1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + X + E + 10 + ... = −0.1 (which equals the negative value of the reciprocal of 10).

Both 10±1 are primes, both (10±2)/2 are primes, both (10±3)/3 are primes, both 10±4 are powers of 2 (also common bases for computer, base 8 and base 14), both 10±5 (also 10±7 and 10±E) are primes (or units), and both 10±6 are regular numbers (i.e. 3-smooth numbers) (i.e. numbers n such that 1/n terminates in dozenal) (also twice a power of 3).

Both 100±1 are semiprimes, both (100±2)/2 are primes, (100−d)/d is prime for all divisors d of 10 except 1 and 4 (note that this number for d = 1 and 4 cannot be primes, because of algebra factors: 100−1 = (10−1) × (10+1), and 100−4 = (10−2) × (10+2), and can only be semiprimes), both 100±5 are primes, both 100±7 are primes, both (100±8)/8 are primes, both 100±E are semiprimes, both (100±10)/10 are primes (Note: (100−3)/3 and (100−6)/6 are primes, but (100+3)/3 and (100+6)/6 are prime squares, also, (100+9)/9 is prime, but (100−9)/9 also has algebra factors (100−9 = (10−3) × (10+3)) and cannot be prime, also, (100−X)/2 is prime, but (100+X)/2 is divisible by E (since its digit sum is E) and cannot be prime)

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 In other words,

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 , 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 , the number of partitions of n (p(n)): , or more formally,

Also the sum of divisors of n (σ(n)):

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 is the k^th generalized pentagonal number. Since for all , 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)

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

R_n² (where R_n 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 n^k+2 ends with the same digit as n^k, thus, if i≥2, j≥2 and i and j have the same parity, then nⁱ and n^j 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 x²−x=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.

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

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

More generally, for every n≥2, the period of the final n digits of powers of a number must be a divisor of 10ⁿ⁻¹ (= λ(10ⁿ)).

The period of the digital roots of powers of a number must be a divisor of X (= λ(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, F_n is the n-th Fibonacci number, and L_n is the n-th Lucas number.

(Note that F_2X begins with L_1X, and F_2E begins with L_1E)

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.

The recurring dozenal of 1/F(n) terminates if and only if n is a divisor of 10. (10 is exactly the base of dozenal)

Factors of the Fibonacci numbers:

F(10)   =         100 = 102

(F(10) is exactly the square of 10, and F(10) is also the largest Fibonacci number which is square)

F(10−1) = F(E)  =  75
F(10+1) = F(11) = 175

75 × 175 = 10001 = 104 + 1

(10±1 are both primes, and F(10±1) are also both primes)
(104 + 1 = 10001 is a semiprime, and 104 − 1 is a member of betrothed number pair, together with 5600)

F(10−2) = F(X)  =  47 =  5 × E
F(10+2) = F(12) = 275 = 25 × 11

5 × 25 = 101 = 102 + 1
E × 11 =  EE = 102 − 1

(10±2 are both semiprimes, and F(10±2) are also both semiprimes)
(102 ± 1 are both semiprimes, and 5, E, 25, and 11 are all primes)

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)

2^1XE = 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 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 and ) ends with 6.

If n ends with 2 and n/2 is prime (or 1), then the denominator of the Bernoulli number 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 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 is 6, but 82/2 = 41 = 7² is neither prime nor 1)

is very close to 1.5, since a near-value for is 15/10 (=N₄/P₄, where N_n is nth NSW number, and P_n is nth Pell number, N_n/P_n is very close to when n is large). Besides, is very close to 2.2X, since a near-value for is 22X/100 (= L₁₀/F₁₀, where L_n is nth Lucas number, and F_n is nth Fibonacci number, L_n/F_n is very close to 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 , where 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 over Q includes coefficients other than (i.e. the 89th cyclotomic polynomial, , is the first with coefficients other than ), besides, the repunit with length k (R_k) (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, if we add zero (0, the additive identity) between 8 and 9, we get 809, the number ≤1000 whose Collatz sequence is longest, besides, the squarefree part of n²−1 for n = 8 and 9 are 7 and 5, respectively (7 and 5 are the only primes ≤10 dividing neither 10 nor 10−1), 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×T_k+1, where T_k 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, (note that 8+4 (centered 8-gonal numbers and non-centered 4-gonal numbers) and 9+3 (centered 9-gonal numbers and non-centered 3-gonal numbers) are both 10), 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 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 (4×60⁴⁶⁰⁰⁸⁹−1) is 78E425-digit prime and 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 is terminating number (where is Euler's totient function) (i.e. is 3-smooth, or 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

