nyuuzyou/wikis huggingface dataset:
Divisor of Integer/Examples/3 divides n(n+1)(n+2)

== Theorem ==
Let $n$ be an integer.
Then:
:$3 \divides n \paren {n + 1} \paren {n + 2}$
where $\divides$ indicates divisibility.
== Proof ==
$n$ is of one of these forms:
for some $k \in \Z$.
Suppose $n = 3 k$.
Then $3 \divides n$ by definition.
Suppose $n = 3 k + 1$.
Then:
:$n + 2 = 3 k + 3 = 3 \paren {k + 1}$
Thus:
:$3 \divides n + 2$.
Suppose $n = 3 k + 2$.
Then:
:$n + 1 = 3 k + 3 = 3 \paren {k + 1}$
Thus:
:$3 \divides n + 1$.
Hence from Divisor Divides Multiple:
:$3 \divides n \paren {n + 1} \paren {n + 2}$
== Sources ==
* : Chapter $2$: Divisibility Theory in the Integers: $2.2$ The Greatest Common Divisor: Problems $2.2$: $5 \ \text {(a)}$
Category:Examples of Divisors of Integers
Divisor of Integer/Examples/3 does not divide 10

== Example of Non-Divisor of Integer ==
:$3 \nmid 10$
where $\nmid$ denotes non-divisibility.
== Proof ==
:$4 = 3 \times 3 + 1$
As the remainder is not $0$, the result follows.
== Sources ==
* : Chapter $2$: Divisibility Theory in the Integers: $2.2$ The Greatest Common Divisor
Category:Examples of Divisors of Integers
Divisor of Integer/Examples/3 does not divide 4

== Example of Non-Divisor of Integer ==
:$3 \nmid 4$
where $\nmid$ denotes non-divisibility.
== Proof ==
:$4 = 3 \times 1 + 1$
As the remainder is not $0$, the result follows.
== Sources ==
* : $\text {2-2}$ Divisibility: Example $\text {2-1}$
Category:Examples of Divisors of Integers
Divisor of Integer/Examples/4 divides -12

== Example of Divisor of Integer ==
:$4 \divides \paren {-12}$
where $\divides$ indicates divisibility.
== Proof ==
:$\paren {-12} = 4 \times \paren {-3}$
== Sources ==
* : Chapter $2$: Divisibility Theory in the Integers: $2.2$ The Greatest Common Divisor
Category:Examples of Divisors of Integers
Divisor of Integer/Examples/63 divides 8^2n - 1

== Theorem ==
Let $n \in \Z_{\ge 0}$ be a positive integer.
Then:
:$63 \divides 8^{2 n} - 1$
where $\divides$ denotes divisibility.
== Proof 1 ==
== Proof 2 ==
Category:Divisor of Integer/Examples/63 divides 8^2n - 1
Category:Examples of Divisors of Integers
Divisor of Integer/Examples/63 divides 8^2n - 1/Proof 1

== Theorem ==
== Proof ==
Proof by induction:
For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition:
:$63 \divides 8^{2 n} - 1$
$\map P 0$ is the case:
Thus $\map P 0$ is seen to hold.
=== Basis for the Induction ===
$\map P 1$ is the case:
Thus $\map P 1$ is seen to hold.
This is the basis for the induction.
=== Induction Hypothesis ===
Now it needs to be shown that if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.
So this is the induction hypothesis:
:$63 \divides 8^{2 k} - 1$
from which it is to be shown that:
:$63 \divides 8^{2 \paren {k + 1} } - 1$
=== Induction Step ===
This is the induction step:
From the induction hypothesis we have that:
:$63 \divides 8^{2 k} - 1$
Hence by definition of divisibility, we have:
:$\exists r \in \Z: 8^{2 k} - 1 = 63 r$
and so:
:$(1): \quad \exists r \in \Z: 8^{2 k} = 63 r + 1$
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\forall n \in \Z_{\ge 0}: 63 \divides 8^{2 n} - 1$
== Sources ==
* : $\text I$. Algebra: The Method of Induction: Exercises $\text {II}$: $6$
Category:Divisor of Integer/Examples/63 divides 8^2n - 1
Category:Proofs by Induction
Divisor of Integer/Examples/63 divides 8^2n - 1/Proof 2

