nyuuzyou/wikis huggingface dataset:
Number of Standard Instances of Categorical Syllogism

== Theorem ==
There are $256$ distinct standard instances of the categorical syllogism.
== Proof ==
Recall the four figures of the categorical syllogism:
:$\begin{array}{r|rl}
\text I & & \\
\hline \\
\text{Major Premise}: & \mathbf \Phi_1 & \tuple {M, P} \\
\text{Minor Premise}: & \mathbf \Phi_2 & \tuple {S, M} \\
\hline \\
\text{Conclusion}: & \mathbf \Phi_3 & \tuple {S, P} \\
\end{array}
\qquad
\begin{array}{r|rl}
\text {II} & & \\
\hline \\
\text{Major Premise}: & \mathbf \Phi_1 & \tuple {P, M} \\
\text{Minor Premise}: & \mathbf \Phi_2 & \tuple {S, M} \\
\hline \\
\text{Conclusion}: & \mathbf \Phi_3 & \tuple {S, P} \\
\end{array}$
:$\begin{array}{r|rl}
\text {III} & & \\
\hline \\
\text{Major Premise}: & \mathbf \Phi_1 & \tuple {M, P} \\
\text{Minor Premise}: & \mathbf \Phi_2 & \tuple {M, S} \\
\hline \\
\text{Conclusion}: & \mathbf \Phi_3 & \tuple {S, P} \\
\end{array}
\qquad
\begin{array}{r|rl}
\text {IV} & & \\
\hline \\
\text{Major Premise}: & \mathbf \Phi_1 & \tuple {P, M} \\
\text{Minor Premise}: & \mathbf \Phi_2 & \tuple {M, S} \\
\hline \\
\text{Conclusion}: & \mathbf \Phi_3 & \tuple {S, P} \\
\end{array}$
By the definition of a categorical syllogism, the following are fixed:
:$(1): \quad$ The order of the categorical statements: major premise, then minor premise, then conclusion.
:$(2): \quad$ The structure of the conclusion: the secondary term then the primary term.
:$(3): \quad$ The content of the two premises: the major premise contains the primary term and the middle term, while the minor premise contains the middle term and the secondary term.
The things that can be varied are:
:$(A): \quad$ The specific nature of each of the categorical statements: each can be any of four types: $\mathbf A$, $\mathbf E$, $\mathbf I$ or $\mathbf O$.
:$(B): \quad$ Within the major and minor premises, the order of the two terms: the middle term can come either first or second.
The order of the terms within the premises determines the figure of the categorical syllogism.
For each of the two premises there are two options for the position of the middle term.
By the Product Rule for Counting it follows that there are $2 \times 2 = 4$ figures.
In each figure, there are $3$ categorical statements.
Each can be of any of $4$ types.
Hence by the Product Rule for Counting, there are $4 \times 4 \times 4 = 64$ possible standard instances for each figure.
For all $4$ figures, by the Product Rule for Counting, it follows that there are $4 \times 64 = 256$ standard instances in total.
== Sources ==
* : Chapter $4$: The Predicate Calculus $2$: $4$ The Syllogism
* 
* 
Category:Categorical Syllogisms
Category:256
Number of Subgroups of Prime Power Order is Congruent to 1 modulo Prime

