7, 8, 9, X, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 22, 23, 24, 26, 28, 2X, 2E, 30, 31, 32, 33, 34, 36, 39, 40, 43, 44, 46, 48, 49, 50, 53, 54, 55, 58, 5X, 60, 61, 62, 64, 66, 68, 69, 70, 71, 76, 77, 7E, 80, 81, 86, 88, 89, 90, 91, 93, 94, 96, 99, 9E, X0, X6, X8, XX, E1, E3, E4, E8, 100, 102, 104, 108, 109, 110, 114, 116, 117, 120, 122, 123, 130, 132, 135, 139, 13X, 140, 141, 142, 143, 150, 154, 156, 160, 162, 163, 165, 166, 168, 170, 176, 17X, 180, 183, 187, 190, 193, 194, 195, 197, 198, 1X2, 1X6, 1X8, 1X9, 1E4, 1E9, 200, 203, 204, 208, 214, 216, 220, 223, 228, 22E, 230, 232, 233, 239, 240, 244, 246, 253, 259, 260, 264, 265, 26X, 276, 278, 280, 282, 284, 286, 293, 299, 2X0, 2X8, 2E0, 300, 301, 304, 306, 30X, 310, 314, 31E, 320, 323, 330, 338, 340, 341, 345, 346, 347, 349, 352, 360, 366, 367, 368, 369, 36X, 372, 374, 384, 390, 394, 395, 396, 3X3, 3X8, 3E3, 3E6, 400, 401, 403, 406, 408, 409, 414, 417, 428, 430, 440, 445, 446, 454, 45X, 460, 464, 466, 469, 473, 475, 476, 480, 487, 488, 490, 4X6, 4X7, 4E6, 500, 508, 509, 50X, 518, 519, 530, 534, 537, 539, 540, 541, 543, 544, 548, 549, 550, 566, 576, 57E, 580, 583, 594, 5X0, 5E3, 600, 602, 608, 609, 610, 618, 620, 628, 63X, 640, 646, 660, 669, 671, 674, 680, 682, 685, 689, 68X, 690, 692, 696, 699, 6X4, 6E3, 700, 710, 712, 714, 716, 718, 724, 728, 739, 748, 753, 760, 768, 76X, 770, 773, 781, 786, 794, 7X6, 7E0, 7E1, 800, 801, 802, 806, 810, 814, 816, 828, 832, 839, 853, 854, 860, 86E, 875, 880, 88X, 890, 891, 8X8, 8E1, 8E8, 8EE, 900, 901, 903, 908, 910, 916, 926, 92X, 930, 940, 947, 952, 954, 959, 960, 969, 977, 990, 992, 9X1, 9E0, X00, X03, X13, X14, X16, X17, X18, X19, X23, X34, X36, X60, X68, X72, X76, X79, X80, X82, X83, X86, X88, X8E, X94, X96, XX0, E10, E27, E30, E3X, E40, E43, E46, E55, E68, E69, E80, E99, EX6, 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 the Dirichlet characters mod 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 is a divisor of 20, then k−1 is prime for all divisors k>2 of n.

If and only if n+1 is a divisor of 20, then $\tbinom{n}{k}=\tfrac{n!}{k!(n-k)!}$ is squarefree for all 0 ≤ kn, i.e. all numbers in the nth row of the Pascal's triangle are squarefree (the topmost row (i.e. the row which contains only one 1) of the Pascal's triangle is the 0th row, not the 1st row). (Note that all such n are primes or 1 or 0, and 20 is the largest number m such that if n+1 is a divisor of m, then n is prime or 1 or 0, besides, if and only if m is a divisor of 20, then m satisfies this condition)

If we only have the numbers 1 to 20 (including 1 and 20), then only the primes dividing 10 (i.e. primes ≤3) can be squared, since 5^2 = 21 > 20, and for the numbers such that the reciprocal of n terminates, they can only have at most 2 digits (which is the case of 8, 9, 14, 16 and 20), since the numbers with terminate reciprocal with >2 digits, they must be divisible by either 2^5 = 28 or 3^3 = 23, but both are >20. (these (prime power) numbers are >20: (prime > 3)^(>1), (odd prime)^(>2), 2^(>4))

If xy≤20, then at least one of x and y is a divisor of 10 (this is not true for xy=21, 5×5=21, but 5 is not a divisor of 10).

googol (mod n) = googolplex (mod n) for all 1 ≤ n ≤ 20 (but not for n = 21).

For all numbers n ≤ 100 (but not for n = 101, and not for n = smallest prime > 100 (i.e. 105)), there is k ≤ 6 such that nk−1 or nk+1 (or both) is prime. (note that for n = 101, k = 7, 8 and 9 also not satisfy this condition, the smallest k satisfying this condition for n = 101 is X)

20 is the GCD of all Fermat-Wilson quotients. (thus, Fermat-Wilson quotients are never primes)

100 is the smallest number whose nth power can be written as the sum of (≥2 and <n) positive nth powers for some n (n=5, the formula is 100^5 = 23^5 + 70^5 + 92^5 + E1^5, only four 5th powers).