== Theorem ==
== Proof ==
From Integer Less One divides Power Less One, we have that:
:$\forall m, n \in \Z: \paren {m - 1} \divides \paren {m^n - 1}$
This result is the special case where $m = 8^2$.
Category:Divisor of Integer/Examples/63 divides 8^2n - 1
Divisor of Integer/Examples/6 divides 7^n - 1

== Theorem ==
Let $n \in \Z_{\ge 0}$ be a positive integer.
Then:
:$6 \divides 7^n - 1$
where $\divides$ denotes divisibility.
== Proof 1 ==
== Proof 2 ==
Category:Divisor of Integer/Examples/6 divides 7^n - 1
Category:Examples of Divisors of Integers
Divisor of Integer/Examples/6 divides 7^n - 1/Proof 1

== Theorem ==
== Proof ==
Proof by induction:
For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition:
:$6 \divides 7^n - 1$
$\map P 0$ is the case:
Thus $\map P 0$ is seen to hold.
=== Basis for the Induction ===
$\map P 1$ is the case:
Thus $\map P 1$ is seen to hold.
This is the basis for the induction.
=== Induction Hypothesis ===
Now it needs to be shown that if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.
So this is the induction hypothesis:
:$6 \divides 7^k - 1$
from which it is to be shown that:
:$6 \divides 7^{k + 1} - 1$
=== Induction Step ===
This is the induction step:
From the induction hypothesis we have that:
:$6 \divides 7^k - 1$
Hence by definition of divisibility, we have:
:$\exists r \in \Z: 7^k - 1 = 6 r$
and so:
:$(1): \quad \exists r \in \Z: 7^k = 6 r + 1$
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\forall n \in \Z_{\ge 0}: 6 \divides 7^n - 1$
== Sources ==
* : $\text I$. Algebra: The Method of Induction: Exercises $\text {II}$: $7$
Category:Divisor of Integer/Examples/6 divides 7^n - 1
Category:Proofs by Induction
Divisor of Integer/Examples/6 divides 7^n - 1/Proof 2

== Theorem ==
== Proof ==
From Integer Less One divides Power Less One, we have that:
:$\forall m, n \in \Z: \paren {m - 1} \divides \paren {m^n - 1}$
This result is the special case where $m = 7$.
Category:Divisor of Integer/Examples/6 divides 7^n - 1
Divisor of Integer/Examples/6 divides n (n+1) (n+2)

== Theorem ==
Let $n$ be an integer.
Then:
:$6 \divides n \paren {n + 1} \paren {n + 2}$
where $\divides$ indicates divisibility.
== Proof 1 ==
== Proof 2 ==
Category:Divisor of Integer/Examples/6 divides n (n+1) (n+2)
Category:Examples of Divisors of Integers
Divisor of Integer/Examples/6 divides n (n+1) (n+2)/Proof 1

== Theorem ==
== Proof ==
From $3$ divides $n \paren {n + 1} \paren {n + 2}$:
:$3 \divides n \paren {n + 1} \paren {n + 2}$
From $2$ divides $n \paren {n + 1}$:
:$2 \divides n \paren {n + 1}$
and so:
:$2 \divides n \paren {n + 1} \paren {n + 2}$
Hence:
:$2 \times 3 = 6 \divides \paren {n + 1} \paren {n + 2}$
Category:Divisor of Integer/Examples/6 divides n (n+1) (n+2)
Divisor of Integer/Examples/6 divides n (n+1) (n+2)/Proof 2