== Theorem ==
Let $G$ be a finite group whose order is $n$.
Let $p$ be a prime number such that $p^k$ is a divisor of $n$.
Then the number of subgroups of order $p^k$ is congruent to $1$ modulo $p$.
== Proof ==
Let $r_k$ denote the number of subgroups of $G$ of order $p^k$.
From Number of Order p Elements in Group with m Order p Subgroups:
:The number of elements of order $p$ is $r_1 (p - 1)$.
Since there exists $mp$ elements whose order divide $p$, for some $m \in \Z$:
:The number of elements of order $p$ is $mp - 1$.
Thus,
:$r_1 \equiv 1 \mod p$.
Let $H_1, H_2, \dots, H_{r_{k - 1}}$ be the $r_{k - 1}$ subgroups of order $p^{k - 1}$.
Let $K_1, K_2, \dots, K_{r_k}$ be the $r_k$ subgroups of order $p^k$.
Suppose $H_i$ is contained in $a_i$ of the subgroups $K_1, K_2, \dots, K_{r_k}$.
Suppose $K_j$ contains $b_j$ of the subgroups $H_1, H_2, \dots, H_{r_{k - 1}}$.
Then:
:$a_1 + a_2 + \dots + a_{k - 1} = b_1 + b_2 + \dots + b_k$
is the number of distinct pairs of subgroups $H_i, K_j$ for which $H_i$ is a subgroup of $K_j$.
Let $H$ be one of the subgroups of order $p^{k - 1}$.
Let $K_1, K_2, \dots, K_a$ be the subgroups of order $p^k$ which contain $H$.
Then from Composition Series of Group of Prime Power Order:
:$a > 0$
and:
:$H$ is normal in $K_1, K_2, ..., K_a$.
Hence, $H$ is also normal $G$.
By correspondence theorem:
:$G / H$ contains the subgroups $K_1 / H, K_2 / H, \dots, K_a / H$.
These are the only subgroups of $G / H$ of order $p$.
Then from above:
:$a \equiv 1 \mod p$.
Since this is the case for each $a_i$, we have:
:$a_1 + a_2 + \dots + a_{r_{k - 1}} \equiv r_{k - 1} \mod p$.
From Number of Order p^(n-1) Subgroups in Order p^n Group, we have also that:
:$b \equiv 1 \mod p$.
and:
:$b_1 + b_2 + \dots + b_{r_k} \equiv r_k \mod p$.
Therefore:
:$r_{k - 1} \equiv r_k \mod p$.
But $r_1 \equiv 1 \mod p$.
So we have, by induction:
:$1 \equiv r_1 \equiv r_2 \equiv \dots \equiv r_k \mod p$.
Hence the result.
== Historical Note ==
This proof is due to Ferdinand Georg Frobenius. 
He showed this result and other generalizations of the Sylow Theorems in a $1895$ paper Verallgemeinerung des Sylow’schen Satzes.
== Sources ==
* 
* : Chapter $2$: The Sylow Theorems: $\S 59 \iota$
Category:Sylow p-Subgroups
Number of Sylow p-Subgroups in Group of Order 15

== Theorem ==
Let $G$ be a group whose order is $15$.
Then:
:the number of Sylow $3$-subgroups is in the set $\set {1, 4, 7, \ldots}$
:the number of Sylow $5$-subgroups is in the set $\set {1, 6, 11, \ldots}$
== Proof ==
Let $G$ be a group of order $15$.
From the Fourth Sylow Theorem:
:the number of Sylow $p$-subgroups is equivalent to $1 \pmod p$
We have that $15 = 3 \times 5$.
Hence the result.
== Sources ==
* : Chapter $11$: The Sylow Theorems: Example $11.7$
Category:Groups of Order 15
Category:Sylow p-Subgroups
Number of Sylow p-Subgroups is Index of Normalizer of Sylow p-Subgroup

== Theorem ==
Let $P$ be a Sylow p-subgroup of a group $G$, $N_G (S)$ its normalizer, and $n_p$ be the number of Sylow p-subgroups in $G$.
Then $[ G : N_G (S) ] = n_p$.
== Proof ==
By Number of Distinct Conjugate Subsets is Index of Normalizer, $[ G : N_G (S) ] = n_p$.
Number of Terms in Homogeneous Polynomial

== Theorem ==
The number of terms in a homogeneous polynomial of degree $n$ in $m$ indeterminates is given by:
:$N = \dbinom {n + m - 1} n = \dfrac {\paren {n + m - 1}!}{n! \, \paren {m - 1}!}$
== Proof ==
== Sources ==
* : $1$ Partial Differentiation: $\S 1$. Introduction: Exercise $12$
Category:Homogeneous Polynomials
Number of Total Orderings on Finite Set