100 is the smallest number whose 5th power can be written as the sum of 4 positive 5th powers, besides, the smallest number whose 5th power can be written as the sum of 5 positive 5th powers is 60, which is exactly half of 100, besides, the smallest number whose 5th power can be written as the sum of 6 positive 5th powers is 10, which is exactly square root of 100

100 is the maximum number of steps of numbers < 4X7 (the smallest number that reach a number > 5414 (the record which gotten by the start value 23), namely 100E74) in Collatz sequence (for the numbers 461, 466, 467 and 477).

The exponents on the right hand side of $(1-x)(1-x^2)(1-x^3) \cdots = 1 - x - x^2 + x^5 + x^7 - x^{10} - x^{13} + x^{1\mathcal{X}} + x^{22} - \cdots.$ are exactly the numbers n such that kn+1 is square for k=20. (note that 10 is one of such numbers)

All negative-Pell solvable numbers (i.e. numbers n such that x2ny2 = −1 is solvable) end with negative-Pell solvable digits (i.e. end with 1, 2, 5 or X).

By Benford's law, the probability for the leading digit d (d ∈ {1, 2, 3, 4, 5, 6, 7, 8, 9, X, E}) occurs (for some sequences, e.g. powers of 2 (1, 2, 4, 8, 14, 28, 54, X8, 194, 368, 714, 1228, 2454, ...) and Fibonacci sequence (0, 1, 1, 2, 3, 5, 8, 11, 19, 2X, 47, 75, 100, ...)) are:

d probability d probability
1 34.2% 7 7.9%
2 1E.6% 8 6.X%
3 14.8% 9 6.1%
4 10.E% X 5.6%
5 X.7% E 5.1%
6 8.E%

(Note: the percentage in the list are also in dozenal, i.e. 20% means 0.2 or $\frac{20}{100}=\frac{1}{6}$, 36% means 0.36 or $\frac{36}{100}=\frac{7}{20}$, 58.7% means 0.587 or $\frac{587}{1000}$)

Star numbers are exactly the numbers obtained as the concatenation of a triangular number followed by 1 (the triangular numbers are 0, 1, 3, 6, X, 13, 19, 24, 30, 39, 47, 56, 66, ..., and the star numbers are 1, 11, 31, 61, X1, 131, 191, 241, 301, 391, 471, 561, 661, ...), thus, all star numbers end with 1. (The star numbers are exactly the centered 10-gonal numbers)

The Hilbert numbers are the numbers end with 1, 5 or 9. (i.e. = 1 mod 4)

The Lagado numbers are the numbers end with 1, 4, 7 or X. (i.e. = 1 mod 3)

The smallest two 4-digit palindromic numbers (1001 and 1111) are both Ziesel numbers, they are also the smallest two palindromic numbers which cannot be prime when read in any base, and the smallest 4-digit palindromic number (1001) is exactly the smallest absolute Euler pseudoprime and the smallest number expressible as the sum of two cubes in two different ways, i.e. 1001 = 1 + 1000 (=13 + 103) = 509 + 6E4 (=93 + X3), besides, 1111 is also the only known n such that both n and n+1 are Quasi-Carmichael numbers (squarefree composites n with the property that for every prime factor p of n, p+b divides n+b positively with b being any integer besides 0), besides, the smallest number needing >2 steps in the "Reverse and Add!" problem is 6E, and the palindrome at which it stops is exactly 1111 (6E + E6 = 165, 165 + 561 = 706, 706 + 607 = 1111), besides, 1111 is the smallest composite repunit number, besides, the square of 1111 is 1234321, another palindrome.

The largest 4-digit number (EEEE) is a member of a betrothed number pair (its betrothed number is 5600 (also a 4-digit number, note that 5600 is E-smooth), and if we calculate EEEE/gcd(EEEE, 5600), we get the 4-digit repunit (1111)).

“the smallest (Fermat) pseudoprime to both base 2 and base 3 that is not Carmichael number” (1691) and “conjectured largest panconsummate number” (1961) are both strobogrammatic numbers (the same upside down). (these two numbers are both composite, however, the strobogrammatic numbers 160091 and 190061 are primes) (of course, the Zeisel numbers 1001 and 1111 are also strobogrammatic numbers)

All prime numbers end with prime digits or 1 (i.e. end with 1, 2, 3, 5, 7 or E), more generally, except for 2 and 3, all prime numbers end with 1, 5, 7 or E (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 relatively low, but the density of primes end with 5, 7 and E are nearly equal. (since all prime squares except 4 and 9 end with 1, no prime squares end with 5, 7 or E)

Except (3, 5), all twin primes end with (5, 7) or (E, 1), and the density of these two types of twin primes are nearly equal.