== Theorem ==
== Proof ==
Proof by induction:
For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition:
:$6 \divides n \paren {n + 1} \paren {n + 2}$
$\map P 0$ is the case:
Thus $\map P 0$ is seen to hold.
=== Basis for the Induction ===
$\map P 1$ is the case:
Thus $\map P 1$ is seen to hold.
This is the basis for the induction.
=== Induction Hypothesis ===
Now it needs to be shown that if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.
So this is the induction hypothesis:
:$6 \divides k \paren {k + 1} \paren {k + 2}$
from which it is to be shown that:
:$6 \divides \paren {k + 1} \paren {k + 2} \paren {k + 3}$
=== Induction Step ===
This is the induction step:
From the induction hypothesis we have that:
:$6 \divides k \paren {k + 1} \paren {k + 2}$
Hence by definition of divisibility, we have:
:$(1): \quad \exists r \in \Z: k \paren {k + 1} \paren {k + 2} = 6 r$
From $2$ divides $n \paren {n + 1}$ we have:
:$(2): \quad \exists s \in \Z: \paren {k + 1} \paren {k + 2} = 2 s$
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\forall n \in \Z_{\ge 0}: 6 \divides n \paren {n + 1} \paren {n + 2}$
== Sources ==
* : $\text I$. Algebra: The Method of Induction: Exercises $\text {II}$: $8$
Category:Divisor of Integer/Examples/6 divides n (n+1) (n+2)
Divisor of Integer/Examples/7 divides 2^3n - 1

== Theorem ==
Let $n$ be an integer such that $n \ge 1$.
Then:
:$7 \divides 2^{3 n} - 1$
where $\divides$ indicates divisibility .
== Proof ==
The proof proceeds by induction.
For all $n \in \Z_{\ge 1}$, let $\map P n$ be the proposition:
:$7 \divides 2^{3 n} - 1$
=== Basis for the Induction ===
$\map P 1$ is the case:
Thus $\map P 1$ is seen to hold.
This is the basis for the induction.
=== Induction Hypothesis ===
Now it needs to be shown that if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.
So this is the induction hypothesis:
:$7 \divides 2^{3 k} - 1$
from which it is to be shown that:
:$7 \divides 2^{3 \paren {k + 1} } - 1$
=== Induction Step ===
This is the induction step:
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\forall n \in \Z_{\ge 1}: 7 \divides 2^{3 n} - 1$
== Sources ==
* : Chapter $2$: Divisibility Theory in the Integers: $2.2$ The Greatest Common Divisor: Problems $2.2$: $6 \ \text {(a)}$
Category:Examples of Divisors of Integers
Divisor of Integer/Examples/80 divides 9^2n - 1

== Theorem ==
Let $n \in \Z_{\ge 0}$ be a non-negative integer.
Then:
:$80 \divides 9^{2 n} - 1$
where $\divides$ denotes divisibility.
== Proof 1 ==
== Proof 2 ==
Category:Divisor of Integer/Examples/80 divides 9^2n - 1
Category:Examples of Divisors of Integers
Divisor of Integer/Examples/80 divides 9^2n - 1/Proof 1

== Theorem ==
== Proof ==
The proof proceeds by induction.
For all $n \in \Z_{\ge 0}$, let $\map P n$ be the equivalent proposition:
:$80 \divides 9^{2 n} - 1$
where $\divides$ denotes divisbility
$\map P 0$ is the case:
Thus $\map P 0$ is seen to hold.
=== Basis for the Induction ===
$\map P 1$ is the case:
Thus $\map P 1$ is seen to hold.
This is the basis for the induction.
=== Induction Hypothesis ===
Now it needs to be shown that if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.
So this is the induction hypothesis:
:$80 \divides 9^{2 k} - 1$
from which it is to be shown that:
:$80 \divides 9^{2 \paren {k + 1} } - 1$
=== Induction Step ===
This is the induction step:
From the induction hypothesis we have that:
:$80 \divides 9^{2 k} - 1$
Hence by definition of divisibility, we have:
:$\exists r \in \Z: 9^{2 k} - 1 = 80 r$
and so:
:$(1): \quad \exists r \in \Z: 9^{2 k} = 80 r + 1$
Hence by definition of divisibility:
:$80 \divides 9^{2 \paren {k + 1} } - 1$
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\forall n \in \Z_{\ge 0}: 80 \divides 9^{2 n} - 1$
== Sources ==
* : $\text I$. Algebra: The Method of Induction: Example $2$
Category:Divisor of Integer/Examples/80 divides 9^2n - 1
Category:Proofs by Induction
Divisor of Integer/Examples/80 divides 9^2n - 1/Proof 2

