properties of the numbers 4 and 20 (all written in dozenal)

[sweety439]

All numbers ≤4 divide 10 (the base of the numeral system of this wiki), but 5 does not. (note that 10 is the smallest such number)

All numbers ≤4 satisfy the property that general polynomial equation of degree n has a solution in radicals, but 5 does not. (all numbers ≥5 do not)

All numbers ≤4 satisfy the property that the Fermat number 2^(2^n)+1 is prime, but 5 does not. (all numbers from 5 to 28 do not)

All numbers ≤4 satisfy the property that the double Mersenne number 2^(2^(nth prime)−1)−1 is prime, but 5 does not. (all numbers from 5 to 15 do not)

All numbers ≤4 satisfy the property that the completed graph K_n is planar graph, but 5 does not. (all numbers ≥5 do not)

All numbers ≤4 satisfy the property that 1/n has terminated dozenal, but 5 does not. (only the 3-smooth numbers do)

All numbers ≤4 satisfy the property that the alternating group A_n is a solvable group, but 5 does not. (all numbers ≥5 do not)

All numbers ≤4 satisfy the property that the Lucas number L(2^n) is prime, but 5 does not. (all numbers from 5 to 20 do not)

All numbers ≤4 divides an n such that there are n consecutive numbers each have exactly n divisors (such n are 1, 2, 10, 20), but 5 does not. (only divisors of 20 do)

10 (the base of the numeral system of this wiki) is the least common multiple of the numbers ≤4. (note that this is true only for “the set of the numbers ≤4”, this is true neither for “the set of the numbers ≤3” nor for “the set of the numbers ≤5”)

The largest minimal prime (start with b+1) problem in bases b = 2, 3, 4 are very small (thus the problems are trivial to solve): 3, 11, 35, respectively, but the largest minimal prime (start with b+1) problem in base b=5 is rather large: 5^2E+8 (with 52 decimal digits)

For the number 4! = 20:

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 {\displaystyle {\tbinom {n}{k}}={\tfrac {n!}{k!(n-k)!}}} 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.

20 is the largest number n such that there are n consecutive numbers each have exactly n divisors (such n are 1, 2, 10, 20, they are {1,2} * {1,10}).

20 is the largest number n such that there are n consecutive harshad numbers.

The exponents on the right hand side of {\displaystyle (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 20n+1 is square. (note that 10 is one of such numbers)

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:Pin+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 {\displaystyle (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.

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.

The original minimal prime problem (i.e. prime > base is not required) in all bases b<=20 have at most one unsolved family, but for base b=21, the problem has as many as 10 unsolved family.

[sweety439]