(20 is double of 10)

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 is squarefree for all 0 ≤ k ≤ n, 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))

20 is the largest integer that is divisible by all natural numbers no larger than its square root.

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)

The exponents on the right hand side of are exactly the numbers n such that 20n+1 is square. (note that 10 is one of such numbers) (these numbers can be used to calculate , the number of partitions of n: ), and , the sum of divisors of n: using the same equation, but if the last term is , then we change it to n instead of 0 or 1)

20 is the Euler characteristic of a K3 surface.

20 is the order of the cyclic group equal to the stable 3-stem in homotopy groups of spheres: Template:Pi_n+3(Sⁿ) = Z/20Z for all n ≥ 5.

20 is the only number whose divisors — 1, 2, 3, 4, 6, 8, 10, 20 — are exactly those numbers n for which every invertible element of the commutative ring Z/nZ is a square root of 1. It follows that the multiplicative group of invertible elements (Z/20Z)^× = {±1, ±5, ±7, ±E} is isomorphic to the additive group (Z/2Z)³. This fact plays a role in monstrous moonshine.

The kissing number is only known in dimensions dividing 20 (1, 2, 3, 4, 6, 8, 10, 20) except 6 and 10

The densest packing is only known in dimensions dividing 20 (1, 2, 3, 4, 6, 8, 10, 20) but not the composite divisors of 10 (i.e. 4, 6, 10)

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).

20-cell is the only one of the six convex regular 4-polytopes which is not the four-dimensional analogue of one of the five regular Platonic solids (i.e. 20-cell is the only 4-dimension regular polytope which does not have a regular analogue in 3 dimensions).

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)

20 is the kissing number in 4-dimensional space: the maximum number of unit spheres that can all touch another unit sphere without overlapping. (The centers of 20 such spheres form the cells of a 20-cell), for 3-dimensional space, the kissing number is 10, the centers of 10 such spheres form the faces of a 10-hedron.

20-cell is the only convex regular 4-polytope which does not have a regular analogue in 3 dimensions. (but with a regular analogue in 2 dimensions: regular hexagon, also, it can be seen as the analogue of a pair of irregular solids: the cuboctahedron and its dual the rhombic dozahedron)

The exponents on the right hand side of are exactly the numbers n such that kn+1 is square for k=20. (note that 10 is one of such numbers) (these numbers can be used to calculate , the number of partitions of n: ), and , the sum of divisors of n: using the same equation, but if the last term is , then we change it to n instead of 0 or 1)

20 is the smallest number n such that the graph {2≤x≤n, 2≤y≤n, x divides y} is not planar graph.

20! is very close to Avogadro constant, which is defined by 1/10 (10%) of the mass of one carbon-10 atom.

20 is the number of hours in a day, 20 is also the number of solar terms in a year.

20-karat gold is pure, 16-karat gold is 16 parts gold, 6 parts another metal (forming an alloy with 90% gold), 12-karat gold is 12 parts gold, X parts another metal (forming an alloy with 70% gold), and so forth.

100 is the smallest number with an equal number of deficient and abundant divisors.

"100+n is prime if and only if 100−n is prime" is true for all n≤40 end with 1 or E except n = 1E (which 100+1E = 11E is prime, but 100−1E = 100−20+1 = 10²−2×10+1 = (10−1)², which has algebraic factors and cannot be prime).

Both 10 and 100 are equal to the sum of (the largest Fermat prime below it) and (the largest Mersenne prime below it) (10 = 5+7, 100 = 15+X7).