== Theorem ==
== Proof ==
From Integer Less One divides Power Less One, we have that:
:$\forall m, n \in \Z: \paren {m - 1} \divides \paren {m^n - 1}$
This result is the special case where $m = 9^2$.
Category:Divisor of Integer/Examples/80 divides 9^2n - 1
Divisor of Integer/Examples/8 divides 3^2n + 7

== Theorem ==
Let $n$ be an integer such that $n \ge 1$.
Then:
:$8 \divides 3^{2 n} + 7$
where $\divides$ indicates divisibility .
== Proof 1 ==
== Proof 2 ==
Category:Divisor of Integer/Examples/8 divides 3^2n + 7
Category:Examples of Divisors of Integers
Divisor of Integer/Examples/8 divides 3^2n + 7/Proof 1

== Theorem ==
== Proof ==
The proof proceeds by induction.
For all $n \in \Z_{\ge 1}$, let $\map P n$ be the proposition:
:$8 \divides 3^{2 n} + 7$
=== Basis for the Induction ===
$\map P 1$ is the case:
Thus $\map P 1$ is seen to hold.
This is the basis for the induction.
=== Induction Hypothesis ===
Now it needs to be shown that if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.
So this is the induction hypothesis:
:$8 \divides 3^{2 k} + 7$
from which it is to be shown that:
:$8 \divides 3^{2 \paren {k + 1} } + 7$
=== Induction Step ===
This is the induction step:
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\forall n \in \Z_{\ge 1}: 8 \divides 3^{2 n} + 7$
== Sources ==
* : Chapter $2$: Divisibility Theory in the Integers: $2.2$ The Greatest Common Divisor: Problems $2.2$: $6 \ \text {(a)}$
Category:Divisor of Integer/Examples/8 divides 3^2n + 7
Category:Proofs by Induction
Divisor of Integer/Examples/8 divides 3^2n + 7/Proof 2

== Theorem ==
== Proof ==
From Integer Less One divides Power Less One, we have that:
:$\forall m, n \in \Z: \paren {m - 1} \divides \paren {m^n - 1}$
Hence we have the special case where $m = 3^2$:
:$8 \divides 3^{2 n} - 1$
from which it follows immediately that:
:$8 \divides 3^{2 n} - 1 + 8 = 3^{2 n} + 7$
Category:Divisor of Integer/Examples/8 divides 3^2n + 7
Divisor of Integer/Examples/Divisors of 6

== Examples of Divisors of Integer ==
The divisors of $6$ are $1$, $2$, $3$ and $6$.
== Proof ==
See :
where $\sigma_0$ denotes the divisor count function.
Then we have:
which accounts for them all.
== Sources ==
* 
* 
Category:Examples of Divisors of Integers
Divisor of One of Coprime Numbers is Coprime to Other

== Theorem ==
Let $a, b \in \N$ be numbers such that $a$ and $b$ are coprime:
:$a \perp b$
Let $c > 1$ be a divisor of $a$:
:$c \divides a$
Then $c$ and $b$ are coprime:
:$c \perp b$
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : Chapter $2$: Divisibility Theory in the Integers: $2.2$ The Greatest Common Divisor: Problems $2.2$: $16 \ \text {(b)}$
Category:Coprime Integers
Category:Divisor of One of Coprime Numbers is Coprime to Other
Divisor of One of Coprime Numbers is Coprime to Other/Proof 1

== Theorem ==
== Proof ==
Let $a \perp b$ and $c > 1: c \divides a$.
 $c \not \perp b$.
So by definition of not coprime:
:$\exists d > 1: d \divides c, d \divides b$
But from Divisor Relation is Transitive:
:$d \divides c, c \divides a \implies d \divides a$
So $d$ is a common divisor of both $a$ and $b$.
So, by definition, $a$ and $b$ are not coprime.
The result follows by Proof by Contradiction.
Category:Divisor of One of Coprime Numbers is Coprime to Other
Divisor of One of Coprime Numbers is Coprime to Other/Proof 2