An interesting thing is 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + X + E + 10 + ... = −1/10 = −0.1, i.e.
$ \sum_{k=1}^{\infty}(k)=-\frac{1}{10}=-0.1 $ (this number is
$ \zeta(-1) $ , which equals the negative value of the reciprocal of 10, or the additive inverse of the multiplicative inverse of 10) i.e.
$ \zeta(-1)=-\frac{1}{10}=-0.1 $
The negative value of the reciprocal of 10 (i.e. −1/10) is the value counter-intuitively ascribed to the series 1+2+3....
10 is a composite number, the smallest positive integer with exactly six divisors, its divisors being 1, 2, 3, 4, 6 and 10. 10 is also a highly composite number, the next one being 20 (=2×10), and all highly composite numbers ≥10 are divisible by 10.
10 is the smallest base such that all of the five simplest fractions (1/2, 1/3, 2/3, 1/4, and 3/4) all only have one digit (0.6, 0.4, 0.8, 0.3, and 0.9).
10 is the smallest abundant number.
10 is the average of a twin prime pair (E and 11), i.e. both 10−1 and 10+1 are primes.
10 is the least common multiple of the n such that anxn+an-1xn-1+...+a1x+a0=0 have algebraic solution: {1, 2, 3, 4}.
10 is the least common multiple of the all-harshad numbers: {1, 2, 4, 6}. (10 itself is a harshad number in all bases except base 8 (octal))
The numbers n such that all primes not dividing n are = ±1 mod n (i.e. either of the form nk+1 or of the form nk−1) are exactly the proper divisors of 10: {1, 2, 3, 4, 6}, and the least common multiple of these numbers are exactly 10.
The numbers n such that cos(2pi/n) is rational (in fact, 2cos(2pi/n) is integer) are exactly the proper divisors of 10: {1, 2, 3, 4, 6}, and the least common multiple of these numbers are exactly 10.
10 is the only number which can be written as x^y+x^(y+1) with x≥2, y≥1 in 2 different ways.
10 is the ratio of 20 and 2, where 20 is the largest n such that lambda(n)=2, and 2 is the largest n such that lambda(n)=1, where lambda is the Carmichael lambda function.
10 is the least common multiple of 1 to 4.
10 is the greatest common divisor of the order of unsolvable groups ≤ 14000.
10 is the greatest common divisor of the composite order of simple groups ≤ 14000.
10 is the product of the n such that there are no Greco-Latin squares of order n: {2, 6}.
The star numbers are exactly the centered n-gonal numbers for n=10.
10 is the value of
$ \frac{\pi^2}{(\ln 2) \cdot (\ln \lambda)} $ , where
$ \lambda $ is the Levy’s constant. 10 is the smallest nonnegative even number n such that the numerator of
$ \frac{\zeta(n)}{\pi^n} $ is not 1.
10 is the smallest nonnegative even number n such that the numerator of Bn has at least one prime factor ≥ n+3, where Bn is the Bernoulli number.
It is known that
$ \zeta(n) $ is irrational for n=3, and for at least one other odd n≤10.
A number n has terminate reciprocal if and only if n is regular to 10, i.e. all prime divisors of n are prime divisors of 10, besides, a number n has reciprocal with only 1 digit after the dozenal point if and only if n is divisor of 10.
The n-th Fibonacci number F(n) has terminate reciprocal if and only if n is divisor of 10.
10 is the smallest abundant number, since it is the smallest integer for which the sum of its proper divisors (1 + 2 + 3 + 4 + 6 = 14) is greater than itself.
10 is the smallest positive integer which is not power of squarefree number (i.e. not in OEIS A072774).
10 is the smallest positive integer which is neither squarefree nor prime power.
10 is the smallest positive integer which is neither squarefree nor perfect power.
10 is the smallest positive integer which is neither squarefree nor powerful.
10 is the smallest number n>1 such that the n-th cyclotomic polynomial has no root mod p for all primes p <= n.
10 is the smallest number k such that k/(largest power of squarefree kernel of k) is larger than 1, i.e. 10/6 = 2 > 1.
The highly composite numbers contains the first 5 multiples of 10 (10, 20, 30, 40, and 50), and no other numbers between them.
10 is the only number n such that for all prime factors p of n, p^(largest r such that p^r divides n) + p^(smallest r such that p^r does not divide n) = n. (i.e. for p=2, 2^2 + 2^3 = 10, and for p=3, 3^1 + 3^2 = 10)
10 is the only number which is both (equal to mn) and (average of 2m and 2n) for some integer pairs (m, n) such that m and n are consecutive integers (i.e. n = m+1). ((m, n) = (3, 4))
10 is the largest number n such that if 0≤k≤n−1 and gcd(k, n) = 1 (i.e. k and n are coprime), then n+k is prime.
Both of 10−1 and 10+1 are primes, in fact, all of 10±1, (10/2)±1, and (10/3)±1 are primes, the next such number is 220990, besides, all such numbers must be divisible by 10.
10 is the sum of the numbers in the first Pythagorean triple: {3, 4, 5}.
10−2 is the smallest noncototient, 10+2 is the smallest nontotient, the average of them is 10.
10 is the average of the smallest nontotient (12) and the smallest noncototient (X).
10 is the first sublime number, a number that has a perfect number (6) of divisors, and the sum of its divisors is also a perfect number (24), there are only two known such numbers (including 10 itself).
10 is the product of the first three factorials (not including 0! = 1).
10 is the smallest base b such that the number of trailing zeros in n! (the factorial of n) does not depend on one of the primes or prime powers dividing b.
The square of any prime >3 is congruent to 1 mod 10 (in fact, congruent to 1 mod 20).
The sum of any pair of twin primes (other than 3 and 5) is divisible by 10. (i.e. end with 0)
The product of any pair of twin primes (other than 3 and 5) is congruent to E mod 10. (i.e. end with E)
10 is the largest number which is a divisor of the sum of all pairs of twin primes other than (3, 5).