The sum of any pair of twin primes (other than (3, 5)) ends with 0.

If n ≥ 3 and n is not divisible by E, 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, E, 11, 111, 131, 141, 171, 181, 1E1, 535, 545, 565, 575, 585, 5E5, 727, 737, 747, 767, 797, E1E, E2E, E6E.

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

Except for 3, all Fermat primes end with 5. (In fact, there are only 5 known Fermat primes (3, 5, 15, 195 and 31E15) and it is conjectured that there are no more Fermat primes, interestingly, all digits of all known Fermat primes are odd)

Except for 3, all Mersenne primes end with 7. (Besides, all Mersenne primes except 3 and 7 end with one of the only two 2-digit Mersenne primes (27 and X7))

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

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

A prime p is Gaussian prime (prime in the ring $Z[i]$, where $i=\sqrt{-1}$) if and only if p ends with 7 or E (or p=3). (i.e. p = 3 mod 4)

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

A prime p can be written as x2 + y2 if and only if p ends with 1 or 5 (or p=2). (i.e. p = 1 or 2 mod 4)

A prime p can be written as x2 + 3y2 if and only if p ends with 1 or 7 (or p=3). (i.e. p = 0 or 1 mod 3)

A number n can be written as x2 + y2 if and only if all prime factors of the squarefree part of n end with 1 or 5 (or 2).

A number n can be written as x2 + 3y2 if and only if all prime factors of the squarefree part of n end with 1 or 7 (or 3).

All numbers ≤ 20 coprime to 10 are either primes or 1 (unit). (this is not true for 21, 21 is the smallest composite coprime to 10)

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. $\left(\frac{10}{p}\right)=-1$, where $\left(\frac{m}{n}\right)$ 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 0.076E45076E45..., 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 186X35 (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, X7, E5, E7, 105, 107, 117, 125, 145, 167, 195, 1X5, 1E5, 1E7, 205, 225, 255, 267, 277, 285, 295, 315, 325, 365, 377, 397, 3X5, 3E5, 3E7, 415, 427, 435, 437, 447, 455, 465, 497, 4X5, 517, 527, 535, 545, 557, 565, 575, 585, 5E5, 615, 655, 675, 687, 695, 6X7, 705, 735, 737, 745, 767, 775, 785, 797, 817, 825, 835, 855, 865, 8E5, 8E7, 907, 927, 955, 965, 995, 9X7, 9E5, X07, X17, X35, X37, X45, X77, X87, X95, XE7, E25, E37, E45, E95, E97, EX5, EE5, EE7. (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 = $\Phi_4(10)$, 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 = E1 = $\Phi_6(10)$, 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 $\frac{a}{p}=0.\overline{a_1a_2a_3...a_n}$, then ai + ai+n/2 = E for every 1 ≤ in/2. e.g. 1/5 = 0.249724972497..., and 24 + 97 = EE, and 1/7 = 0.186X35186X35..., and 186 + X35 = EEE, all primes (other than 2 and 3) ≤ 37 except E, 1E and 31 have even period length, thus they can use Midy theorem to get an E-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 E, 11, 111, E0E1, EE01, 11111, 24727225, E0E0E0E0E1, E00E00EE0EE1, 100EEEXEXEE000101, 1111111111111111111, EEEE0000EEEE0000EEEE0001, 100EEEXEE0000EEEXEE000101, 10EEEXXXE011110EXXXE00011, EEEEEEEE00000000EEEEEEEE00000001, EEE000000EEE000000EEEEEE000EEEEEE001, and the period length of their reciprocals are 1, 2, 3, X, 10, 5, 18, 1X, 19, 50, 17, 48, 70, 5X, 68, 53.

If p is a safe prime other than 5, 7 and E, then the period length of 1/p is (p-1)/2. (this is not true for all primes ends with E (other than E itself), the first counterexample is p = 2EE, 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. (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, ..., E 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 E, and 10 is quadratic residue for all primes end with E. (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)

