nyuuzyou/wikis huggingface dataset:
Smallest Set of Weights for Two-Pan Balance

== Classic Problem ==
Consider a balance for determining the weight of a physical object.
Let this balance be such that weights may be placed in either of the two pans.
What is the smallest set of weights needed to weigh any given integer weight up to a given amount?
== Solution ==
A set of weights up to $3^m$ in the sequence $\sequence {3^n}$:
:$1, 3, 9, 27, \ldots$
allows one to weigh any given integer weight up to $\dfrac {3^{m + 1} - 1} 2$.
== Proof ==
Place the item to be weighed in the left hand pan of the balance.
Let it weigh $n$.
Let $n$ be expressed in balanced ternary representation.
From Representation of Integers in Balanced Ternary, $n$ can be uniquely so represented.
With $m$ digits, we can count up to $\dfrac {3^{m + 1} - 1} 2$.
We use the balanced ternary representation to model how to place the weights.
Let the digits of $n$ in such a representation be numbered $0, 1, \ldots, m$ from the to the .
The $k$th digit represents the weight which weighs $3^k$.
When the $k$th digit is $1$, place weight $3^k$ in the right hand pan.
When the $k$th digit is $\underline 1$, place weight $3^k$ in the left hand pan.
When the $k$th digit is $0$, place weight $3^k$ in neither pan.
This will make the balance level.
== Examples ==
== Also see ==
* Smallest Set of Weights for One-Pan Balance
== Historical Note ==
== Sources ==
* 
* 
* : $31$
* : Bachet: $108$
* : $31$
Category:Classic Problems
Category:Ternary Notation
Category:Smallest Set of Weights for Two-Pan Balance
Smallest Set of Weights for Two-Pan Balance/Examples/40

== Examples of Smallest Set of Weights for Two-Pan Balance ==
Consider a balance such that weights may be placed in either or both of the pans.
Let $S$ be the smallest set of weights needed to weigh any given integer weight up to $40$ units.
Then $\size S = 4$.
== Proof ==
From Smallest Set of Weights for Two-Pan Balance, a set of $4$ weights in the sequence $\sequence {3^n}$:
:$1, 3, 9, 27$
allows one to weigh any given integer weight up to $\dfrac {3^4 - 1} 2 = 40$.
Hence the result.
== Sources ==
* 
* : $\text {1-2}$ The Basis Representation Theorem: Exercise $5$
* : Bachet: $108$
Category:Smallest Set of Weights for One-Pan Balance
Smallest Set of Weights for Two-Pan Balance/Mistake

== Source Work ==
:
:The Dictionary
::$31$
== Mistake ==
:Using both pans, the solution is similar, but now relies on expressing the weight as the sum and difference of powers of $3$. With the weights $1$, $3$, $9$ and $27$ it is possible to weight up to $40$. In general the weights up to $3$ will weigh up to a maximum of $\frac 1 2 \paren {3^{n + 1} - 1}$.
== Correction ==
In that last sentence there is a typo.
It should say:
: ... the weights up to $3^n$ will weigh ...
This typo does not exist in the first edition of .
== Sources ==
* : $31$
Category:Mistakes/Curious and Interesting Numbers
Smallest Seventh Power which is Sum of 8 other Seventh Powers

== Theorem ==
The smallest seventh power that can be expressed as the sum of $8$ other seventh powers is $102^7$:
:$102^7 = 12^7 + 35^7 + 53^7 + 58^7 + 64^7 + 83^7 + 85^7 + 90^7$
== Proof ==
== Sources ==
* : $102$
* : $102$
Category:Seventh Powers
Category:102
Smallest Simple Graph with One Vertex Adjacent to All Others

== Theorem ==
Let $G = \struct {V, E}$ be a simple graph of order $n$.
Let $G$ be the simple graph with the smallest size such that one vertex is adjacent to all other vertices of $G$.
Then $G$ is the star graph of order $n$ and is of size $n - 1$.
== Proof ==
In order for the one vertex in question to be adjacent to all the others, it needs to be incident to $n - 1$ edges.
This is the smallest number of edges required.
Hence $G$ is a star graph.
From Size of Star Graph, the size of $G$ is thus $n - 1$.
== Sources ==
* : Chapter $1$: Mathematical Models: $\S 1.3$: Graphs: Problem $21$
Category:Graph Theory
Smallest Square Inscribed in Two Pythagorean Triangles