10 is the largest number which is a divisor of the sum of all but finitely many pairs of twin primes (assuming that there are infinitely many pairs of twin primes).
10 is the largest number which is a divisor of xy for all Pythagorean triples {x, y, z} (x≤y≤z).
10 is the largest base such that both "all squares end with square digits" and "all primes not dividing the base end with either prime digits not dividing the base or 1" are true.
10 is the largest base such that all these three properties are true:
Property of the numerical system Property of the base (b) Bases (Note: 1 cannot be the base of the system)
All squares end with square digits Numbers n such that all x^2 mod n are squares 1, 2, 3, 4, 5, 8, 10, 14
All primes not dividing the base end with either prime digits not dividing the base or 1 (unit) Numbers n such that reduced residue system of n consists of only primes and 1 1, 2, 3, 4, 6, 8, 10, 16, 20, 26
All squares of the primes not dividing the base end with 1 Numbers n with the property m^2 == 1 (mod n) for all integer m coprime to n 1, 2, 3, 4, 6, 8, 10, 20 (exactly the divisors of 20)
10 is the largest base such that both "all squares end with square digits" and "all primes end with prime digits or 1" are true.
10 is the largest number n with the property that every number m in the range n < m < 2n that is coprime to n is also prime.
10 is the largest number n with the property that every number m in the range 1 < m ≤ 2n that is coprime to n is also prime.
10 is the largest natural number n such that for all even numbers k dividing n, both k−1 and k+1 are noncomposite (including 0).
10 is the largest natural number n such that for all numbers k dividing n, both k−1 and k+1 are either prime or power of 2 (or 0).
10 is the largest n such that all even divisors >2 of n are average of a twin prime pair.
10 is the largest number k such that k+j is prime for every j, where 1 <= j < k and gcd(j,k) = 1.
10 is the only number n ≤ 220000 such that n±1, n/2±1 and n/3±1 are all primes. (the next such n is 220990, note that this number contains 3 distinct digits all twice) (in fact, all such n except 10 is divisible by 890)
10!+1 is divisible by 121 = 11^2, and 10 is the only known non-semiprime n such that n!+1 is divisible by (n+1)^2.
10 is the smallest integer b such that more than one prime factor p of b attains the maximum of (p-1)*v_p(b) where v_p(b) is the valuation of b at p. (the other such b≤100 are 39, 68, 76, and 100) (Given b, the number of trailing zeros at the end of the base-b representation of x! is asymptotic to x/M where M is the maximum over p|b of (p-1)*v_p(b). Usually only one prime p attains the maximum and then the number is v_p(x!)/v_p(b) for all but finitely many x. But for b = 10, 39, 68, 76, 100, ..., at least two v_p(x!) must be computed)
10 is the smallest base n for which there exists k≤n (in fact, k≤n2) such that k! (where ! is the factorial) is not harshad number base n (k=7).
The Euclid number n#+1 (where # is the primorial) is prime for all n≤10 (but not for n=11).
Regular 10-gon is the highest polygon in the tessellation.
The numbers n ≤ 10 with terminate reciprocal are {1, 2, 3, 4, 6, 8, 9, 10}, and the divisors of 10 are {1, 2, 3, 4, 6, 10}, the difference of these two sets are 8 and 9 (of course, the second set is a subset of the first set), (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, 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 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)
If m and n are integers ≥3 such that centered m-gonal numbers are always n-gonal numbers, then (m,n) = (8,4) or (9,3), both of these (m,n) pairs have m+n=10. (note that there are no pairs of integers (m,n) with m≥3, n≥3 such that m-gonal numbers are always centered n-gonal numbers)
If k=10, then k * prod{ p | k } = 60, which is the smallest Achilles number.
In base 10 (dozenal, the numeral system of this wiki), by Midy's theorem, if 0<a<p and the period of the dozenal representation of a/p is 2n, so that
$ \frac{a}{p}=0.\overline{a_1a_2a_3\dots a_na_{n+1}\dots a_{2n}} $
then the digits in the second half of the repeating dozenal period are the Es complement of the corresponding digits in its first half (e.g. 1/5 = 0.24 97, 1/7 = 0.186 X35, and 24 + 97 = EE, 186 + X35 = EEE), and the Es complement of the digits are:
0 & E (the zero digit & the largest digit)
1 & X (the unit digit & the only digit whose reciprocal is neither terminating nor purely repeating)
2/3 & 8/9 (the prime factors of 10 & the difference of "the numbers n ≤ 10 with terminate reciprocal" and "the divisors of 10" (also the only digits whose reciprocal has exactly 2 digits, also the numbers in the Catalan's conjecture, also 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, see above))
4/6 & 5/7 (the only composite proper factors of 10 & the only primes < 10−1 not dividing 10)
(note that E is the largest number which is not sum of two composites, and E is also the smallest number >3 which is not sum of two primes) (if you regard 1 as prime, then E is the smallest number >1 which is not sum of two primes)
(also, note that the only primes < 10−1 not dividing 10 are 5 and 7 (and the value of 5+7 is exactly 10, and (5,7) are twin primes, and all sums of twin primes ≥(5,7) are also divisible by 10, and 10 is the largest number with this property), and 1/5 = 0.24 97, 1/7 = 0.186 X35, all digits except 0 and E (the zero digit and the largest digit) appear in the periodic part of either 1/5 or 1/7 (but not both) and exactly once in this number (1/5 or 1/7))
Except 0 and 1, 10 is the only natural number whose square is a Fibonacci number. (i.e. 0, 1 and 100 are the only three square Fibonacci numbers)
If (and only if) there are no Wall-Sun-Sun primes, then 10 is the largest n such that the Pisano period of n2 is the same as the Pisano period of n (both of them are 20).