== Theorem ==
Let $S$ be a finite set with $n$ elements.
Then there are $n!$ different total orderings that can be applied to $S$.
== Proof ==
A total ordering on $S$ is by definition a permutation on $S$ in the sense of an ordered selection.
The result follows from Number of Permutations.
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 14$: Orderings: Exercise $14.5$
Category:Total Orderings
Number of Twin Primes less than Powers of 10

== Sequence ==
For each $n \in \N$, the number of pairs of twin primes below $10^n$ is given by the following sequence:
:$2, 8, 35, 205, 1224, 8169, 58 \, 980, 440 \, 312, 3 \, 424 \, 506, 27 \, 412 \, 679, \ldots$
== Sources ==
* 1976: Tables Concerning Irregularities in the Distribution of Primes and Twin Primes to $10^{11}$
* : $8169$
* : $3,424,506$
Category:Twin Primes
Number of Type Rational r plus s Root 2 is Irrational

== Theorem ==
Let $r, s \in \Q$ be rational numbers.
Then $r + s \sqrt 2$ is irrational.
== Proof ==
 $t = r + s \sqrt 2$ be rational.
Then:
:$\sqrt 2 = \dfrac {t - r} s$ is also rational.
This contradicts the fact that Square Root of 2 is Irrational.
Hence the result by Proof by Contradiction.
== Sources ==
* : $\S 1$: Real Numbers: Exercise $\S 1.20 \ (4)$
Category:Real Analysis
Number of Ways of Seating People at Circular Table

== Theorem ==
Let there be $n$ people to be seated at a round table.
Let $N$ be the number of different ways to seat those $n$ people.
Note that if person $A$ is seated between person $B$ and person $C$, then having $B$ on the left and $C$ on the right is considered different from having $B$ on the right and $C$ on the left.
Then:
:$N = \paren {n - 1}!$
== Proof ==
From Number of Permutations of All Elements, the number of different ways of arranging $n$ people in line is $n!$.
However, a round table has no beginning and end.
Hence the vital factor is the arrangement relative to an arbitrary given person.
So we fix one person, and arrange the other $n - 1$ people relative to that person.
From Number of Permutations of All Elements, there are $\paren {n - 1}!$ ways to arrange those people. 
== Examples ==
== Sources ==
* : $\text I$. Algebra: Permutations and Combinations: The number of ways of seating $n$ people at a circular table
Category:Number of Ways of Seating People at Circular Table
Category:Combinatorics
Number of Ways of Seating People at Circular Table/Examples/8

== Example of Use of Number of Ways of Seating People at Circular Table ==
$8$ people can be seated at a circular table in $5040$ ways.
== Proof ==
From Number of Ways of Seating People at Circular Table, $n$ people can be so seated in $\paren {n - 1}!$ ways, where $!$ denotes the factorial function.
Hence $8$ people can be seated at a circular table in $\paren {8 - 1}! = 5040$ ways.
== Sources ==
* : $\text I$. Algebra: Permutations and Combinations: Exercises $\text I$: $1$
Category:Number of Ways of Seating People at Circular Table
Number of Ways of Threading Beads on a Loop of Wire

== Theorem ==
Let there be $n$ beads to be threaded on a circular loop of wire.
Let $N$ be the number of different ways to thread those $n$ beads.
Then:
:$N = \dfrac {\paren {n - 1}!} 2$
== Proof ==
This is the same as Number of Ways of Seating People at Circular Table, except that having threaded the beads in one arrangement, you can get another arrangement by flipping the ring of beads over.
From Number of Ways of Seating People at Circular Table, the number of arrangements of $n$ beads on a ring, without flipping it over, is $\paren {n - 1}!$.
But every arrangement can be paired with exactly one other arrangement such that those two arrangements are effectively the same.
Hence the result.
== Sources ==
* : $\text I$. Algebra: Permutations and Combinations: The number of ways of threading $n$ beads on a wire
Category:Combinatorics
Number of Ways to Tile Standard Chessboard with Dominoes