== Theorem ==
The smallest square with integer sides that can be inscribed within two different Pythagorean triangles so that one side of the square lies on the hypotenuse has side length $780$.
The two Pythagorean triangles in question have side lengths $\tuple {1443, 1924, 2405}$ and $\tuple {1145, 2748, 2977}$.
== Proof ==
By Inscribed Squares in Right-Angled Triangle/Side Lengths/Side Lies on Hypotenuse:
:For a Pythagorean triangle with side lengths $a, b, c$, the required inscribed square has side length given by:
::$\dfrac {abc}{ab + c^2}$
For primitive Pythagorean triples, $a, b, c$ are pairwise coprime, so the above fraction is in canonical form.
In other words, if the required side length is an integer, the triangle cannot be primitive, and this side length would be equal to some multiple of the product $abc$ of its primitive version.
Therefore in order to find the smallest such square, we would need to compare and find two sets of primitive Pythagorean triples such that their product would have a sufficiently small lowest common multiple.
The $\tuple {3, 4, 5}$ triple has a product of $60$.
The $\tuple {5, 12, 13}$ triple has a product of $780$.
These two products have a lowest common multiple of $780$.
Since $\sqrt [3] {780} < 10$, there is no need to search any further for triangles with smallest side length greater than $10$.
The ones remaining are:
:$7, 24, 25$
:$8, 15, 17$
:$9, 40, 41$
which all have products greater than $780$.
Therefore the solution must be generated with the two smallest triangles.
The inscribed square side length for the $\tuple {3, 4, 5}$ triangle is:
:$\dfrac {60} {3 \times 4 + 5^2} = \dfrac {60} {37}$
so it must be enlarged $481$ times to have a side length of $780$.
The inscribed square side length for the $\tuple {5, 12, 13}$ triangle is:
:$\dfrac {780} {5 \times 12 + 13^2} = \dfrac {780} {229}$
so it must be enlarged $229$ times to have a side length of $780$.
After enlargement, we get the triples $\tuple {1443, 1924, 2405}$ and $\tuple {1145, 2748, 2977}$, which is the result we have.
== Historical Note ==
== Sources ==
* : $780$
Category:Pythagorean Triangles
Smallest Square which is Sum of 3 Fourth Powers