If n is even and n divides F(n) (where F(n) is the nth Fibonacci number), then n is divisible by 10. (the n’s such that n divides F(n) includes infinitely many multiples of 10, all powers of 5 and other numbers (including some factors of Fib(5^k), e.g. 37501))
The Frampton-Kephart primes are the primes p such that p−1 (or
$ \phi(p) $ , where
$ \phi $ is Euler's totient function) divides 10, these primes are exactly "1 + (the even divisors of 10)" plus the only even prime (2), they (and the number 1) are also the k such that such that d(k) < 3 and d(k − 1) = Pi(k) (where d(k) the divisor count function and Pi(k) the prime counting function), they (and the number 1) are also the possible solutions of x to U(x) = ±1 for some Lucas sequence, they are also the prime factors of 63X00, which is the largest number k satisfying that “for any positive integers x,y coprime to k, x^x == y (mod k) iff y^y == x (mod k)”, they are also the prime factors of 95900, the number of vertices of the 20-dimensional Leech lattice. (Note that 95900 =
$ \prod_{p \text{ prime },\ p-1|10}(\text{ if } p|10 \text{ then } p^{6-p},\ \text{ otherwise } p) $ ), and the product of these primes is 16E6, which is the denominator of B10 (where Bn is the nth Bernoulli number), (note that B10 is also the first Bernoulli number with even index whose numerator has a prime factor p ≥ n+3 (p=497) (i.e. 10 is the smallest number n such that there exists prime p such that (p, n) is an irregular pair)) besides, the dozenal representation of 1/p (with p over these primes), is terminate when p divides 10 (i.e. p = 2 or 3), and for the other three values of p, they are 0.2497, 0.186X35, and 0.0E, the period of them uses all the digits 0, 1, 2, ..., E exactly once.
The numbers n have this property: “all primes not divide n are congruent to 1 or −1 mod n” are {1, 2, 3, 4, 6}, they are exactly the proper divisors of 10. (and the LCM of them is 10)
The set of the proper divisors of 10 is the complete set of k in N such that
$ 2\cos\frac{2\pi}{k} $ is in Z.
10 is the largest natural number n such that the two sets are completely the same: {a | 0<=a<n, either a=1 or a is prime not dividing n} and {a | 0<=a<n, n+a is prime}. (for n=10, the two sets are both {1, 5, 7, E}, both of the two sets are exactly the natural numbers <10 coprime to 10) (the natural numbers n such that the two sets are completely the same are 1, 2, 4, 6, and 10, all of them are divisors of 10)
A number k satisfying that “for any positive integers x,y coprime to k, x^x == y (mod k) iff y^y == x (mod k)” if and only if k is divisible by
$ \lambda(k) $ (where
$ \lambda(n) $ is the Carmichael lambda function and k is a divisor of 63X00, note that 63X00 is the largest k such that
$ \lambda(k)=20 $ , and for all these k’s,
$ \lambda(k) $ also divides 20, besides, the prime factors of these numbers are exactly the Frampton-Kephart primes, i.e. the primes p such that p−1 (or
$ \phi(p) $ , where
$ \phi $ is Euler's totient function) divides 10. (equivalently, p−1 divides 20, since there is no prime p such that p−1 divides 20 but p−1 does not divide 10 (both 9 and 21 are composite (9=32, 21=52, both of them are squares of primes)), in fact, the primes p such that p−1 divides 10 (also 20) are exactly "1 + (the even divisors of 10)" plus the only even prime (2).
14060 is the smallest number divisible by all natural numbers from 1 to 10, it is also the smallest number cannot be written in primorial base using only the dozenal digits (i.e. the digits 0, 1, 2, ..., E) (for factorial base, the smallest such number is 1145000000).
10 is the largest known even number expressible as the sum of two primes in only one way (5+7).
10 is the smallest (and the only known) n>2 such that there exist n consecutive integers with n divisors (at least start with 507711EE945E5199770213X692X2X3, there are 11 consecutive integers with 10 divisors). (The only other two possible such n are 20 and X0)
All numbers ≤10 are panconsummate numbers, but the next number (11) is not, besides, the square of 10 (100) is conjectured to be the largest panconsummate number divisible by either 4 or 6 (i.e. not only for divisible by 10).
10 is the smallest n>1 such that the n-th cyclotomic polynomial has no root mod p for all primes p≤n.
Consider the prime race mod q (where q≥2) between qn+1 and qn-1. 10 is conjectured to be the n such that the q where qn+1 first takes lead over qn-1 is largest. (such q is known for all n<=1000 except n=10 and n=20)
10 is the smallest even number n such that there exists a prime p ≥ n+3 (p = 497) such that p divides the numerator of Bn (where Bn is the nth Bernoulli number). (i.e. 10 is the smallest number n such that there exists prime p such that (p, n) is an irregular pair) (note that the 10th prime (31) is also the smallest irregular prime)
10^2−1 (=EE) is the only number which is product of twin primes but not brilliant number, it is also the smallest composite n coprime to 10 (i.e. not divisible by 2 or 3) such that Fibonacci(n-1) is congruent to (1 - Legendre(n,5))/2 modulo n.
10 is the smallest even number n such that “n^k−1 and n^k+1 are always both prime or both composite for every integer k≥1” is conjectured to be true. (for odd number n, n^k−1 and n^k+1 are both even, and hence both composite (if n^k>3))
The ratio of the first two weird numbers (598/5X = E.E39310...) is very close to 10.
The only two known base 10 Wieferich primes are 1685 and 5E685, both of them end with 685, and it is conjectured that all base 10 Wieferich primes end with 685.