== Theorem ==
The number of ways to tile a standard ($8 \times 8$) chessboard with dominoes is $12 \, 988 \, 816$.
== Proof ==
Follows directly from Kasteleyn's Formula for $m = n = 8$:
:$\ds \prod_{j \mathop = 1}^{\ceiling {\frac 8 2} } \prod_{k \mathop = 1}^{\ceiling {\frac 8 2} } \paren {4 \cos^2 \frac {\pi j} {8 + 1} + 4 \cos^2 \frac {\pi k} {8 + 1} } = 12 \, 988 \, 816$
== Sources ==
* : $12,988,816$
* : Chapter $1$: $1.1$ Chessboard Problems
Category:Chessboard Tilings
Category:Chessboard Puzzles
Category:Dominoes
Category:Recreational Mathematics
Number of form 28000...0007 is Divisible by 7

== Theorem ==
Let $x$ be a natural number in the form:
:$\sqbrk {28 \underbrace {000 \cdots 0}_n 7}_{10}$ 
Then $x$ is divisible by $7$.
== Proof 1 ==
== Proof 2 ==
Category:Number of form 28000...0007 is Divisible by 7
Category:Divisibility by 7
Number of form 4666...6669 is Divisible by 7

== Theorem ==
Let $x$ be a natural number in the form:
:$\sqbrk {4 \underbrace {666 \cdots 6}_n 9}_{10}$ 
Then $x$ is divisible by $7$.
== Proof ==
We have:
The sum of the digits in $14 \times 10^{n + 1} + 7$ is in fact $12 = 1 + 4 + 7$ which is a multiple of $3$.
Hence, by Divisibility by $3$, $\dfrac {14 \times 10^{n + 1} + 7} 3$ is indeed an integer.
Then:
Category:Divisibility by 7
Number of k-Cycles in Symmetric Group

== Theorem ==
Let $n \in \N$ be a natural number.
Let $S_n$ denote the symmetric group on $n$ letters.
Let $k \in N$ such that $k \le n$.
The number of elements $m$ of $S_n$ which are $k$-cycles is given by:
:$m = \paren {k - 1}! \dbinom n k = \dfrac {n!} {k \paren {n - k}!}$
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : Chapter $2$: The Symmetric Groups: $\S 79 \alpha$
Category:Cyclic Permutations
Category:Symmetric Groups
Category:Combinatorics
Category:Number of k-Cycles in Symmetric Group
Number of k-Cycles in Symmetric Group/Proof 1

== Theorem ==
== Proof ==
Let $m$ be the number of elements of $S_n$ which are $k$-cycles.
From Cardinality of Set of Subsets, there are $\dfrac {n!} {k! \paren {n - k}!}$ different ways to select $k$ elements of $\set {1, 2, \ldots, n}$.
From Number of k-Cycles on Set of k Elements, each of these $\dfrac {n!} {k! \paren {n - k}!}$ sets with $k$ elements has $\paren {k - 1}!$ $k$-cycles.
It follows from Product Rule for Counting that:
Category:Number of k-Cycles in Symmetric Group
Number of k-Cycles in Symmetric Group/Proof 2

== Theorem ==
== Proof ==
Suppose $n \ge k$, and consider the number of $k$-cycles in $S_n$.
A $k$-cycle can be represented by a selection of $k$ elements from $n$ without any repeats.
From Number of Permutations, the number of permutations of $k$ elements from $n$ possible elements is $\dfrac {n!} {\paren {n - k}!}$.
However, each such string is merely a representation of an $k$-cycle; the $k$-cycle itself does not depend on the starting elements in the string.
Since there are $k$ possible starting elements, we must divide this number by $k$.
Hence, the number of $m$-cycles is
:$\dfrac {n!} {k \paren {n - k}!}$
Category:Number of k-Cycles in Symmetric Group
Number of k-Cycles on Set of k Elements