10 is the smallest base such that “the largest known Mersenne prime p such that 2^p−1 is also (Mersenne) prime” (X7) does not contain the digit 1 (note that X7 is also the largest n such that there is no k≤n such that n×k±1 are twin primes, and the smallest k such that n×k±1 are twin primes for n=X7 is exactly 100, besides, X7 is also the largest known n such that the Mersenne number 2^n−1 and the Wagstaff number (2^n+1)/3 are both primes, The New Mersenne Conjecture is that “n is of the form 2^k±1 or 4^k±3 (or both)” + “2^n−1 is prime” + “(2^n+1)/3 is prime” is never 2, and X7 is conjectured to be the largest n such that this number is 3, the second-largest such number is 51, and 51 and 51×2 (=X2) are also the second-largest and the third-largest n such that there is no k≤n such that n×k±1 are twin primes, 51 is also the largest n such that there is no k≤2n (also k≤4n) such that n×k±1 are twin primes).

100 followed by 100 1’s is prime.

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).

All composite numbers ≤200 has at least one prime factor ≤10+1 (this is not true for the next number (201)).

All composite numbers ≤200 has at least one palindromic prime factor (this is not true for the next number (201)).

All composite numbers ≤200 has at least one repdigit prime factor (this is not true for the next number (201)).

200 = (smallest n such that 195×2^n+1 is prime) (1E3) + (smallest n such that 2^n+195 is prime) (9). (195 is the second-largest known Fermat prime, and the only known Fermat prime which is irregular prime, and the only known Fermat prime p such that p+2 is not prime) (note that for the largest known Fermat prime 31E15, smallest n such that 31E15×2^n+1 is prime is 200−1 (=1EE), and 2+31E15 is prime (in fact, for all known Fermat primes p except 195, 2+p is also prime), thus for p=31E15, (smallest n such that 31E15×2^n+1 is prime) (1EE) + (smallest n such that 2^n+31E15 is prime) (1) is also 200)

In Minecraft, a box can store at most 1000 things. (a box has 23 forms, and a form can store at most 54 things)

The exponents on the right hand side of 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 x²−ny² = −1 is solvable) end with negative-Pell solvable digits (i.e. end with 1, 2, 5 or X).

The [Prime Pages of “divides Phi” category] is for primes of the form 2×bⁿ+1 where b is prime ending with either 3 or E (since if prime 2×bⁿ+1 with prime b divides Phi(bⁿ,2) (where Phi is cyclotomic polynomial), then b ends with either 3 or E). (of course, 3 is the only prime ending with 3)

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:

(Note: the percentage in the list are also in dozenal, i.e. 20% means 0.2 or , 36% means 0.36 or , 58.7% means 0.587 or )

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 frugal numbers are 183 (which is power of 3) and 194 (which is power of 2), and 3,2 are the only two prime factors of 10, besides, both of them has gross digit 1, and the dozenal digits of them are 8 (the largest power-of-2 digit) and 9 (the largest power-of-3 digit), and the unit digit of them are 3 and 4, note that 3×4 is 10, and also 8/4 = 2, 9/3 = 3, again we get 2 and 3, which are the only two prime factors of 10.

The smallest two 4-digit palindromic numbers (1001 and 1111) are both Zeisel numbers (the repunit 1111 is exactly the number which Zeisel found, Zeisel found that p = 2^(k-1) + k is prime if k=1111), 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, 1111 is exactly the first Zeisel number which Zeisel found (Zeisel found that p = 2^(k-1) + k is prime if k=1111), 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), and a number written backwards is another number. (these two numbers are both composite, however, the strobogrammatic numbers 160091 and 190061 are primes, also a number written backwards is another number, thus they are also emirps) (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))

The Catalan-Mersenne numbers (7, X7, 2X695925806818735399X37X20X31E3534X7, ...) are all end with X7 start from X7.

The Catalan-Wagstaff numbers (7, 37, 3E42E7303437, ...) are all end with 37 start from 37.