The sum of this 4 fifth powers is 10X, the Xth power of 10 (i.e. 10,000,000,000): 235 + 705 + 925 + E15 = 10X, this is a counterexample of Euler's sum of powers conjecture, since 10X equals 1005, it is also a fifth power, but it only require 4 fifth powers (less than the conjectured 5) to add. (Note that 100 is the smallest number whose fifth power is the sum of 4 or less fifth powers)
10 is the smallest weight for which a cusp form exists. This cusp form is the discriminant Δ(q) whose Fourier coefficients are given by the Ramanujan τ-function and which is (up to a constant multiplier) the 20th power of the Dedekind eta function. This fact is related to a constellation of interesting appearances of the number 10 in mathematics ranging from the value of the Riemann zeta function at −1 i.e. ζ(−1) = −1/10, the fact that the abelianization of SL(2,Z) has 10 elements, and even the properties of lattice polygons.
10 is the smallest admirable number.
10 is a Pell number, a pronic number, and a pentagonal number (5-gonal number).
10 is the number of pentominoes (5-ominoes).
1/10 = 10% is the property such that a Rubik's Cube is solvable.
10 is the least common multiple of 3 and 4, the number of sides of the first two regular polygons (equilateral triangle (3-gon) and square (4-gon)), 10 is also the least common multiple of 4 and 6, the number of faces of the first two regular polyhedrons (tetrahedron (4-hedron) and cube (6-hedron)).
The centered 10-gonal numbers (centered dozagonal numbers) are exactly the star numbers. (thus, the star numbers are exactly the numbers obtained as the concatenation of a triangular number followed by a 1)
The known generalized Fermat primes (primes of the form b2n+1) with bases b≤10 are: (this number can be prime only if b is even (since if it is odd then b2n+1 is always even, and thus can’t be prime), thus we only consider even bases)
base (b) n
2 0, 1, 2, 3, 4
4 0, 1, 2, 3
6 0, 1, 2
8 (impossible since 8=23)
X 0, 1
10 0
Except the case b=8 (which has algebraic factors: 82n+1 = (22n+1) × (42n−22n+1)), the n for these b are exactly (the nonnegative integers ≤ m for m = 4 to 0) in order, besides, "if b2n+1 is composite, then b2n+1+1 is also composite" is conjectured to be true for all even b≤10 (but not for b = next even number (12)).
There is a known generalized Cullen prime for all bases b ≤ 10 (but not for 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 b = 101). (no matter whether you require n ≥ b−1 or not)
A 10-sided polygon is a dozagon. A 10-faced polyhedron is a dozahedron. Regular cubes (hexahedrons, 6-faced) and octahedrons (8-faced) both have 10 edges, while regular octadozahedrons (18-faced) have 10 vertices.
The evenish numbers (numbers coprime to 10 and 10's digit is even) are exactly the numbers n such that
$ \left(\frac{n}{20}\right)=1 $ , and the oddish numbers (numbers coprime to 10 and 10's digit is odd) are exactly the numbers n such that
$ \left(\frac{n}{20}\right)=-1 $ . (thus, there are no oddish numbers that are squares)
Two dozagons (10-gons) and one triangle can fill a plane vertex, all solutions using at least one dozagon are {3, 10, 10}, {4, 6, 10}, {3, 3, 4, 10}, and {3, 4, 3, 10}, but only the first two solutions can fill the plane.
(there are 19 solutions for filling a plane vertex, but only E of them can fill the plane, the solutions are:
{3, 7, 36}, {3, 8, 20}, {3, 9, 16}, {3, X, 13}, {3, 10, 10}, {4, 5, 18}, {4, 6, 10}, {4, 8, 8}, {5, 5, X}, {6, 6, 6}
{3, 3, 4, 10}, {3, 4, 3, 10}, {3, 3, 6, 6}, {3, 6, 3, 6}, {3, 4, 4, 6}, {3, 4, 6, 4}, {4, 4, 4, 4}
{3, 3, 3, 3, 6}, {3, 3, 3, 4, 4}, {3, 3, 4, 3, 4}
{3, 3, 3, 3, 3, 3}
bold for the solutions that can fill the plane, note that dozagon is the highest regular polygon in convex uniform tiling, i.e. 10 is the largest bold number in the list)
10 and 18,E27,099,E93,490,727,709,9E9,1X4,841,59E,264,E0X,X13,59X,268,567,863,258,910,E74,98X,682,454 (=(2X6)(251 − 1)(227 − 1)(217 − 1)(27 − 1)(25 − 1)(23 − 1)) are the only two known sublime numbers. (A sublime number is a positive integer which has a perfect number of positive factors (including itself), and whose positive factors add up to another perfect number, i.e. a positive integer n is sublime number if and only if both
$ \sigma_0(n) $ and
$ \sigma_1(n) $ are perfect numbers (where
$ \sigma_k(n) $ is the sum of the kth powers of the divisors of n, i.e.
$ \sigma_k(n)=\sum_{d\mid n} d^k $ ), e.g. for n=10 we have
$ \sigma_0(10)=1^0+2^0+3^0+4^0+6^0+10^0=6 $ and
$ \sigma_1(10)=1^1+2^1+3^1+4^1+6^1+10^1=24 $ , and both 6 and 24 are perfect numbers, thus 10 is a sublime number)
Although 6 is a divisor of 10, there exists a group of order 10 (A4) without a subgroup with order 6, it is the smallest such example (i.e. 10 is the smallest number n such that there exists k dividing n and a group of order n such that this group has no subgroup with order k)
All orders of non-solvable groups (thus all orders of non-cyclic simple groups) are divisible by either 10 or 18, and all orders of non-solvable groups ≤ 14000 are divisible by 10, (the smallest order of non-solvable groups not divisible by 10 is 14X28) the first two orders of non-cyclic simple groups are 50 and 120, and the greatest common divisor of them is indeed 10.
All odd perfect numbers (if exist) end with 1, 09, 39, 69, or 99, and if an odd perfect number ends with 1 (i.e. = 1 mod 10), then it has at least 10 distinct prime factors.
10 is the kissing number in three dimensions (for a long time, people did not know that whether the answer is 11, this was the subject of a famous disagreement between mathematicians Isaac Newton and David Gregory). (In two demensions, it is 6 = 10÷2, and in four dimensions, it is 20 = 10×2. The values of the kissing numbers are only known in 1, 2, 3, 4, 8 and 20 dimensions (note that all such numbers of dimensions are divisors of 20), and the kissing numbers in these numbers of dimensions are in order 2, 6, 10, 20, 180 and 95900, and the 8 dimension case and the 20 dimension case are in order the E8 lattice and the Leech lattice, however, there are also the upper bounds and the lower bounds for all other number of dimensions ≤ 20, see the list below, the dimensions in which the kissing number is known are listed in boldface)