== Theorem ==
Let $k \in \N$ be a natural number.
Let $S_k$ denote the symmetric group on $k$ letters.
The number of elements of $S_k$ which are $k$-cycles is $\paren {k - 1}!$.
== Proof ==
Let $\N_k$ denote the set $\set {1, 2, \ldots, k}$.
Let $\pi$ be a $k$-cycle in $S_k$.
By definition, $\pi$ moves all $k$ elements of $\set {1, 2, \ldots, k}$.
Let $j \in \N_k$ be an arbitrary element of $\N_k$.
By definition of $k$-cycle, $k$ is the smallest integer such that $\map {\pi^k} j = j$.
Thus for each $r \in \set {1, 2, \ldots, k}$, $\map {\pi^r} j$ is a different element of $\N_k$.
That is:
:$\set {\map {\pi^r} j: r \in \N_k} = \N_k$
Let $\pi$ and $\rho$ be $k$-cycles in $S_k$ such that $\pi \ne \rho$.
From the above, the distinguishing feature of $\rho$ and $\pi$ is the order of elements of $\N_k \setminus \set j$ that $\map {\pi^r} j$ and $\map {\rho^r} j$ move $j$ to for $1 \le r < k$.
Each different ordered $\paren {k - 1}$-tuple of $\N_k \setminus \set j$ corresponds to a different $k$-cycle of $S_k$
Thus there are as many $k$-cycles of $S_k$ as there are different ordered $\paren {k - 1}$-tuples of $\N_k \setminus \set j$.
From Number of Permutations, there are $\paren {k - 1}!$ different ordered $\paren {k - 1}$-tuples of $\N_k \setminus \set j$.
Hence the result.
Category:Cyclic Permutations
Number times Recurring Part of Reciprocal gives 9-Repdigit

== Theorem ==
Let a (strictly) positive integer $n$ be such that the decimal expansion of its reciprocal has a recurring part of period $d$ and no non-recurring part.
Let $m$ be the integer formed from the $d$ digits of the recurring part.
Then $m \times n$ is a $d$-digit repdigit number consisting of $9$s.
=== Generalization ===
== Proof ==
Let $x = \dfrac 1 n = \sqbrk {0. mmmm \dots}$.
Then:
:$10^d x = \sqbrk {m.mmmm \dots}$
Therefore:
which is the $d$-digit repdigit number consisting of $9$s.
== Examples ==
== Sources ==
* : $142,857$
* : $142,857$
Category:Number times Recurring Part of Reciprocal gives 9-Repdigit
Category:Repdigit Numbers
Category:Reciprocals
Number times Recurring Part of Reciprocal gives 9-Repdigit/Examples/37

== Example of Use of Number times Recurring Part of Reciprocal gives 9-Repdigit ==
== Proof ==
Let $x = \dfrac 1 {37} = 0 \cdotp \dot 0 2 \dot 7$.
Then $n = 37$; $m = 27$ and $d = 3$
Therefore:
:$27 \times 37 = 10^3 - 1$
Category:37
Category:Number times Recurring Part of Reciprocal gives 9-Repdigit
Number times Recurring Part of Reciprocal gives 9-Repdigit/Examples/41

== Example of Use of Number times Recurring Part of Reciprocal gives 9-Repdigit ==
== Proof ==
Let $x = \dfrac 1 {41} = 0 \cdotp \dot 0 243 \dot 9$.
Then $n = 41$; $m = 2439$ and $d = 5$
Therefore:
:$2439 \times 41 = 10^5 - 1$
Category:41
Category:Number times Recurring Part of Reciprocal gives 9-Repdigit
Number times Recurring Part of Reciprocal gives 9-Repdigit/Generalization