pperiod length of 1/ppperiod length of 1/ppperiod length of 1/ppperiod length of 1/ppperiod length of 1/ppperiod length of 1/ppperiod length of 1/ppperiod length of 1/ppperiod length of 1/ppperiod length of 1/ppperiod length of 1/ppperiod length of 1/p
20111326726641E20E5913X767766927926E1E56E1107110612E512E414E114E16X716X6
30117116271274216359E2XE76E395955954E2157011151114130176014E514E416E5188
5411E6E2772764274265E115977113295E48EE25E2411251124131750614EE85E16E716E6
7612512427E13E4311095E55E4775774965964E2E575112E6751337512150E8651705398
E112E752852844354345E76677E39E971172E3111611351134133E77E15171XX1711493
11213176291834374365EE2EE78578498732XE37E36114E68513451344152143617151714
151413E7E2952944474466111637913X6995994E45E44115111513511661525152417271726
176141202X11504554546156147977969X79X6E6116116511641365136415471E217351734
1EE1451442XE15545715X6172067X11389XE4E5E673X21167421367136615618917451744
254147562E12645E22E61E30E7EE3EE9E19EE6E59511851184136E795156E9717471746
2726157122EE374654646372128011409E59E4E71596118E6X5137710615771576175132X
3191671663019046E763E31E80E4059EE4EEE912E31197472138E7X5157E89E17551754
353416E9530710247113647216817816X07X06E95E9411X16E13913E31585158417571756
37361719630E16548124655654825824X0E505E97E9611X511X41395139415872X176E995
3E1E17583153144854466117682E415X1156EX5EX411X711X613X17E015912E617814E0
454417E9E32117048E245665138835834X17X16EE5EE411XE6E513X753615XE8E517851784
4E25181X032532449749666E33584184X27156EE7EE611E711E613E113E15EE8EE178E9X5
511318EX532710X4X54X467567485114XX35X3410051004120170013E513E4160116017971796
575619519432E1754E19X687686855854X37X36101173120E705140514041615161417X19E0
5E2E19EXE33E17E4EE25E68E34585E42EX3E51E1017101612117061407326162191017X517X4
61301X51X434746507182695694865864X411881021610121E70E142514241625162417EE9EE
67221X74634E2E51126669E34E8672X2X45X44102710261231123142E81516352741807682
6E351E1E635711X5175166X76X6871152X4E5251041620123E71E1431286164727618151814
7581E51E435E18E51E456E16E881440X5E52E10471046124512441437143616551654181EX0E
81141E71E636536452752670136088E445X6E535104E6251255114143E81E165761X18251824
8584205204375345312767057048X598X77X761051313125749X14451444165E92E1831509
87862178637737653553470E3658X72E6X87X86106116125E72E145714561667622183EX1E
8E4521E10E3911X654154711718XE455X91283106E63512617301461381671936184EX25
91462216639739654554471E36E8E58E4X95X94107E63E126E73514651464167720X1861269
95942252243X53X45575567213708E78E6X9E54E10871086127E73E1467562167E93E18651864
X7X6237923XE1E556556472724X901230XX7376109E64E128174014714191681140186EX35
XE552411203E53E4575574735734905198XXE55510E1329129594147514741685168418751874
E5E424E1253E73E6577116737736907906XE7XE610E710E612971296147E83E168E9451877146
E7E62517340110058558474574490E465XEE55E10EE65E12X175148E84516971696189EX4E
10510425525440E2055871XX7479X91E46EE111X2110122012X52E814951494169E94E18X1210
10710625E12E41541458E2X57511X3921236E152281105110412X712X6149E84E16X195018XEX55
period length primes period length primes
1 E 11 1E0411, 69X3901
2 11 12 157, 7687
3 111 13 51, 471, 57E1
4 5, 25 14 15, 81, 106X95
5 11111 15 X9X9XE, 126180EE0EE
6 7, 17 16 E61, 1061
7 46E, 2X3E 17 1111111111111111111
8 75, 175 18 24727225
9 31, 3X891 19 E00E00EE0EE1
X E0E1 1X E0E0E0E0E1
E 1E, 754E2E41 1E 3E, 78935EX441, 523074X3XXE
10 EE01 20 141, 8E5281

The period level of a prime p ≥ 5 is (p−1)/(period length of 1/p), e.g., $\frac{1}{17}$ has period level 3, thus the numbers $\frac{a}{17}$ with integer 1 ≤ a ≤ 16 from 3 different cycles: 076E45 (for a = 1, 7, 8, E, 10, 16), 131X8X (for a = 2, 3, 5, 12, 14, 15) and 263958 (for a = 4, 6, 9, X, 11, 13). Besides, $\frac{1}{15}$ has period level 1, thus this number is a cyclic number and 15 is a full-reptend prime, and all of the numbers $\frac{a}{15}$ with integer 1 ≤ a ≤ 14 from the cycle 08579214E36429X7.