== Theorem ==
The smallest positive integer whose square is the sum of $3$ fourth powers is $481$:
:$481^2 = 12^4 + 15^4 + 20^4$
== Proof ==
The smallest solution must be fourth powers of coprime integers, otherwise dividing by their GCD would yield a smaller solution.
By Fermat's Right Triangle Theorem, $x^4 + y^4 = z^2$ has no solutions.
Thus none of the fourth powers is $0$.
Consider the equation $x^4 + y^4 + z^4 = n^2$.
By Square Modulo 4, since fourth powers are squares, at most one of $x, y, z$ is odd.
If none of them are odd, the numbers are not coprime.
Therefore there is exactly one odd number.
By Fermat's Little Theorem, for $3 \nmid a$:
:$a^2 \equiv 1 \pmod 3$
Therefore by Congruence of Powers:
:$a^4 \equiv 1^2 \equiv 1 \pmod 3$
For $3 \divides a$:
:$a^2 \equiv a^4 \equiv 0 \pmod 3$
Therefore we cannot have two of $x, y, z$ not divisible by $3$.
By our coprimality condition we cannot have $3 \divides x, y, z$.
Hence either $3 \nmid x, y, z$ or exactly one of $x, y, z$ is not divisible by $3$.
Similarly, by Fermat's Little Theorem, for $5 \nmid a$:
:$a^4 \equiv 1 \pmod 5$
For $5 \divides a$:
:$a^4 \equiv 0 \pmod 5$
By Square Modulo 5:
:$a^2 \equiv 0, 1, 4 \pmod 5$
Therefore there is exactly one number not divisible by $5$.
Since $\sqrt {481} < 22$, we only need to check fourth powers of numbers less than $22$, satisfying the criteria above.
Since there are exactly two multiples of $2$ and $5$ among $x, y, z$, at least one of them must be a multiple of $10$.
Suppose $5 \nmid x$, $5 \divides y$ and $10 \divides z$.
=== Case $1$: $z = 10$ ===
The case $y = 20$ will be equivalent to Case $2$.
Suppose $y = 15$.
By the criteria above, $x$ is divisible by $2$ and $3$.
Also $x < 22$, so we only need to check $x = 6, 12, 18$.
Suppose $y = 10$.
By the criteria above, $x$ is not divisible by $2$ or $3$.
Also $x < 22$, so we only need to check $x = 1, 7, 11, 13, 17, 19$.
Suppose $y = 5$.
By the criteria above, $x$ is divisible by $2$ but not $3$.
Also $x < 22$, so we only need to check $x = 2, 4, 8, 14, 16$.
=== Case $2$: $z = 20$ ===
Suppose $y = 20$. Then $x^4 + y^4 + z^4 > 20^4 + 20^4 = 320 \, 000 > 481^2$.
Suppose $y = 15$.
By the criteria above, $x$ is divisible by $2$ and $3$.
Also $x < \sqrt [4] {481^2 - 20^4 - 15^4} = 12$, so we only need to check $x = 6$.
Suppose $y = 10$.
By the criteria above, $x$ is not divisible by $2$ or $3$.
Also $x < \sqrt [4] {481^2 - 20^4} < 17$, so we only need to check $x = 1, 7, 11, 13$.
Suppose $y = 5$.
By the criteria above, $x$ is divisible by $2$ but not $3$.
Also $x < \sqrt [4] {481^2 - 20^4} < 17$, so we only need to check $x = 2, 4, 8, 14, 16$.
And we see that none of the above cases yield any smaller solutions.
Hence the result.
== Sources ==
* : $481$
Category:Square Numbers
Category:Fourth Powers
Smallest Strictly Positive Rational Number does not Exist

== Theorem ==
There exists no smallest element of the set of strictly positive rational numbers.
== Proof ==
 $x = \dfrac p q$ is the smallest strictly positive rational number.
By definition of strictly positive:
:$0 < \dfrac p q$
Let us calculate the mediant of $0$ and $\dfrac p q$:
:$\dfrac 0 1 < \dfrac {0 + p} {1 + q} < \dfrac p q$
The inequalities follow from Mediant is Between.
Thus $\dfrac p {1 + q}$ is a strictly positive rational number which is smaller than $\dfrac p q$.
Thus $\dfrac p q$ cannot be the smallest strictly positive rational number.
The result follows by Proof by Contradiction.
== Sources ==
* : $\S 1.2.2$: Numbers, Powers, and Logarithms: Exercise $1$
Category:Rational Numbers
Smallest Strong Fibonacci Pseudoprime of Type I

== Theorem ==
The smallest strong Fibonacci pseudoprime of type I is $443 \, 372 \, 888 \, 629 \, 441$.
== Proof ==
Let $N := 443 \, 372 \, 888 \, 629 \, 441$, to save writing it out in full each time.
From the definition of a strong Fibonacci pseudoprime of type I:
We have that:
:$N = 17 \times 31 \times 41 \times 43 \times 89 \times 97 \times 167 \times 331$
Of these, we see that:
Thus there are $4$ (an even number) of prime factors of $N$ of the form $4 n - 1$.
We note that:
Thus for all $p \divides N$, we have that $\paren {p^2 - 1} \divides \paren {N - 1}$.
From Difference of Two Squares we have that:
:$p^2 - 1 = \paren {p + 1} \paren {p - 1}$
and so for all $p \divides N$:
:$\paren {p - 1} \divides \paren {N - 1}$
We also have that $N$ is square-free.
Thus by Korselt's Theorem, $N$ is a Carmichael number.
Now we also have that:
Thus we have that:
:$\forall p \divides N: 2 \paren {p + 1} \divides \paren {N - 1}$
This is actually stronger than the conditions for which $N$ is a strong Fibonacci pseudoprime of type I:
It can be established by an exhaustive search that there are no smaller Carmichael numbers with this property.
Hence the result.
== Sources ==
* 
* : $443,372,888,629,441$
Category:Strong Fibonacci Pseudoprimes
Category:443,372,888,629,441
Smallest Sum of 2 Lucky Numbers in n Ways