== Theorem ==
Let a (strictly) positive integer $n$ be such that the decimal expansion of its reciprocal has a recurring part of period $d$ and no non-recurring part.
Let $m$ be the integer formed from the $d$ digits of the recurring part.
Let $M$ be an arbitrary integer.
Then:
:$M \equiv \sqbrk {mmm \dots m} \pmod {10^c}$
for some positive integer $c$, :
:$M \times n \equiv -1 \pmod {10^c}$
In other words, the last $c$ digits of $M$ coincide with that of $\sqbrk {mmm \dots m}$ the last $c$ digits of $M \times n$ are all $9$s.
== Proof ==
$\sqbrk {mmm \dots m}$ can be expressed as:
:$\ds \sum_{k \mathop = 0}^{K - 1} m 10^{k d}$
for some sufficiently large $K > \dfrac c d$.
=== Sufficient Condition ===
Suppose:
:$M \equiv \sqbrk {mmm \dots m} \pmod {10^c}$
We have:
which is the $c$-digit repdigit number consisting of $9$s.
Therefore the last $c$ digits of $M \times n$ are all $9$'s.
=== Necessary Condition ===
It is observed that all implications but the first one:
:$M \equiv \sqbrk {mmm \dots m} \pmod {10^c} \implies M \times n \equiv \sqbrk {mmm \dots m} \times n \pmod {10^c}$
trivially reverses.
By Common Factor Cancelling in Congruence/Corollary 2, the converse of the above is true if $n$ and $10^c$ are coprime.
We have:
Hence the result.
== Example ==
:$\dfrac 1 {27} = 0 \cdotp \dot 0 3 \dot 7$
The last $4$ digits of $17 \, 037$ coincides with that of $037 \, 037$.
We have:
:$17 \, 037 \times 27 = 459 \, 999$
It is observed that the last $4$ digits of $459 \, 999$ are all $9$'s.
Category:Number times Recurring Part of Reciprocal gives 9-Repdigit
Number times Recurring Part of Reciprocal gives 9-Repdigit/Mistake/First Edition

== Source Work ==
:
:The Dictionary
::$142,857$
== Mistake ==
:This is a property of all the periods of repeating decimals. If the period of $n$ is multiplied by $n$, the result is as many $9$s as there are digits in $n$.
== Correction ==
This should read:
:This is a property of all the periods of reciprocals of (strictly) positive integers. If the period of $1 / n$ is multiplied by $n$, the result is as many $9$s as there are digits in the period of $1 / n$.
In of $1997$, this has been partially corrected to:
:This is a property of all the periods of repeating decimals. If the period of $n$ is multiplied by $n$, the result is as many $9$s as there are digits in the period of $1 / n$.
But even then, this appears not to be true.
The number $6$, for example, does not have this property.
:$\dfrac 1 6 = 0 \cdotp 1 \dot 6$
but:
:$6 \times 6 = 36$
which is not equal to $9$ as the theorem states ought to be the case.
Research is ongoing as to what the theorem should actually say.
== Sources ==
* : $142,857$
Category:Mistakes/Curious and Interesting Numbers
Number times Recurring Part of Reciprocal gives 9-Repdigit/Mistake/Second Edition

== Source Work ==
:
:The Dictionary
::$142,857$
== Mistake ==
:This is a property of all the periods of repeating decimals. If the period of $n$ is multiplied by $n$, the result is as many $9$s as there are digits in the period of $1/n$.
In of $1986$, this statement is even more incorrect:
:This is a property of all the periods of repeating decimals. If the period of $n$ is multiplied by $n$, the result is as many $9$s as there are digits in $n$.
so clearly some work has been done on it. However, it is still not completely accurate.
== Correction ==
This should read:
:This is a property of all the periods of reciprocals of (strictly) positive integers. If the period of $1/n$ is multiplied by $n$, the result is as many $9$s as there are digits in the period of $1/n$.
But even then, this appears not to be true.
The number $6$, for example, does not have this property.
:$\dfrac 1 6 = 0 \cdotp 1 \dot 6$
but:
:$6 \times 6 = 36$
which is not equal to $9$ as the theorem states ought to be the case.
Research is ongoing as to what the theorem should actually say.
== Sources ==
* : $142,857$
Category:Mistakes/Curious and Interesting Numbers
Number to Reciprocal Power is Decreasing

== Theorem ==
The real sequence $\sequence {n^{1/n} }$ is decreasing for $n \ge 3$.
== Proof ==
We want to show that $\paren {n + 1}^{1 / \paren {n + 1} } \le n^{1/n}$.
Thus:
But from One Plus Reciprocal to the Nth:
:$\paren {1 + \dfrac 1 n}^n < 3$
Thus the reversible chain of implication can be invoked and we see that $\paren {n + 1}^{1 / \paren {n + 1} } \le n^{1/n}$ when $n \ge 3$.
So $\sequence {n^{1 / n} }$ is decreasing for $n \ge 3$.
== Sources ==
* : $\S 4$: Convergent Sequences: Exercise $\S 4.20 \ (6)$
Category:Real Analysis
Category:Reciprocals
Number which is Square and Cube Modulo 7