There are only 9 repunit primes below R1000: R2, R3, R5, R17, R81, R91, R225, R255 and R4X5 (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, 4X5, 5777, 879E, 198E1, 23175, 311407, ..., note that 879E is the smallest (and the only known) such number ends with E)

By Fermat's little theorem, if p is a prime other than 2, 3 and E, 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, E1, 101, 187, 275, 4X7, 777, 781, E55, they are exactly the Fermat pseudoprimes for base 10 (composite numbers c such that 10c-1 = 1 mod c) which are not divisible by E, 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+X+E+10
0+ 110101010110100110010010101111111111110
10+ 1110010101010010111100100110101001011111111111011101100
20+ 11011111010001001010110101010111000111111111110101110101101100
30+ 10010011001011010100111001100101010011111111111101010011101010001111100
40+ 1011110111011101011101101100101100010111011111111001001001010101010111010
50+ 10010101101011010010010001111111111111110110010110111011101010110101100111100
60+ 10101101100100101101110100101011101111111110101010111010010000111001010011100110010
70+ 11010011010010101011111111111001000110100100111101010101101000111101011011000
80+ 111011101111010111111111110011011101100011011101010011111001011010110010101001111000
90+ 10101111111011101111111010010010100100101010011001011010100110101100101011010110001110100
X0+ 11111111110110010101011100101010110111000011001001100001100001010010111101111011111111111111110
E0+ 1001110010101010001011100101011111010110101111011010100011110110011101111111111111111111111100
 n 1 5 7 E 11 15 17 1E 21 25 27 2E smallest k such that k×n is a repunit 1 275 1X537 123456789E 1 92X79E43715865 8327 69E63848E 634X159788253X72E1 55 509867481E793XX5X1243628E317 45X3976X7E the length of the repunit k×n 1 4 6 E 2 14 6 E 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, 19E, 275 and 46E, and only 19E and 46E 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 = 46E is 2X3E, which makes k×n = R7 = 1111111, a repunit semiprime, and this k for n = 19E is a X8-digit prime number, with k×n = RXE, another repunit semiprime)

For every prime p except E, 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, E1, 101, 187, 1E0, 275, 444, 4X7, 777, 781, E55, 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 E. (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 E) 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 E, 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 E (of course, if n is divisible by E, then such number does not exist).

Smallest multiple of n with digit sum n are:

1, 2, 3, 4, 5, 6, 7, 8, 9, X, E, 1E0, 20E, 22X, 249, 268, 287, 2X6, 45X, 488, 4E6, 1EX, 8E4, 3EX0, 3EE, 23EX, 1899, XX8, 2E79, 4E96, 1EX9, 4XX8, 2EE9, 3XEX, 799X, 5EE90, ..., such number is conjectured to exist for all n.

45 is the smallest prime that produces prime reciprocal magic square, i.e. write the recurring digits of 1/45 (=0.Template:Overline, which has period 44) to 44/45, we get a 44×44 prime reciprocal magic square (its magic number is 1EX), 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)/E = (309×10n−1)/E and it can be factored to ((19×10n/2−1)/E) × (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{1E78}, it has E7X 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 EX, since all numbers obtained as the concatenation of a power of EX (>1) followed by a 1 are divisible by either E or 11 and thus composite. However, the smallest prime obtained as the concatenation of a power of 58 (>1) followed by a 1 is 10×582781E5+1, it has 459655 digits.

All numbers of the form 1{5}1 are composite (proof: 1{5n}1 = (14×10n+1−41)/E and it can be factored to (4×10(n+1)/2−7) × ((4×10(n+1)/2−7)/E) for odd n and divisible by 11 for even n).

The emirps below 1000 are 15, 51, 57, 5E, 75, E5, 107, 117, 11E, 12E, 13E, 145, 157, 16E, 17E, 195, 19E, 1X7, 1E5, 507, 51E, 541, 577, 587, 591, 59E, 5E1, 5EE, 701, 705, 711, 751, 76E, 775, 785, 7X1, 7EE, E11, E15, E21, E31, E61, E67, E71, E91, E95, EE5, EE7.

The non-repdigit permutable primes below 1010100 are 15, 57, 5E, 117, 11E, 5EEE (the smallest representative prime of the permutation set).

The non-repdigit circular primes below 1010100 are 15, 57, 5E, 117, 11E, 175, 1E7, 157E, 555E, 115E77 (the smallest representative prime of the cycle).

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

The only known Smarandache–Wellin primes are 2 and 2357E11.

There are exactly 15 minimal primes, and they are 2, 3, 5, 7, E, 11, 61, 81, 91, 401, X41, 4441, X0X1, XXXX1, 44XXX1, XXX0001, XX000001.

The smallest weakly prime is 6E8XE77.

The largest left-truncatable prime is 28-digit 471X34X164259EX16E324XE8X32E7817, and the largest right-truncatable prime is X-digit 375EE5E515.

The only two base 10 Wieferich primes up to 1010 are 1685 and 5E685, 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 = 010001000100(2) = 444(14) is a repdigit in base 14, and 2046 = 110110110110(2) = 6666(8) is also a repdigit in base 8, see Wieferich prime#Binary periodicity of p − 1)

For the numbers between 5X0 (the smallest number divisible by all of the numbers 1 to 8) and 630 (the square of 26 = 5#) and end with 1, 5, 7 or E (the digits coprime to 10), all numbers whose dozen digit is odd are primes, and all numbers whose dozen digit is even are composites.