Dimension Lower
bound Upper
bound
1 2
2 6
3 10
4 20
5 34 38
6 60 66
7 X6 E2
8 180
9 216 264
X 358 3X2
E 406 606
10 5X0 951
11 802 1,245
12 E1X 1,X13
13 1,598 2,996
14 2,600 4,30E
15 3,116 6,4X8
16 4,346 9,710
17 6,210 12,438
18 X,0X0 19,338
19 14,060 27,708
1X 24,X60 3E,798
1E 45,XX6 60,000
20 95,900
(note that k(1)=2 divides k(2)=3, k(2)=3 divides k(3)=10, k(3)=10 divides k(4)=20, and the three quotients are all primes, however, k(4)=20 does not divide k(5), since 34≤k(5)≤38)
10 is the difference of "the largest number n such that x2 + x + n is prime for all 0≤x≤n−2" (35) and "the largest number n such that 2x2 + n is prime for all 0≤x≤n−1" (25).
10 is the difference of "the smallest number n such that (define an: a0 = 1, for k > 0, ak = (1+a02+a12+...+ak−12)/k) an is not integer" (37) and "the smallest number n such that ʃ0∞(cos(x)cos(x/2)cos(x/3)...cos(x/n)) ≠ π/2" (27).
10 is the difference of the first two numbers not of the form pm×qn with p, q primes, m≥0, n≥0 (36−26).
10 is the number of Latin squares of order 3.
10 is the smallest number such that it is unknown whether there are n−1 mutually orthogonal Latin squares of order n. (Note that there cannot be n mutually orthogonal Latin squares of order n, if n > 1)
10 is the smallest possible order of magic square with 1 and consecutive primes starting with 3. (the smallest possible order of magic square with consecutive primes starting with 3 is 2E)
10 appears in this form of these three almost integers related to the three largest Heegner numbers (117, 57 and 37):
$ e^{\pi\sqrt{37}}=10^3(9^2-1)^3+520-4.74596469\mathcal{E}07\ldots\times 10^{-4} $
$ e^{\pi\sqrt{57}}=10^3(19^2-1)^3+520-3.\mathcal{E}\mathcal{E}15385\mathcal{X}402\ldots\times 10^{-6} $
$ e^{\pi\sqrt{117}}=10^3(173^2-1)^3+520-6.82\mathcal{X}19322127\ldots\times 10^{-10} $
(the numbers with scientific notation are rounded to 10 significant figures)
All of them contains 103, besides, the three negative exponents of 10 of them are −10, −6 and −4 (the negative exponent of 10 of the largest number (eπ√117) is −10, etc.), the absolute values of them are exactly 10/n for n = 1, 2 and 3, besides, the numbers of the consecutive E around the dozenal point are also 10, 6 and 4 (exactly 10/n for n = 1, 2 and 3), since all of the integer part of these three numbers end with 51E (the reason is all of them are a multiple of 103 (=1000) plus the number 520 minus a number <1, see the formulas above), which already has one E before the dozenal point.
In music, 10 is the number of pitch classes in an octave (an octave has 10 semitones), not counting the duplicated (octave) pitch. Also, the total number of major keys, (not counting enharmonic equivalents) and the total number of minor keys (also not counting equivalents). This applies only to 10 tone equal temperament (10-TET), the most common tuning used today in western influenced music.
10 is used for timekeeping, e.g. one year has 10 months, one day has 20 hours, one hour has 50 minutes, and one minute has 50 seconds (all of these numbers are multiples of 10). Also, there are 10 signs of the zodiac (astrological signs), and the Chinese use a 10-year cycle for time-reckoning called Earthly Branches (Chinese zodiac).
10 is the number of numbers on a clock.
10 is the number of yellow stars on the Flag of Europe.
10 is the largest possible points for rolling two dices.
10 appears in the natural logarithm of Levy’s constant:
$ \frac{\pi^2}{10\ln 2} $ .
1/10 (the reciprocal of 10) also appears in the limit of Glaisher-Kinkelin constant:
$ A=\lim_{n\rightarrow\infty} \frac{K(n+1)}{n^{n^2/2+n/2+1/10} e^{-n^2/4}} $ .
The series 1 + 2 + 3 + 4 + ... is divergent, however, we can use the Riemann zeta function to define its sum: −1/10 (= −0.1), since zeta(−1) = −0.1 (thus, 10 + 20 + 30 + 40 + ... (the sum of all positive multiples of 10) = −1).
0 = Aries, 1 = Taurus, 2 = Gemini, 3 = Cancer, 4 = Leo, 5 = Virgo, 6 = Libra, 7 = Scorpio, 8 = Sagittarius, 9 = Capricorn, X = Aquarius, E = Pisces. (4k = fire, 4k+1 = earth, 4k+2 = air, 4k+3 = water; 3k = cardinal, 3k+1 = fixed, 3k+2 = mutable; 2k = yang, 2k+1 = yin)
0 = C, 1 = C#/Db, 2 = D, 3 = D#/Eb, 4 = E, 5 = F, 6 = F#/Gb, 7 = G, 8 = G#/Ab, 9 = A, X = A#/Bb, E = B. ({k, k+4, k+7} = k major chord, {k, k+3, k+7} = k minor chord, {k, k+4, k+8} = k augmented triad, {k, k+3, k+6} = k diminished triad, {k, k+4, k+7, k+E} = k major seventh chord, {k, k+3, k+7, k+X} = k minor seventh chord, {k, k+3, k+7, k+E} = k minor major seventh chord, {k, k+4, k+7, k+X} = k dominant seventh chord, {k, k+3, k+6, k+9} = k diminished seventh chord, {k, k+3, k+6, k+X} = k half-diminished seventh chord)
The digits in 1/5 = {2,4,7,9}, plus the digit 0 (the smallest digit) is the musical notes in the C-major pentatonic scale, plus the digit E (the largest digit) is the musical notes in the G-major pentatonic scale.
The digits in 1/7 = {1,3,5,6,8,X}, plus the digit 0 (the smallest digit) is the musical notes in the C#/Db-major diatonic scale, plus the digit E (the largest digit) is the musical notes in the F#/Gb-major diatonic scale.
(note that (C,F#/Gb) and (G,C#/Db) are tritones, and tritone is equal to 6 semitones, and 6 is equal to half of 10)
0 = unison, 1 = minor second, 2 = major second, 3 = minor third, 4 = major third, 5 = perfect fourth, 6 = tritone, 7 = perfect fifth, 8 = minor sixth, 9 = major sixth, X = minor seventh, E = major seventh, 10 = octave.
(they are the elements in the cyclic group Z10)