== Sequence ==
The sequence of positive integers which can be expressed as the sum of $2$ distinct lucky numbers in $n$ different ways begins:
:$4, 10, 16, 34, 46, 144, 76, 112, 100, 148, 166, 136, 202, 226, 238, 268, 298, 304, 310, 352, 400, 430, 490, \ldots$
== Proof ==
The technique is to create all sums of $2$ distinct lucky numbers and count the number of times each sum occurs.
To reach the term $46$, for example, is it sufficient to go as high as building the sums up to the lucky number immediately less than $46$.
The sequence of lucky numbers begins:
:$1, 3, 7, 9, 13, 15, 21, 25, 31, 33, 37, 43, 49, \ldots$
and so:
The first $5$ terms of the sequence are seen to appear.
Category:Lucky Numbers
Smallest Titanic Palindromic Prime

== Theorem ==
The smallest titanic prime that is also palindromic is:
:$10^{1000} + 81 \, 918 \times 10^{498} + 1$
which can be written as:
:$1 \underbrace {000 \ldots 000}_{497} 81918 \underbrace {000 \ldots 000}_{497} 1$
== Proof ==
It remains to be demonstrated that it is the smallest such palindromic prime with $1000$ digits or more.
By 11 is Only Palindromic Prime with Even Number of Digits, there are no palindromic primes with exactly $1000$ digits.
Hence such a prime must be greater than $10^{1000}$.
We need to check all numbers of the form:
:$1 \underbrace {000 \ldots 000}_{497} abcba \underbrace {000 \ldots 000}_{497} 1$
with $\sqbrk {abc} < 819$.
Using the Alpertron integer factorization calculator and the argument:
 x=0;x=x+1;x<820;10^1000+x*10^500+RevDigits(x/10+10^499,10)
it is verified that there are no primes in the range $\sqbrk {abc} < 819$.
Therefore the number above is the smallest titanic palindromic prime.
== Sources ==
* 
* : $10^{1000} + 81,918 \times 10^{498} + 1$
Category:Prime Numbers
Smallest Triplet of Consecutive Integers Divisible by Cube

== Theorem ==
The smallest sequence of triplets of consecutive integers each of which is divisible by a cube greater than $1$ is:
:$\tuple {1375, 1376, 1377}$
== Proof ==
We will show that:
is the smallest such triplet.
Each number in such triplets of consecutive integers is divisible by a cube of some prime number.
Only $2, 3, 5, 7, 11$ are less than $\sqrt [3] {1377}$.
Since the numbers involved are small, we can check the result by brute force.
For general results one is encouraged to use the Chinese Remainder Theorem.
=== Case $1$: a number is divisible by $11^3$ ===
The only multiple of $11^3$ less than $1377$ is $1331$, and:
Since neither $1330$ nor $1332$ are divisible by a cube of some prime number, $1331$ is not in a triplet.
=== Case $2$: a number is divisible by $7^3$ ===
The only multiples of $7^3$ less than $1377$ are $343, 686, 1029, 1372$, and:
Hence none of these numbers is in a triplet.
=== Case $3$: the numbers are divisible by $2^3, 3^3, 5^3$ respectively ===
Let $n = k \times 5^3$.
We show that $k$ cannot be divisible by $3$ or $4$.
Suppose $3 \divides k$.
Then none of $n \pm 1, n \pm 2$ are divisible by $3$, and consequently $3^3$.
Suppose $4 \divides k$.
Then none of $n \pm 1, n \pm 2$ are divisible by $4$, and consequently $2^3$.
The only multiples of $5^3$ less than $1377$ are $125, 250, 375, 500, 625, 750, 875, 1000, 1125, 1250, 1375$, and we eliminate $375, 500, 750, 1000, 1125$ due to the reasons above.
Now:
Hence none of these numbers is in a triplet.
== Historical Note ==
== Sources ==
* : $1375$
* : $1375$
Category:Cube Numbers
Smallest Triplet of Consecutive Integers each Divisible by Fourth Power