For all odd composites c up to 1000, there exists integer a such that GCD(a, c) = 1 and a(c−1)/2 is congruent to neither 1 nor −1 mod c (i.e. c is not an Euler pseudoprime base a), however, this is not true for c = 1001, 1001 is Euler pseudoprime to all bases coprime to itself, i.e. 1001 is an absolute Euler pseudoprime.

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, X, E, 25, X5, 577, 668, X83, 14765, 938X4, 369862, X2394X, ..., the largest of which is 43-digit 15079346X6E3E14EE56E395898E96629X8E01515344E4E0714E. (see Template:Oeis)

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, 2X, 88, X8, 118, 56, 51, 22}, {18, 55, 42}, {68, 84}, or one of the only two fixed points other than 1: 25 and X5.

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, 52E, X20, 700, 247, 2X7, 947, 7X8, 10X7, 940, 561, 246, 200}. (for the example of the famous Hardy–Ramanujan number 1001 = 93 + X3, we know that this sequence with initial term 9X is 9X, 1001, 2, 8, 368, 52E, X20, 700, 247, 2X7, 947, 7X8, 10X7, 940, 561, 246, 200, 8, 368, 52E, X20, 700, 247, 2X7, 947, 7X8, 10X7, 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}, {X}, {E} 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1
2 {1}, {5, 21}, {8, 54, 35, 2X, 88, X8, 118, 56, 51, 22}, {18, 55, 42}, {25}, {68, 84}, {X5} 1, 2, X, 3, 1, 2, 1
3 {1}, {8, 368, 52E, X20, 700, 247, 2X7, 947, 7X8, 10X7, 940, 561, 246, 200}, {577}, {668}, {6E5, E74, 100X}, {X83}, {11XX} 1, 12, 1, 1, 3, 1, 1
4 {1}, {X6X, 103X8, 8256, 35X9, 9EXE, 22643, E69, 1102X, 596X, X842, 8394, 6442, 1080, 2455}, {206X, 6668, 4754}, {3X2E, 12396, 472E, X02X, E700, 9X42, 98X9, 13902} 1, 12, 3, 8

The harshad numbers up to 200 are 1, 2, 3, 4, 5, 6, 7, 8, 9, X, E, 10, 1X, 20, 29, 30, 38, 40, 47, 50, 56, 60, 65, 70, 74, 80, 83, 90, 92, X0, X1, E0, 100, 10X, 110, 115, 119, 120, 122, 128, 130, 134, 137, 146, 150, 153, 155, 164, 172, 173, 182, 191, 1X0, 1E0, 1EX, 200, although the sequence of factorials begins with harshad numbers, not all factorials are harshad numbers, after 7! (=2E00, with digit sum 11 but 11 does not divide 7!), 8X4! is the next that is not (8X4! has digit sum 8275 = E×8E7, thus not divide 8X4!). 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, E, 56, 66, EE, 444, 778, EEE, 12XX, 1640, 2046, 2929, 3333, 4973, 5E60, 6060, 7249, 8889, 9293, 9E76, X580, X912, EEEE.

The Kaprekar's routine of any four-digit number which is not repdigit converges to either the cycle {3EE8, 8284, 6376} or the cycle {4198, 8374, 5287, 6196, 7EE4, 7375}, and the Kaprekar map of any three-digit number which is not repdigit converges to the fixed point 5E6, and the Kaprekar map of any two-digit number which is not repdigit converges to the cycle {0E, X1, 83, 47, 29, 65}.

n Cycles for Kaprekar's routine for n-digit numbers Length of these cycles Number of these cycles
1 {0} 1 1
2 {00}, {0E, X1, 83, 47, 29, 65} 1, 6 2
3 {000}, {5E6} 1, 1 2
4 {0000}, {3EE8, 8284, 6376}, {4198, 8374, 5287, 6196, 7EE4, 7375} 1, 3, 6 3
5 {00000}, {64E66, 6EEE5}, {83E74} 1, 2, 1 3
6 {000000}, {420X98, X73742, 842874, 642876, 62EE86, 951963, 860X54, X40X72, X82832, 864654}, {65EE56} 1, X, 1 3
7 {0000000}, {841E974, X53E762, 971E943, X64E652, 960EX53, E73E741, X82E832, 984E633, 863E754}, {962E853} 1, 9, 1 3
8 {00000000}, {4210XX98, X9737422, 87428744, 64328876, 652EE866, 961EE953, X8428732, 86528654, 6410XX76, X92EE822, 9980X323, X7646542, 8320X984, X7537642, 8430X874, X5428762, 8630X854, X540X762, X830X832, X8546632, 8520X964, X740X742, X8328832, 86546654}, {873EE744}, {X850X632} 1, 20, 1, 1 4

The self numbers up to 600 are 1, 3, 5, 7, 9, E, 20, 31, 42, 53, 64, 75, 86, 97, X8, E9, 10X, 110, 121, 132, 143, 154, 165, 176, 187, 198, 1X9, 1EX, 20E, 211, 222, 233, 244, 255, 266, 277, 288, 299, 2XX, 2EE, 310, 312, 323, 334, 345, 356, 367, 378, 389, 39X, 3XE, 400, 411, 413, 424, 435, 446, 457, 468, 479, 48X, 49E, 4E0, 501, 512, 514, 525, 536, 547, 558, 569, 57X, 58E, 5X0, 5E1.