== Theorem ==
== Proof ==
Let $A, B$ be two numbers which are prime to one another.
Let $C$ be any number greater than $1$ which measures $A$.
:
Suppose $C$ and $B$ are not prime to one another.
Then some number $D$ will measure them both.
We have that $D$ measures $C$ and $C$ measures $A$.
So $D$ measures $A$.
But $D$ also measures $B$.
So $D$ measures $A$ and $B$ which are prime to one another.
By , this is a contradiction.
Therefore there can be no such $D$ that measures both $B$ and $C$.
That is, $B$ and $C$ are prime to one another.
== Sources ==
* : Book $\text{VII}$. Propositions
Category:Divisor of One of Coprime Numbers is Coprime to Other
Divisor of Perfect Number is Deficient

== Theorem ==
Let $n$ be a perfect number.
Let $n = r s$ where $r$ and $s$ are positive integers such that $r > 1$ and $s > 1$.
Then $r$ and $s$ are both deficient.
== Proof ==
, consider $r$.
We have by definition of divisor sum function and perfect number that:
:$\dfrac {\map {\sigma_1} {r s} } {r s} = 2$
But from Abundancy Index of Product is greater than Abundancy Index of Proper Factors:
:$\dfrac {\map {\sigma_1} {r s} } {r s} > \dfrac {\map {\sigma_1} r} r$
That is:
:$\dfrac {\map {\sigma_1} r} r < 2$
Mutatis mutandis:
:$\dfrac {\map {\sigma_1} r} s < 2$
Hence the result by definition of deficient.
== Sources ==
* : $12$
* : $12$
Category:Deficient Numbers
Category:Perfect Numbers
Divisor of Polynomial/Examples/Arbitrary Example 3

== Examples of Divisors of Polynomials ==
The polynomial:
:$x^2 - 2$
can be expressed in terms of its divisors as:
:$\paren {x + \sqrt 2} \paren {x - \sqrt 2}$
== Sources ==
* 
* 
Category:Examples of Divisors of Polynomials
Divisor of Product

== Theorem ==
Let $a, b, c \in \Z$ be integers.
Let the symbol $\divides$ denote the divisibility relation.
Let $a \divides b c$.
Then there exist integers $r, s$ such that:
:$a = r s$, where $r \divides b$ and $s \divides c$.
=== Corollary ===
== Proof ==
Let $r = \gcd \set {a, b}$.
By Integers Divided by GCD are Coprime:
:$\exists s, t \in \Z: a = r s \land b = r t \land \gcd \set {s, t} = 1$
So we have written $a = r s$ where $r$ divides $b$.
We now show that $s$ divides $c$.
Since $a$ divides $b c$ there exists $k$ such that $b c = k a$.
Substituting for $a$ and $b$:
:$r t c = k r s$
which gives:
:$t c = k s$
So $s$ divides $t c$.
But we have that:
:$s \perp t$
Hence by Euclid's Lemma $s \divides c$ as required.
Category:Divisors
Divisor of Product may not be Divisor of Factors

== Theorem ==
Let $a, b, c \in \Z_{>0}$ be (strictly) positive integers.
Let:
:$c \divides a b$
where $\divides$ expresses the relation of divisibility.
Then it is not necessarily the case that either $c \divides a$ or $c \divides b$.
== Proof ==
Proof by Counterexample:
Let $c = 6, a = 3, b = 4$.
Then $6 \times 2 = 12$ so $c \divides a b$.
But neither $6 \divides 4$ nor $6 \divides 3$.
== Sources ==
* : Appendix $1$: Some Results from the Elementary Theory of Numbers
Category:Number Theory
Divisor of Sum of Coprime Integers

== Theorem ==
Let $a, b, c \in \Z_{>0}$ such that:
:$a \perp b$ and $c \divides \paren {a + b}$.
where:
:$a \perp b$ denotes $a$ and $b$ are coprime
:$c \divides \paren {a + b}$ denotes that $c$ is a divisor of $a + b$.
Then $a \perp c$ and $b \perp c$.
That is, a divisor of the sum of two coprime integers is coprime to both.
== Proof ==
Let $d \in \Z_{>0}: d \divides c \land d \divides a$.
Then:
A similar argument shows that if $d \divides c \land d \divides b$ then $d \divides a$.
It follows that:
:$\gcd \set {a, c} = \gcd \set {b, c} = 1$
Hence the result.
== Sources ==
* : Chapter $2$: Some Properties of $\Z$: Exercise $2.2$
* : Chapter $2$: Divisibility Theory in the Integers: $2.2$ The Greatest Common Divisor: Problems $2.2$: $16 \ \text {(d)}$
Category:Number Theory
Category:Divisors
Category:Coprime Integers
Divisor of Unit is Unit