== Theorem ==
This triplet of consecutive integers has the property that each of them is divisible by a fourth power:
:$33 \, 614, 33 \, 615, 33 \, 616$
This is the smallest such triplet.
== Proof ==
Each number in such triplets of consecutive integers is divisible by a fourth power of some prime number.
Only $2, 3, 5, 7, 11, 13$ are less than $\sqrt [4] {33 \, 616}$.
=== Case $1$: a number is divisible by $13^4$ ===
The only multiple of $13^4$ less than $33 \, 616$ is $28 \, 561$, and:
Since neither $28 \, 559$ nor $28 \, 562$ are divisible by a fourth power of some prime number, $28 \, 561$ is not in a triplet.
=== Case $2$: a number is divisible by $11^4$ ===
The only multiples of $11^4$ less than $33 \, 616$ are $14 \, 641$ and $29 \, 282$, and:
Hence none of these numbers is in a triplet.
=== Case $3$: a number is divisible by $7^4$ ===
There are $14$ multiples of $7^4$ less than $33 \, 616$, and:
Hence the smallest valid triplet is $\tuple {33 \, 614, 33 \, 615, 33 \, 616}$.
=== Case $4$: the numbers are divisible by $2^4, 3^4, 5^4$ respectively ===
We utilise Chinese Remainder Theorem.
We are to solve the system of linear congruences:
:$x \equiv b_1 \pmod {2^4}$
:$x \equiv b_2 \pmod {3^4}$
:$x \equiv b_3 \pmod {5^4}$
where $\set {b_1, b_2, b_3} = \set {1, 0, -1}$.
First note the linear congruences:
:$3^4 5^4 x \equiv 1 \pmod {2^4}$
:$2^4 5^4 x \equiv 1 \pmod {3^4}$
:$2^4 3^4 x \equiv 1 \pmod {5^4}$
have solutions $1, 46, 231$ respectively.
Thus our system of linear congruences has the solution:
Now we assign $\set {b_1, b_2, b_3}$ to $\set {1, 0, -1}$.
The solutions are:
:$\tuple {0, 1, -1}: 160 \, 624$
:$\tuple {0, -1, 1}: -160 \, 624 \equiv 649 \, 376$
:$\tuple {-1, 0, 1}: 248 \, 751$
:$\tuple {1, 0, -1}: -248 \, 751 \equiv 561 \, 249$
:$\tuple {-1, 1, 0}: 409 \, 375$
:$\tuple {1, -1, 0}: -409 \, 375 \equiv 400 \, 625$
and none of these solutions are less than $33 \, 616$.
== Historical Note ==
== Sources ==
* : $33,614$
Category:Fourth Powers
Smallest Triplet of Integers whose Product with Divisor Count are Equal

== Theorem ==
Let $\map {\sigma_0} n$ denote the divisor count function: the number of divisors of $n$.
The smallest set of $3$ integers $T$ such that $m \, \map {\sigma_0} m$ is equal for each $m \in T$ is:
:$\set {168, 192, 224}$
== Proof ==
== Sources ==
* 
* : $168$
Category:Divisor Count Function
Category:168
Category:192
Category:224
Smallest Triplet of Primitive Pythagorean Triangles with Same Area

== Theorem ==
The smallest set of $3$ primitive Pythagorean triangles which all have the same area are:
:the $4485-5852-7373$ triangle
:the $3059-8580-9109$ triangle
:the $1380-19 \, 019-19 \, 069$ triangle.
That area is $13 \, 123 \, 110$.
== Proof ==
We have that:
:the $4485-5852-7373$ triangle $T_1$ is Pythagorean
:the $3059-8580-9109$ triangle $T_2$ is Pythagorean
:the $1380-19 \, 019-19 \, 069$ triangle $T_3$ is Pythagorean.
Then from Area of Triangle, their areas $A_1$, $A_2$ and $A_3$ respectively are given by:
== Historical Note ==
== Sources ==
* : $13,123,110$
Category:Specific Numbers
Category:13,123,110
Smallest Triplet of Primitive Pythagorean Triangles with Same Area/Historical Note