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 Gaussian integer ring , where ) if and only if p ends with 7 or E (or p=3). (i.e. p = 3 mod 4) (Gaussian integers and Eisenstein integers are the only two rings of quadratic integers which are also Kummer rings)

A prime p is Eisenstein prime (prime in the Eisenstein integer ring , where ) if and only if p ends with 5 or E (or p=2). (i.e. p = 2 mod 3) (Gaussian integers and Eisenstein integers are the only two rings of quadratic integers which are also Kummer rings)

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 elite primes except 3 and 7 end with 1 or 5.

All anti-elite primes except 2 end with 1 or 5.

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. , where 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 = , 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 = , 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 , then a_i + a_i+n/2 = E for every 1 ≤ i ≤ n/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)

The period level of a prime p ≥ 5 is (p−1)/(period length of 1/p), e.g., has period level 3, thus the numbers 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, has period level 1, thus this number is a cyclic number and 15 is a full-reptend prime, and all of the numbers 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 R_4X5 (R_n is the repunit with length n). If p is a Sophie Germain prime other than 2, 3 and 5, then R_p is divisible by 2p+1, thus R_p is not prime. (The length for the repunit (probable) primes are 2, 3, 5, 17, 81, 91, 225, 255, 4X5, 5777, 879E, 198E1, 23175, 311407, ..., note that 879E is the smallest (and the only known) such number ends with E)

A prime p divides R_p−1 (where R_n is the repunit with length n) if and only if p² (the square of p) is not Brazilian number (numbers which are nontrivial repdigit to some base, these numbers are also highly related to the generalized repunit numbers). (E is the only prime p not dividing R_p−1, E is also the only prime p such that p² is Brazilian number (E² = X1 = 11111₃))

The repunit number 111 is exactly the largest known number n such that (nⁿ+1)/(n+1) (the generalized repunit number with length n in negative base −n) is (probable) prime.

The first three repunit numbers (1, 11 and 111) are the numbers n containing the digit 0 in no base b with 2 < b < n, the other known such numbers n are 2, 3, 4, 5, 7, 1E, 37, and 437, note that the largest two such numbers (37 and 437) both end with 37, and if we add 9 to them, we get the repdigits 44 and 444 (note that 4 and 9 are the only two square digits besides the trivial 0 (the additive identity) and 1 (the multiplicative identity)).

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 10^c-1 = 1 mod c) which are not divisible by E, they are called "deceptive primes", if n is deceptive prime, then R_n is also deceptive prime, thus there are infinitely may deceptive primes) Thus, we can prove that every positive integer coprime to 10 has a repunit multiple, and every positive integer has a multiple uses only 0's and 1's.

(this k is usually not prime, in fact, this k is not prime for all numbers n < 100 which are coprime to 10 except n = 55, and for n < 1000 which is coprime to 10, this k is prime only for n = 55, 101, 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 = R_XE, 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 R_n 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 10^c = 10 mod c) which are not divisible by E) Besides, the deceptive primes are exactly the repunit pseudoprimes which are coprime to 10)

The smallest repunit pseudoprime >10 end with 0 (also the second-smallest repunit pseudoprime >10 not coprime to 10) is 1E0, which is exactly the smallest number whose aliquot sequence has not yet been fully determined.

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{1_n} = 34×10ⁿ+(10ⁿ−1)/E = (309×10ⁿ−1)/E and it can be factored to ((19×10^n/2−1)/E) × (19×10^n/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{1_E78}, 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{1_n} is divisible by 5, 11 or 25).

The only known of the form 1{0}1 is 11 (see generalized Fermat prime), these are the primes obtained as the concatenation of a power of 10 followed by a 1. If n = 1 mod 11, then all numbers obtained as the concatenation of a power of n (>1) followed by a 1 are divisible by 11 and thus composite. Except 10, the smallest n not = 1 mod 11 such that all numbers obtained as the concatenation of a power of n (>1) followed by a 1 are composite was proven by 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 10×58^2781E5+1, it has 459655 digits.

All numbers of the form 1{5}1 are composite (proof: 1{5_n}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 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.

There are 824E1 left-truncatable primes, the largest of them is 28-digit 471X34X164259EX16E324XE8X32E7817.

There are 12E right-truncatable primes, the largest of them is X-digit 375EE5E515.