== Theorem ==
Let $\struct {D, +, \circ}$ be an integral domain whose unity is $1_D$.
Let $\struct {U_D, \circ}$ be the group of units of $\struct {D, +, \circ}$.
Then:
:$x \in D, u \in U_D: x \divides u \implies x \in U_D$
That is, if $x$ is a divisor of a unit, $x$ must itself be a unit.
== Proof ==
Let $x \in D, u \in U_D$ such that $x \divides u$.
By definition:
:$\exists t \in D: u = t \circ x$
Thus:
:$1_D = u^{-1} \circ u = u^{-1} \circ t \circ x$
Also, as $D$ is an integral domain and hence a commutative ring, we have:
:$u^{-1} \circ t \circ x = 1_D = x \circ u^{-1} \circ t$
It follows by definition that $x$ is a unit, as it has an inverse, namely $u^{-1} \circ t$.
== Sources ==
* : $\S 62.1$ Factorization in an integral domain: $\text{(iii)}$
Category:Integral Domains
Category:Units of Rings
Divisors obey Distributive Law

== Theorem ==
In modern algebraic language:
:$a = \dfrac 1 n b, c = \dfrac 1 n d \implies a + c = \dfrac 1 n \paren {b + d}$
== Euclid's Proof ==
== Modern Proof ==
Category:Algebra
Category:Divisors
Category:Divisors obey Distributive Law
Divisors obey Distributive Law/Proof 1

== Theorem ==
== Proof ==
Let the (natural) number $A$ be an aliquot part of the (natural) number $BC$.
Let the (natural) number $D$ be the same aliquot part of another (natural) number $EF$ that $A$ is of $BC$.
We need to show that $A + D$ is the same aliquot part of $BC + EF$.
:
We have that whatever aliquot part $A$ is of $BC$, then $D$ is also the same aliquot part of $EF$.
Therefore as many numbers as there are in $BC$ equal to $A$, so many numbers are there in $EF$ equal to $D$.
Let $BC$ be divided into the numbers equal to $A$, that is, $BG, GC$.
Let $EF$ be divided into the numbers equal to $D$, that is, $EH, HF$.
Then the multitude of $BG, GC$ will be equal to the multitude of $EH, HF$.
Since $BG = A$ and $EH = D$, then $BG + EH = A + D$.
For the same reason, $GC + HF = A + D$.
Therefore, as many numbers as there are in $BC$ equal to $A$, so many are there also in $BC + EF$ equal to $A + D$.
Therefore, whatever multiple $BC$ is of $A$, the same multiple also is $BC + EF$ of $A + D$.
Therefore, whatever aliquot part $A$ is of $BC$, the same aliquot part also is $A + D$ of $BC + EF$.
== Sources ==
* : Book $\text{VII}$. Propositions
Category:Divisors obey Distributive Law
Divisors of Factorial

== Theorem ==
Let $n \in \N_{>0}$.
Then all natural numbers less than or equal to $n$ are divisors of $n!$:
:$\forall k \in \left\{{1, 2, \ldots, n}\right\}: n! \equiv 0 \pmod k$
== Proof ==
From the definition of factorial:
:$n! = 1 \times 2 \times \cdots \times \left({n-1}\right) \times n$
Thus every number less than $n$ appears as a divisor of $n!$.
The result follows from definition of congruence.
Category:Factorials
Category:Divisors
Divisors of One