[sweety439]

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, 9 1, 2, 3 3
2 none none 0
3 121, 484 11, 22 2
4 none none 0
5 10201, 12321, 14641, 16661, 16E61, 40804, 41414, 44944 101, 111, 121, 12E, 131, 202, 204, 212 8
6 160061 42E 1
7 1002001, 102X201, 1093901, 1234321, 148X841, 4008004, 445X544, 49XXX94 1001, 1015, 1047, 1111, 1221, 2002, 2112, 2244 8
8 none none 0
9 100020001, 102030201, 104060401, 1060E0601, 121242121, 123454321, 125686521, 1420E0241, 1444X4441, 1468E8641, 14X797X41, 1621E1261, 163151361, 1XX222XX1, 400080004, 404090404, 410212014, 4414X4144, 4456E6544, 496787694, 963848369 10001, 10101, 10201, 10301, 11011, 11111, 11211, 11E21, 12021, 12121, 1229E, 1292E, 12977, 14685, 20002, 20102, 20304, 21012, 21112, 22344, 31053 19
X 1642662461 434X5 1
E 10000200001, 10221412201, 10444X44401, 12102420121, 12345654321, 141E1E1E141, 14404X40441, 16497679461, 40000800004, 40441X14404, 41496869414, 44104X40144, 49635653694 100001, 101101, 102201, 110011, 111111, 11E13E, 120021, 12X391, 200002, 201102, 204204, 210012, 223344 11
10 none none 0
It is conjectured that if n is divisible by 4, then there are no n-digit palindromic squares.
Rn2 (where Rn is the repunit with length n) is a palindromic number for n ≤ 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, ...
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 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.
unit digit of nk
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
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.
n Fn digit root of Fn Ln digit root of Ln n Fn digit root of Fn Ln digit root of Ln
1 1 1 1 1 21 37501 5 81101 E
2 1 1 3 3 22 5X301 8 111103 7
3 2 2 4 4 23 95802 2 192204 7
4 3 3 7 7 24 133E03 X 2X3307 3
5 5 5 E E 25 209705 1 47550E X
6 8 8 16 7 26 341608 E 758816 2
7 11 2 25 7 27 54E111 1 1012125 1
8 19 X 3E 3 28 890719 1 176X93E 3
9 2X 1 64 X 29 121E82X 2 2780X64 4
X 47 E X3 2 2X 1XE0347 3 432E7X3 7
E 75 1 147 1 2E 310EE75 5 6XE0647 E
10 100 1 22X 3 30 5000300 8 E22022X 7
11 175 2 375 4 31 8110275 2 16110875 7
12 275 3 5X3 7 32 11110575 X 25330XX3 3
13 42X 5 958 E 33 1922082X 1 3E441758 X
14 6X3 8 133E 7 34 2X3311X3 E 6477263E 2
15 E11 2 2097 7 35 47551X11 1 X3EE4197 1
16 15E4 X 3416 3 36 75882EE4 1 148766816 3
17 2505 1 54E1 X 37 101214X05 2 23075X9E1 4
18 3XE9 E 8907 2 38 176X979E9 3 379305607 7
19 6402 1 121E8 1 39 2780E0802 5 5X9X643E8 E
1X X2EE 1 1XE03 3 3X 432E885EE 8 967169X03 7
1E 14701 2 310EE 4 3E 6XE079201 2 13550121EE 7
20 22X00 3 50002 7 40 E22045800 X 2100180002 3
(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 the square of 10), thus, 10 is the only base such that 100 is a Fibonacci number (since 100 in a base is just the square of this base, and 0 and 1 cannot be the base of numeral system), and thus we can make the near value of the golden ratio: F11/F10 = 175/100 = 1.75 (since the ratio of two connected Fibonacci numbers is close to the golden ratio, as the numbers get large). Besides, the only cube Fibonacci number is 8 = F6.
n 2n n 2n n 2n n 2n n 2n n 2n
1 2 21 E2X20X8 41 5317E5804588X8 61 256906X1X93096E8934X8 81 11X12X743504482569888538X0X8 X1 65933E8691303X448E712227X7E11448X8
2 4 22 1X584194 42 X633XE408E5594 62 4E161183966171E566994 82 238259286X08944E17554X758194 X2 10E667E51626078895E22445393X2289594
3 8 23 38E48368 43 190679X815XXE68 63 9X302347710323XE11768 83 4744E6551815689X32XX992E4368 X3 21E113XX305013556EX4488X76784556E68
4 14 24 75X94714 44 361137942E99E14 64 1786046932206479X23314 84 9289E0XX342E15786599765X8714 X4 43X2279860X026XE1E88955931348XE1E14
5 28 25 12E969228 45 702273685E77X28 65 3350091664410937846628 85 16557X198685X2E350E7730E95228 X5 878453750180519X3E556XE6626959X3X28
6 54 26 25E716454 46 120452714EE33854 66 66X0163108821673491054 86 30XE3837514E85X6X1E3261E6X454 X6 15348X72X0340X3787XXE19E10516E787854
7 X8 27 4EE2308X8 47 2408X5229EX674X8 67 111803062154431269620X8 87 619X7472X29E4E9183X6503E188X8 X7 2X695925806818735399X37X20X31E3534X8
8 194 28 9EX461594 48 48158X457E912994 68 2234061042X886251704194 88 103792925857X9E634790X07X35594 X8 5916E64E41143526X777873841863X6X6994
9 368 29 17E8902E68 49 942E588E3E625768 69 446810208595504X3208368 89 20736564E4E397E06936181386XE68 X9 E631E09X82286X5193335274835079191768
X 714 2X 33E5605E14 4X 1685XE55X7E04E314 6X 891420414E6XX0986414714 8X 41270E09X9X773X116703427519E14 XX 1E063X179445518X36666X52946X136363314
E 1228 2E 67XE00EX28 4E 314E9XXE93X09X628 6E 1562840829E1981750829228 8E 82521X179793278231206852X37X28 XE 3X107833688XX358711118X56918270706628