== Historical Note on Smallest Triplet of Primitive Pythagorean Triangles with Same Area ==
The smallest triplet of primitive Pythagorean triangles with the same area was reported by as having been discovered by in $1945$.
== Sources ==
* : $13,123,110$
Category:Pythagorean Triples
Smallest n for which 2^n-3 is Divisible by n

== Theorem ==
The smallest positive integer $n$ such that $2^n - 3$ is divisible by $n$ is $4 \, 700 \, 063 \, 497$.
== Proof ==
== Sources ==
* : $4,679,307,774$
* : $\mathbf {A 10}$: Residues of powers of two.
Category:Number Theory
Smallest n for which 3 over n produces 3 Egyptian Fractions using Greedy Algorithm when 2 Sufficient

== Theorem ==
Consider proper fractions of the form $\dfrac 3 n$ expressed in canonical form.
Let Fibonacci's Greedy Algorithm be used to generate a sequence $S$ of Egyptian fractions for $\dfrac 3 n$.
The smallest $n$ for which $S$ consists of $3$ terms, where $2$ would be sufficient, is $25$.
== Proof ==
We have that:
But then we have:
By Condition for 3 over n producing 3 Egyptian Fractions using Greedy Algorithm when 2 Sufficient, we are to find the smallest $n$ such that:
:$n \equiv 1 \pmod 6$
:$\exists d: d \divides n$ and $d \equiv 2 \pmod 3$
The first few $n \ge 4$ which satisfies $n \equiv 1 \pmod 6$ are:
:$7, 13, 19, 25$
of which $7, 13, 19$ are primes, so they do not have a divisor of the form $d \equiv 2 \pmod 3$.
We see that $5 \divides 25$ and $5 \equiv 2 \pmod 3$.
Hence the result.
== Sources ==
* : Egyptian Fractions: $5$
Category:Fibonacci's Greedy Algorithm
Category:25
Smallest n needing 6 Numbers less than n so that Product of Factorials is Square

== Theorem ==
Let $n \in \Z_{>0}$ be a positive integer.
Then it is possible to choose at most $6$ positive integers less than $n$ such that the product of their factorials is square.
The smallest $n$ that actually requires $6$ numbers to be chosen is $527$.
== Proof ==
Obviously the product cannot be a square if $n$ is a prime.
For $n$ composite, we can write:
:$n = a b$
where $a, b \in \Z_{>1}$.
Then:
which is a square.
Hence no more than $6$ factorials is required.
To show that $527$ is the smallest that actually requires $6$, observe that:
If $n$ is itself square, then so is $n! \paren {n - 1}!$.
If $n$ is not square-free, write $n = a^2 b$, where $b$ is square-free.
Then $n! \paren {n - 1}! b! \paren {b - 1}!$ is square.
If $n$ is divisible by $2$, write $n = 2 m$.
Then $\paren {2 m}! \paren {2 m - 1}! \paren {m!} \paren {m - 1}! \paren {2!}$ is square.
If $n$ is divisible by $3$, write $n = 3 m$.
Then $\paren {3 m}! \paren {3 m - 1}! \paren {2 m}! \paren {2 m - 1}! \paren {3!}$ is square.
If $n$ is divisible by $5$, write $n = 5 m$.
Then $\paren {5 m}! \paren {5 m - 1}! \paren {m!} \paren {m - 1}! \paren {6!}$ is square.
If $n$ is divisible by $7$, write $n = 7 m$.
Then $\paren {7 m}! \paren {7 m - 1}! \paren {5 m}! \paren {5 m - 1}! \paren {7!}$ is square.
If $n$ is divisible by $11$, write $n = 11 m$.
Then $\paren {11 m}! \paren {11 m - 1}! \paren {7 m}! \paren {7 m - 1}! \paren {11!}$ is square.
The remaining numbers less than $527$ that are not of the above forms are:
:$221, 247, 299, 323, 377, 391, 403, 437, 481, 493$
Each of the following is a product of $5$ factorials which is square:
:$221! \, 220! \, 18! \, 11! \, 7!$
:$247! \, 246! \, 187! \, 186! \, 20!$
:$299! \, 298! \, 27! \, 22!$
:$323! \, 322! \, 20! \, 14! \, 6!$
:$377! \, 376! \, 29! \, 23! \, 10!$
:$391! \, 389! \, 24! \, 21! \, 17!$
:$403! \, 402! \, 33! \, 30! \, 14!$
:$437! \, 436! \, 51! \, 49! \, 28!$
:$481! \, 479! \, 38! \, 33! \, 22!$
:$493! \, 491! \, 205! \, 202! \, 7!$
== Sources ==
* 
* : $527$
* : $527$
* : $\mathbf {B 23}$: Equal products of factorials.
Category:527
Category:Factorials
Category:Square Numbers
Smallest n such that 6 n + 1 and 6 n - 1 are both Composite