== Theorem ==
Let $n \in \Z$ be an integer.
Let $n$ be both a square and a cube at the same time.
Then either:
:$n \equiv 0 \pmod 7$
or:
:$n \equiv 1 \pmod 7$
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : Chapter $2$: Divisibility Theory in the Integers: $2.1$ The Division Algorithm: Problems $2.1$: $5$
Category:Number which is Square and Cube Modulo 7
Category:Cube Numbers
Category:Square Numbers
Category:Modulo Arithmetic
Number which is Square and Cube Modulo 7/Proof 1

== Theorem ==
== Proof ==
Let $n = r^2 = s^3$ for some $r, s \in \Z$.
Then:
:$n = \paren {m^2}^3 = \paren {m^3}^2 = m^6$
for some $m \in \Z$
There are $7$ cases to consider:
:$(0): \quad m \equiv 0 \pmod 7$: we have $m = 7 k$
:$(1): \quad m \equiv 1 \pmod 7$: we have $m = 7 k + 1$
:$(2): \quad m \equiv 2 \pmod 7$: we have $m = 7 k + 2$
:$(3): \quad m \equiv 3 \pmod 7$: we have $m = 7 k + 3$
:$(4): \quad m \equiv 4 \pmod 7$: we have $m = 7 k + 4$
:$(5): \quad m \equiv 5 \pmod 7$: we have $m = 7 k + 5$
:$(6): \quad m \equiv 6 \pmod 7$: we have $m = 7 k + 6$
Using Congruence of Powers throughout, we make use of:
:$x \equiv y \pmod 7 \implies x^j \equiv y^j \pmod 7$
for $j \in \Z_{>0}$.
First the easy cases:
It is sufficient to investigate the congruence modulo $7$ of the integers from $2$ to $6$.
We have:
Hence the result.
Category:Number which is Square and Cube Modulo 7
Number which is Square and Cube Modulo 7/Proof 2

== Theorem ==
== Proof ==
Let $n = r^2 = s^3$ for some $r, s \in \Z$.
Then:
:$n = \paren {m^2}^3 = \paren {m^3}^2 = m^6$
for some $m \in \Z$
For $m \equiv 0 \pmod 7$:
From Fermat's Little Theorem, we have our prime number $p = 7$.
We also have our $m \in \Z_{>0}$ such that $7$ is not a divisor of $m$.
Then:
Hence the result.
Category:Number which is Square and Cube Modulo 7
Number which is Sum of Subfactorials of Digits

== Theorem ==
The only integer which is the sum of the subfactorials of its digits is $148 \, 349$:
:$148 \, 349 = \mathop !1 + \mathop !4 + \mathop !8 + \mathop !3 \mathop + \mathop !4 \mathop + \mathop !9$
== Proof ==
We have:
A computer search can verify solutions under $10^6$ (that is, with no more than $6$ digits) in seconds.
Let $n$ be a $k$-digit number, for $k \ge 7$.
Then the sum of the subfactorials of its digits is not more than $\mathop !9 \times k$.
But we have:
So no more numbers have this property.
== Historical Note ==
== Sources ==
* : $148,349$
* : $148,349$
Category:Subfactorials
Category:148,349
Number whose Half is Odd is Even-Times Odd

== Theorem ==
Let $a \in \Z$ be an integer such that $\dfrac a 2$ is an odd integer.
Then $a$ is even-times odd.
== Proof ==
By definition:
:$a = 2 r$
where $r$ is an odd integer.
Thus:
:$a$ has an even divisor
and:
:$a$ has an odd divisor.
Hence the result by definition of even-times odd integer.
As $r$ is an odd integer it follows that $2 \nmid r$.
Thus $a$ is not divisible by $4$.
Hence $a$ is not even-times even.
== Sources ==
* : Book $\text{IX}$. Propositions
Category:Euclidean Number Theory