10 2454 30 1139X01E854 50 629E799E678179054 70 2E05481457X37432X1456454 90 144X4383373665344624114X5873854 E0 78213467155986E52222358E1634521211054
11 48X8 31 2277803E4X8 51 1057E377E1343360X8 71 5X0X9428E3872865828E08X8 91 2898874672710X6890482298E5274X8 E1 1344269122XE751XX44446E5X3068X424220X8
12 9594 32 4533407X994 52 20E3X733X268670194 72 E8196855X752550E455X1594 92 557552912522191560944575XX52994 E2 26885162459E2X39888891XE86115884844194
13 16E68 33 8X668139768 53 41X792678515120368 73 1E43714XE92X4XX1X8XE82E68 93 XE2XX5624X44362E01688E2E98X5768 E3 5154X3048E7X58775555639E5022E549488368
14 31E14 34 159114277314 54 839365134X2X240714 74 3X872299E6589983959E45E14 94 19X598E049888705X03155X5E758E314 E4 X2X986095E38E532XXXE077XX045XX96954714
15 63X28 35 2E6228532628 55 147670X269858481228 75 79524577E0E577476E7X8EX28 95 378E75X09755520E8062XE8EE2E5X628 E5 185975016EX75XX65999X1339808E99716X9228
16 107854 36 5E0454X65054 56 29312185174E4942454 76 136X48E33X1XE32931E395E854 96 735E2E8172XXX41E41059E5EX5XE9054 E6 34E72X031E92E990E77782677415E7723196454
17 2134X8 37 EX08X990X0X8 57 5662434X329X96848X8 77 271895X67839X65663X76EE4X8 97 126EX5E432599883X820E7XEE8E9E60X8 E7 69E258063E65E761E3334513282EE32463708X8
18 426994 38 1E81597618194 58 E104869865797149594 78 52356E91347790E107931EX994 98 251E8EX864E775479441E39EE5E7E0194 E8 117X4E4107E0EE303X6668X26545EX6490721594
19 851768 39 3E42E73034368 59 1X20951750E372296E68 79 X46E1E62693361X213663E9768 99 4X3E5E9509E32X936883X77EXEE3X0368 E9 23389X8213X1EX60791115850X8EE90961242E68
1X 14X3314 3X 7X85E26068714 5X 38416X32X1X724571E14 7X 1891X3E051667038427107E7314 9X 987XEE6X17X659671547933E9EX780714 EX 467579442783E90136222E4X195EE61702485E14
1E 2986628 3E 1394EX50115228 5E 74831865839248E23X28 7E 356387X0X3112074852213E2628 9E 17539EE183390E7122X93667E7E9341228 EE 912E36885347E60270445X9836EEE0320494EX28
20 5751054 40 2769E8X022X454 60 12946350E476495X47854 80 6E075381862241294X4427X5054 X0 32X77EX346761E2245967113E3E6682454 100 1625X7154X693E0052088E97471EEX0640969E854
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)
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 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 ᘔ and n/2 is prime). (if the denominator of the Bernoulli number $ B_n $ is 6, then n ends with 2 or ᘔ, 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, 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 ≤ 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))
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)
100 is the smallest number whose nth power can be written as the sum of (≥2 and ≤n) positive nth powers for sum n (n=5, the formula is 100^5 = 23^5 + 70^5 + 92^5 + E1^5, only four 5th powers).
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 20n+1 is square. (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).
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).
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)
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 ≤ 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)
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+ 1 10 10 10 101 10 1001 100 100 1010 11111111111 10
10+ 11 10010 1010 100 10111 100 1001 1010 10010 111111111110 11101 100
20+ 110111 110 1000 10010 101 1010 101011 1000 111111111110 101110 101101 100
30+ 1001001 10010 110 10100 111001 10010 101001 111111111110 10100 111010 10001111 100
40+ 10111101 1101110 101110 110 1100101 1000 1011101111111 100100 10010 1010 101011 1010
50+ 10010101 1010110 100100 1000 1111 111111111110 1100101 101110 111010 1011010 1100111 100
60+ 10101101 10010010 1101110 10010 1011101111111 110 10101011 10100 10000 1110010 100111001 10010
70+ 1101001 1010010 1010 1111111111100 10001 10100 1001 111010 1010110 100011110 101101 1000
80+ 111011 101111010 1111111111100 1101110 110001 101110 10100111 1100 1011010 11001010 100111 1000
90+ 1010111111 10111011111110 10010010 100100 10101001 10010 110101001 1010 1100 1010110 101100011 10100
X0+ 111111111101 100101010 1110010 1010110 11100001 100100 1100001 10000 1010010 11110 1111011111 111111111110
E0+ 1001 11001010 101000 1011100 101011 111010 11010111 1011010 100011110 11001110 1111111111111111111111 100
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 n ≥ b−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 n ≥ b−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, 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.
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))