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, 1E, 22, 2X, 31, 33, 44, 49, 55, 62, 66, 77, 88, 93, 99, XX, EE, 125, 215, 24X, 405, 42X, 654, 80X, 8X3, X59.

There are totally 71822 polydivisible numbers, the largest of which is 24-digit 606890346850EX6800E036206464. However, there are no E-digit polydivisible numbers contain the digits 1 to E exactly once each. (hence there are also no 10-digit polydivisible numbers using all the digits 0 to E exactly once, since if a number with digits abcdefghijkl is a 10-digit polydivisible number using all the digits 0 to E exactly once, then {a, b, c, d, e, f, g, h, i, j, k, l} = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, X, E}, and then abcdefghijkl is divisible by 10, thus we have l = 0 (by divisibility rule of 10), and {a, b, c, d, e, f, g, h, i, j, k} = {1, 2, 3, 4, 5, 6, 7, 8, 9, X, E}, thus a number with digits abcdefghijk is an E-digit polydivisible numbers using all the digits 1 to E exactly once). (proof: if a number with digits abcdefghijk is an E-digit polydivisible numbers using all the digits 1 to E exactly once, then {a, b, c, d, e, f, g, h, i, j, k} = {1, 2, 3, 4, 5, 6, 7, 8, 9, X, E}, and we have:

f = 6 (since abcdef is divisible by 6) (by divisibility rule of 6)

{d, h} = {4, 8} (since abcd is divisible by 4 and abcdefgh is divisible by 8 (thus by 4)) (by divisibility rule of 4)

{c, i} = {3, 9} (since abc is divisible by 3 and abcdefghi is divisible by 9 (thus by 3)) (by divisibility rule of 3)

{b, j} = {2, X} (since ab is divisible by 2 and abcdefghij is divisible by X (thus by 2)) (by divisibility rule of 2)

thus, we have {a, e, g, k} = {1, 5, 7, E}

Since abcdefgh is divisible by 8, thus gh is divisible by 8 (by divisibility rule of 8), and since {a, e, g, k} = {1, 5, 7, E}, thus g is odd, and h must be 4 (if h = 8 and g is odd, then gh is not divisible by 8), and since abcdefghi is divisible by 9, thus hi is divisible by 9 (by divisibility rule of 9), however, h = 4 and i is either 3 or 9, but neither 43 nor 49 is divisible by 9.

If we do not require the number formed by its first 8 digits divisible by 8, then there are 2 solutions: 1X98265E347 and 7298X65E341 (neither satisfies that the number formed by its first 8 digits is divisible by 4).

If we do not require the number formed by its first 9 digits divisible by 9, then there are 4 solutions: 1X38E694725, 7X981634E25, 7X98E654321, and EX987634125 (only 7X98E654321 satisfies that the number formed by its first 9 digits is divisible by 3).

The candidate Lychrel numbers up to 1000 are 179, 1E9, 278, 2E8, 377, 3E7, 476, 4E6, 575, 5E5, 674, 6E4, 773, 7E3, 872, 8E2, 971, 9E1, X2E, X3E, X5E, X70, XXE, XE0, E2X, E3X, E5X, EXX. The only suspected Lychrel seed numbers up to 1000 are 179, 1E9, X3E and X5E. However, it is unknown whether any Lychrel number exists. (Lychrel numbers only known to exist in these bases: E, 15, 18, 22 and all powers of 2)

Most numbers that end with 2 are nontotient (in fact, all nontotients < 58 except 2X end with 2), except 2 itself, the first counterexample is 92, which equals φ(X1) = φ(E2) and φ(182) = φ(2×E2), next counterexample is 362, which equals φ(381) = φ(1E2) and φ(742) = φ(2×1E2), 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 E.

There is a known generalized Cullen prime for all bases b ≤ 10 (but not for b = 11). (no matter whether you require nb−1 or not)

There is a known generalized Woodall prime for all bases b ≤ 100 (but not for b = 101). (no matter whether you require nb−1 or not)

There is a known generalized Carol prime for all even bases b ≤ (100×2 + 100÷2) (=260) (but not for b = next even number (262)).

There is a known generalized Kynea prime for all even bases b ≤ 200 (but not for b = next even number (202)).