== Theorem ==
The only divisors of $1$ are $1$ and $-1$.
That is:
:$a \divides 1 \iff a = \pm 1$
== Proof ==
=== Necessary Condition ===
Let $a = \pm 1$.
From Integer Divides Itself‎ we have that $1 \divides 1$.
From Integer Divides its Negative we have that $-1 \divides 1$.
=== Sufficient Condition ===
Let $\exists a \in \Z: a \divides 1$.
Then $\exists c \in \Z: a c = 1$.
From Absolute Value Function is Completely Multiplicative we have that:
:$\size a \cdot \size c = \size 1$
Neither $a$ nor $c$ can be zero, from Integers form Integral Domain.
So $\size a \ge 1$ and $\size c \ge 1$.
But if $\size a > 1$ then:
:$\size a \cdot \size c > \size c$
and so:
:$\size a \cdot \size c > 1$
So:
:$\size a = 1$
that is:
:$a = 1$ or $a = -1$
== Sources ==
* : Chapter $3$: The Integers: $\S 10$. Divisibility: Theorem $17 \ \text{(i)}$
* : Chapter $2$: Divisibility Theory in the Integers: $2.2$ The Greatest Common Divisor: Theorem $2 \text{-} 2 \ (2)$
Category:Divisors
Category:1
Divisors of One More than Power of 10/Number of Zero Digits Congruent to 2 Modulo 3

== Theorem ==
Let $N$ be a natural number of the form:
:$N = 1000 \ldots 01$
where the number of zero digits between the two $1$ digits is of the form $3 k - 1$.
Then $N$ has divisors:
::$1 \underbrace {00 \ldots 0}_{\text {$k - 1$ $0$'s} } 1$
:where the number of zero digits between the two $1$ digits is $k - 1$
::$\underbrace {99 \ldots 9}_{\text {$k$ $9$'s} } \underbrace {00 \ldots 0}_{\text {$k - 1$ $0$'s} }1$
== Proof ==
By definition, $N$ can be expressed as:
:$N = 10^{3 k} + 1$
Let $a := 10^k$.
Then we have:
where it is noted that:
Hence the result.
== Examples ==
== Also see ==
* Henry Ernest Dudeney: Modern Puzzles: $62$ -- Factorizing
== Sources ==
* : Solutions: $62$. -- Factorizing
* : Answers: $113$. Factorizing
Category:Divisors of One More than Power of 10
Divisors of One More than Power of 10/Number of Zero Digits Congruent to 2 Modulo 3/Examples

== Examples of Divisors of One More than Power of 10: Number of Zero Digits Congruent to 2 Modulo 3 ==
=== $101$ Zero Digits ===
== Demonstration ==
By long multiplication:
 99990001
 x 10001
 --------
 99990001
 999900010000
-------------
1000000000001
== Also see ==
* Henry Ernest Dudeney: Modern Puzzles: $62$ -- Factorizing
== Sources ==
* : Solutions: $62$. -- Factorizing
* : Answers: $113$. Factorizing
Category:Divisors of One More than Power of 10
Divisors of One More than Power of 10/Number of Zero Digits Congruent to 2 Modulo 3/Examples/101 Zero Digits

== Example of Divisors of One More than Power of 10: Number of Zero Digits Congruent to 2 Modulo 3 ==
== Proof ==
$101$ is of the form $3 \times k - 1$ where $k = 34$.
The result follows directly from Divisors of One More than Power of 10: Number of Zero Digits Congruent to 2 Modulo 3.
== Also see ==
* Henry Ernest Dudeney: Modern Puzzles: $62$ -- Factorizing
== Sources ==
* : Solutions: $62$. -- Factorizing
* : Answers: $113$. Factorizing
Category:Divisors of One More than Power of 10
Divisors of One More than Power of 10/Number of Zero Digits Even

== Theorem ==
Let $N$ be a natural number of the form:
:$N = 1 \underbrace {000 \ldots 0}_{\text {$2 k$ $0$'s} } 1$
that is, where the number of zero digits between the two $1$ digits is even.
Then $N$ can be expressed as:
:$N = 11 \times \underbrace {9090 \ldots 90}_{\text {$k - 1$ $90$'s} } 91$
== Proof ==
By definition, $N$ can be expressed as:
:$N = 10^{2 k + 1} + 1$
Then we have:
== Examples ==
== Also see ==
* Henry Ernest Dudeney: Modern Puzzles: $62$ -- Factorizing
== Sources ==
* : Solutions: $62$. -- Factorizing
* : Answers: $113$. Factorizing
Category:Divisors of One More than Power of 10