== Theorem ==
The smallest positive integer $n$ such that $6 n + 1$ and $6 n - 1$ are both composite is $20$.
== Proof ==
Running through the positive integers in turn:
== Sources ==
* : $120$
* : $120$
Category:Prime Numbers
Category:20
Smirnov Metrization Theorem

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Then:
:$T$ is metrizable $T$ is paracompact and locally metrizable.
== Proof ==
== Sources ==
* : Chapter $6$: Metrization Theorems and Paracompactness: $\S41$: The Smirnov Metrization Theorem: Theorem: $42.1$
Category:Metrization Theorems
Category:Metrizable Topologies
Smith Numbers are Infinite in Number

== Theorem ==
There are infinitely many Smith numbers.
=== Lemma ===
=== Algorithm ===
They can be generated from a $9$-repdigit as follows:
:$(1): \quad$ Choose any $n \ge 2$ and factor the $9$-repdigit $m = 10^n - 1$.
:$(2): \quad$ Compute $\map {S_p} m$ and set $h = 9 n - \map {S_p} m$.
:$(3): \quad$ Solve $x = h - 7 b$ for $b, x \in \Z$, $2 \le x \le 8$.
:$(4): \quad$ Find $t \in \set {2, 3, 4, 5, 8, 7, 15}$ such that $\map {S_p} t = x$.
:$(5): \quad$ Then $M = t m \times 10^b$ is a Smith number.
== Proof ==
First we prove the the algorithm above does generate Smith numbers.
Let $n \ge 2$.
We have:
:$m = 10^n - 1 = 3 \times 3 \times R_n$
where $R_n$ is the repunit with $n$ digits.
We apply the Lemma, taking note that $r \ge 3$:
:$\map {S_p} m < 9 \map N m - 0.54 \times 3 = 9 n - 1.62$
Since both $\map {S_p} m$ and $9 n$ are integers, the inequality can be rewritten as:
:$h = 9 n - \map {S_p} m \ge 2$
By the Division Algorithm:
:$\exists! a, b \in \Z: \paren {h - 2} = 7 b + a, 0 \le a < 7$
Since $h - 2 \ge 0$, we must also have $b \ge 0$.
Take $x = a + 2$.
Then $2 \le x \le 8$ and:
:$h - 7 b = a + 2 = x$
Hence both $b, x$ exist and are within the desired range.
Note that:
:$\map {S_p} {\set {2, 3, 4, 5, 8, 7, 15} } = \set {2, 3, 4, 5, 6, 7, 8}$
so we can the corresponding value of $t \in \set {2, 3, 4, 5, 8, 7, 15}$ for each $2 \le x \le 8$ such that $\map {S_p} t = x$.
Since $b \ge 0$, $M = t m \times 10^b$ is an integer.
To show that $M$ is a Smith number, we need to show:
:$\map {S_p} M = \map S M$
We have:
Note that:
:$t \le 15 < 99 = 10^2 - 1 \le 10^n - 1 = m$
Hence we can apply Generalization of Multiple of Repdigit Base minus $1$:
and thus $\map {S_p} M = \map S M = 9 n$, showing that $M$ is indeed a Smith number.
Note that for each $n$, we can generate a Smith number that is greater than $m = 10^n - 1$.
To generate an infinite sequence of Smith numbers, we choose $n$ equal to the number of digits of $M$ previously generated.
Then the next Smith number will be strictly greater than the previous one, thus forming a strictly increasing infinite sequence of Smith numbers.
== Sources ==
* 
* : $4,937,775$
* 
Category:Smith Numbers
Category:Smith Numbers are Infinite in Number