The generalized minimal primes problem has at most one unsolved family for all bases b ≤ 20 (but not for b = 21). (there is one unsolved family for b = 15, 17 and 19, and there are no unsolved families for all other b ≤ 20, but for b = 21, there are 10 unsolved families)

There are no n≤100 which is nontotient, noncototient, and untouchable. (the smallest such n is indeed the smallest even number > 100, i.e. 102)

By sieve of Eratosthenes, we can cross out every composites ≤ 20 by sieve the primes dividing 10 (i.e. the primes ≤3) (i.e. the primes 2 and 3). (however, we cannot cross out the composite 21 by sieve the primes dividing 10 (i.e. the primes ≤3) (i.e. the primes 2 and 3))

By sieve of Eratosthenes, we can cross out every composites ≤ 200 by sieve to the prime 10+1 (=11). (however, we cannot cross out the composite 201 by sieve to the prime 10+1 (=11))

For all odd composites c ≤ 1000, there exists integer b coprime to c such that b(c−1)/2 ≠ ±1 (mod c) (i.e. c is not Euler pseudoprime base b). (this is not true for the composite c = 1001, 1001 is the smallest absolute Euler pseudoprime)

The Wagstaff numbers $\frac{2^p+1}{3}$ is prime for all odd primes p ≤ 20 (but not for p = next odd prime (25)).

There is a known odd generalized Wieferich prime for all prime bases p ≤ 20 (but not for p = next prime (25)).

The smallest Perrin pseudoprime is a near-repunit 111101 (note that this number is square: 3752), this number only contains five 1's and one 0 (no any digit >1), and this number plus 10 is the repunit with length 6, i.e. 111111, and this number is exactly the final 6 (half of 10) digits of 123456789XE×E (a 10-digit number).

The smallest amicable number pair is (164,1E8), and the next number of them in the home prime problem are the same: 225E. (both of these two numbers get to the 9E-digit prime 51712E1X75561E754940E168440995028213825E76175409310343166811361353638X487E880858E7E8E024666872X98X3096E504E110578059E65 after 58 steps)

The “beast number” (666), when add a digit “1” before it and add another digit “1” after it, it become a palindromic square 16661, it is the smallest palindromic square whose square root is not palindrome (12E), it is also the smallest palindromic square depending on base.

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=X, B=E (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 minor third, x and x+4 are major third, x and x+7 are perfect fifth (thus, we can use 7x for x = 0 to E to get the five degree cycle), etc. (since an octave is 10 semitones, a minor third is 3 semitones, a major third is 4 semitones, and a perfect fifth 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 $\frac{\pi}{6}$ 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+E}, and the x minor 7th chord (xm7) is {x, x+3, x+7, x+X}, and the x dominant 7th chord (x7) is {x, x+4, x+7, x+X}, 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) (also, x and x+6 are tritone, which is not harmonic). Besides, x major scale uses the notes {x, x+2, x+4, x+5, x+7, x+9, x+E}, and x minor scale uses the notes {x, x+2, x+3, x+5, x+7, x+8, x+X}. 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 = X9.8E5809 is very close to 27 = X8, since 217 = 2134X8 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 $\sqrt[10]{2}$ (=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) = X/9 (=1.14), and f(5)/f(4) = 14/13 (this number is very close to $\sqrt[10]{2}$), and thus we have that f(5)/f(0) = (9/8) × (X/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 14X28, 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)

The probability for rolling a 6 on a dice is 0.2 or 20%, and the probability for rolling at least one 6 on a dice in 3 rolls is 0.508 (less than one half or 60%), and the probability for rolling at least one 6 on a dice in 4 rolls is 0.6268 (more than one half or 60%), and the probability for rolling a "double 6" on two dices is 0.04 or 4%, and the probability for rolling at least one "double 6" on two dices in 20 rolls is 0.5X9190... (less than one half or 60%), and the probability for rolling at least one "double 6" on two dices in 21 rolls is 0.609685... (more than one half or 60%).

0, 1, 2, 3, 4, 5, 6, 7, 8, 9, X, E

appear in the repeating digits of 1/5 (exactly the even digits ≤ 5 (except 0 (the smallest digit)) and the odd digits ≥ 6 (except E (the largest digit)))

appear in the repeating digits of 1/7 (exactly the odd digits ≤ 5 and the even digits ≥ 6)

appear in the repeating digits of 1/11 (exactly the smallest digit (0) and the largest digit (E))

(note that the number of digits ≤5 is equal to the number of digits ≥6 (both are 6, which is equal to half of the base we used in this wiki (10)), and all digits are either ≤5 or ≥6, but not both)

(note that all of 5, 7, and 11 are primes, and they are the only three primes ≤ 10+1 (=11) and divides neither 10 nor 10−1 (=E))

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