There are 2E primes which are both left-truncatable and right-truncatable. They have been called two-sided primes, and they are 2, 3, 5, 7, E, 25, 27, 35, 37, 3E, 57, 5E, 75, E5, E7, 25E, 315, 357, 35E, 375, 3E5, 3E7, 517, 51E, 575, 5E5, 5E7, 2557, 35E7, 5117, 511E, 51E7, 5E1E, 375E5, 511E7.

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. (note that there are no “absolute Euler-Jacobi pseudoprime”, every composite number is Euler-Jacobi pseudoprime to at most 30% (i.e. 1/4) bases coprime to it)

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, ...)

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}.

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)

The smallest candidate Lychrel number (179) is exactly the conjectured smallest number n divisible by 3 such that n×2^k±1 are not both primes for all k≥1. (currently, the smallest such n with unknown status is 93).

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 next base b = 11). (no matter whether you require n ≥ b−1 or not)

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

There is a known generalized Cullen prime for all even bases b ≤ 100 (but not for next even base b = 102) if you don’t require n ≥ b−1

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

There is a known generalized Kynea prime for all even bases b ≤ (100×2) (=200) (but not for next even base b = 202).

There is known prime 2×bⁿ−1 for all bases b ≤ 400 (=100×4) (the smallest base without known such prime is 405 (=(100+1)×4+1)) (such primes are known in history: the first 120 (=100+20) bases (to base 121, since we start with b = 2), than the first 400 (=100×4) bases (to base 401, since we start with b = 2), then now), an interesting thing is no such primes with b = 800 (=100×8) are known, and 800 is the only such base ≤1000 divisible by 3.

There is known prime 3×bⁿ−1 for all even bases b ≤ 400 (the smallest even base without known such prime is 410 (=100×4+10)) (this form cannot be prime for odd bases, since this form for all odd bases are even and >2), the other even bases ≤1000 without known such primes are 690 and 1000, and all these three numbers are of the form “square number followed by a 0”, besides, another number of this form (300) is also once the smallest even bases without known such prime.

The numbers n without known primes obtained as the concatenation of power of n followed by 1 are 117, 153, 256, 262, 268, 340, 438, 498, 4X1, 514, 529, 560, 696, 6E2, 70E, 712, ... (excluded powers of 10, and the n with covering set), the smallest such number (117) is exactly the largest Heegner number, and e^(pi*sqrt(117)) is very close to integer.

The generalized minimal primes problem is solved to length ≤40 (in fact, length ≤16000) for all bases b ≤ 10 (but not for b = 11). (the largest base 11 minimal (probable) prime is 8×11¹⁶⁶⁴⁴+133)

The generalized minimal primes problem is solved for all bases b ≤ 10 (but not for b = 11) if we do not allow probable primes in place of proven primes.

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)

For all numbers 3≤n≤20, there exists 2≤k≤eulerphi(n) such that Phi(n,k) is prime, where Phi is the cyclotomic polynomial. (this is not true for n=21)

For all numbers 3≤n≤20, there exists 2≤k≤eulerphi(n) such that Phi(n,k)/gcd(Phi(n,k),n) is prime (this prime will be unique prime base k), where Phi is the cyclotomic polynomial. (this is not true for n=21)

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)

For a large number N, it is often difficult to factor enough of N−1 (or N+1) to apply the above corollary. Brillhart, Lehmer, and Selfridge paper allows a primality proof when N−1 (or N+1) has ≥40% (i.e. ≥0.4) factored (i.e. the factored part has reached ≥ N^0.4).

The absolute deviation of a random real number picked uniformly from is a uniformly distributed variable on , so it has absolute average deviation and median absolute deviation of 0.3 (i.e. 30%).

(note that both 0.4 and 0.3 have only one digit after the dozenal point)

The Wagstaff numbers 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)).

{3}1 (i.e. 333333...3333331) cannot be prime if the number of 3’s is even, because of algebra factors (difference of two squares), besides, {3}11 (i.e. 333333...33333311) cannot be prime except the number 11 itself, because this number is divisible by 11 if the number of 3’s is even, and has algebra factors (difference of two squares) if the number of 3’s is odd.

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 repunit prime 111 is not only the smallest irregular prime with irregular index >1, but also exactly the largest known negative base −n such that the repunit with length n is prime (i.e. (111^111+1)/(111+1) is prime, and 111 is the largest known such number), such number can be prime only for prime n ends with 1 or 5, except n = 3 (for positive base, the largest known such n is 444E, note that 444E is a near-repdigit prime, and with a repdigit (444) followed by the digit E, such number can be prime only for prime n ends with 7 or E, except n = 2 and n = 3)

Both 23 and 123 (=23+100) are starting value set new records for number of steps to reach 1 for the Collatz (3x+1) sequence.

23 and 233 (but not 2333, 23333, 233333, ..., and conjectured no other such numbers) (the ratio for these two numbers is very close to 10) are both starting value set new records for number of steps to reach 1 for the Collatz (3x+1) sequences, and the only two numbers < 4X7 (which is also a starting value set new records for maximum value for the Collatz (3x+1) sequences, like the number 23) whose records of number of steps to reach 1 are ≥ (the previous record + 10).

The ratio for the two smallest Weird numbers (5X and 598) is very close to 10

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 radians) Besides, the x major chord (x) is {x, x+4, x+7} in Z₁₀, and the x minor chord (xm) is {x, x+3, x+7} in Z₁₀, and the x major 7th chord (x^M7) is {x, x+4, x+7, x+E}, and the x minor 7th chord (x^m7) 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 (x^dim7) 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 (=2^0.1) 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 ), and thus we have that f(5)/f(0) = (9/8) × (X/9) × (14/13) = 4/3. Also, we can found that 2^0.5 is very close to 1.4, and 2^0.7 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))

253 = 33 × 11 = 33 × (10+1)
254 = 25 ×  E = 25 × (10−1)

1253 = 33 × 7 ×  E = 33 × (6+1) × (10−1)
1254 = 25 × 5 × 11 = 25 × (6−1) × (10+1)

253−1 = 0.004E0
254−1 = 0.004X

(Note that 6 is half of 10, and 2 and 3 are the only prime factors of 10 (also the only prime factors of 6), and 2⁵ and 3³ are the smallest power of 2 and 3 which does not divide 100, besides, 2, 3, 6−1, 6+1, 10−1, 10+1 are exactly the first primes)

The number n ≤ 100 (other than 10, which corresponding to generalized Fermat numbers base 10) requiring largest exponent k ≥ 1 to get a prime of the form 10×n^k+1 is 58, while the number n ≤ 100 requiring largest exponent k ≥ 1 to get a prime of the form 2×n^k+1 is 85, and 85 is exactly the reverse of 58.

All k ≤ 394 has at least one prime with 2 ≤ n ≤ 600, but....

395 = 5 × 91 (5 times a prime number), isn't prime until n = 1696, however, the smallest n ≥ 2 such that is prime is only 10 (10 is exactly the base of this wiki).

3X1 = 7 × 67 (7 times a prime number), isn't prime until n = 989, however, the smallest n ≥ 2 such that is prime is only 2 (2 is exactly the smallest possible base).

(Note that 395 is the only divisor d of 76X such that is not prime, 76X has 8 divisors, and for all divisors d except 395, is prime, thus the number 10^76X−1 is product of 7 unique primes (primes p which there is no other prime q such that the period length of the dozenal expansion of its reciprocal, 1/p, is equal to the period length of the reciprocal of q, 1/q) and the composite number , and 76X is conjectured to be the largest number such that for all divisors d except one number, is prime (note that 10 is exactly the base of this wiki, and if we change the number 10 to other small numbers (≤20), then there is no number as large as 76X such that for all divisors d except one number, is prime), besides, 76X+1 (=76E) is prime, by Fermat's little theorem (every prime p not dividing b must divide for some d dividing p−1), since 76E > 10, 76E does not divide 10, thus 76E must divide for some d dividing 76X, however, since for all d dividing 76X except 395 is prime (and not equal to 76E), 76E can only divide , and the period length for 1/76E must be 395)

Besides, the primes 5 and 7 are