avewright/proofwiki-math huggingface dataset:
Theorem: Second Principle of Recursive Definition

Let $\N$ be the natural numbers.
Let $T$ be a set.
Let $a \in T$.
For each $n \in \N_{>0}$, let $G_n: T^n \to T$ be a mapping.
Then there exists exactly one mapping $f: \N \to T$ such that:
:$\forall x \in \N: \map f x = \begin{cases} a & : x = 0 \\ \map {G_n} {\map f 0, \ldots, \map f n} & : x = n + 1 \end{cases}$

Proof:
Define $T^*$ to be the Kleene closure of $T$:
:$T^* := \ds \bigcup_{i \mathop = 1}^\infty T^i$
Note that, for convenience, the empty sequence is excluded from $T^*$.
Now define a mapping $\GG: T^* \to T^*$ by:
:$\map \GG {t_1, \ldots, t_n} = \tuple {t_1, \ldots, t_n, \map {G_n} {t_1, \ldots, t_n} }$
that is, extending each finite sequence $\tuple {t_1, \ldots, t_n}$ with the element $\map {G_n} {t_1, \ldots, t_n} \in T$.
By the Principle of Recursive Definition applied to $\GG$ and the finite sequence $\sequence a$, we obtain a unique mapping:
:$\FF: \N \to T^*: \map \FF x = \begin{cases} \sequence a & : x = 0 \\ \map \GG {\map \FF n} & : x = n + 1 \end {cases}$
Next define $f: \N \to T$ by:
:$\map f n = \text {the last element of $\map \FF n$}$
We claim that this $f$ has the sought properties, which will be proven by the Principle of Mathematical Induction.
We prove the following assertions by induction:
:$\map \FF n = \tuple {\map f 0, \map f 1, \ldots, \map f {n - 1}, \map {G_n} {\map f 0, \ldots, \map f {n - 1} } }$
:$\map f n = \map {G_n} {\map f 0, \ldots, \map f {n - 1} }$
For $n = 0$, these statements do not make sense, however it is immediate that $\map f 0 = \map {\operatorname {last} } {\sequence a} = a$.
For the base case, $n = 1$, we have:
{{begin-eqn}}
{{eqn | l = \map \FF 1
      | r = \map \GG {\sequence a}
}}
{{eqn | r = \tuple {a, \map {G_1} a}
}}
{{eqn | ll= \leadsto
      | l = \map f 1
      | r = \map {G_1} a
}}
{{end-eqn}}
Now assume that we have that:
:$\map \FF n = \tuple {\map f 0, \map f 1, \ldots, \map f {n - 1}, \map {G_n} {\map f 0, \ldots, \map f {n - 1} } }$
:$\map f n = \map {G_n} {\map f 0, \ldots, \map f {n - 1} }$
Then:
{{begin-eqn}}
{{eqn | l = \map \FF {n + 1}
      | r = \map \GG {\map \FF n}
}}
{{eqn | r = \map \GG {\map f 0, \ldots, \map f {n - 1}, \map {G_{n - 1} } {\map f 0,\ldots, \map f {n - 1} } }
      | c = Induction hypothesis on $\FF$
}}
{{eqn | r = \map \GG {\map f 0, \ldots, \map f {n - 1}, \map f n}
      | c = Induction hypothesis on $f$
}}
{{eqn | r = \tuple {\map f 0, \ldots, \map f n, \map {G_n} {\map f 0, \ldots, \map f n} }
      | c = Definition of $\GG$
}}
{{eqn | ll= \leadsto
      | l = \map f {n + 1}
      | r = \map {\operatorname {last} } {\map \FF {n + 1} }
}}
{{eqn | r = \map {G_n} {\map f 0, \ldots, \map f n}
}}
{{end-eqn}}
The result follows by the Principle of Mathematical Induction.
{{qed}}
Theorem: Condition for Repunits to be Coprime

Let $R_p$ and $R_q$ be repunit numbers with $p$ and $q$ digits respectively.
Then $R_p$ and $R_q$ are coprime {{iff}} $p$ and $q$ are coprime.

Proof:
=== Necessary Condition ===
We show the contrapositive.
Suppose $p$ and $q$ are not coprime.
Then there exists some $d > 1$ such that $p = d m$ and $q = d n$.
Then by Divisors of Repunit with Composite Index:
:$R_d \divides R_p$
and:
:$R_d \divides R_q$
where $\divides$ denotes divisibility.
That is, $R_d$ is a common divisor of $R_p$ and $R_q$.
Therefore $R_p$ and $R_q$ are not coprime.
Thus we have shown the contrapositive of the necessary condition.
{{qed|lemma}}
Theorem: Divisors of Repunit with Composite Index

Let $R_n$ be a repunit number with $n$ digits.
Let $n$ be composite such that $n = r s$ where $1 < r < n$ and $1 < s < n$.
Then $R_r$ and $R_s$ are both divisors of $R_n$.

Proof:
Let $n = r s$.
Then:
{{begin-eqn}}
{{eqn | l = R_n
      | r = \sum_{k \mathop = 0}^{n - 1} 10^k
      | c = Basis Representation Theorem
}}
{{eqn | r = \sum_{j \mathop = 0}^{s - 1} \paren {\sum_{k \mathop = 0}^{r - 1} 10^k} 10^{r j}
      | c = 
}}
{{eqn | r = \paren {\sum_{k \mathop = 0}^{r - 1} 10^k} \paren {\sum_{j \mathop = 0}^{s - 1}  10^{r j} }
      | c = 
}}
{{eqn | r = R_r \paren {\sum_{j \mathop = 0}^{s - 1}  10^{r j} }
      | c = 
}}
{{end-eqn}}
Similarly:
:$R_n = R_s \paren {\ds \sum_{j \mathop = 0}^{r - 1}  10^{s j} }$
{{qed}}
Thus, for example:
{{begin-eqn}}
{{eqn | l = R_6 = 111 \, 111
      | r = 11 \times 10 \, 101
}}
{{eqn | r = 111 \times 1001
}}
{{eqn | l = R_{10} = 1 \, 111 \, 111 \, 111
      | r = 11 \times 101 \, 010 \, 101
}}
{{eqn | r = 11111 \times 100 \, 001
}}
{{end-eqn}}
The pattern is clear.
Category:Repunits
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Theorem: Repunit in Base 9 is Triangular

Let $m$ be a repunit base $9$.
Then $m$ is a triangular number.

Proof:
Let $m$ be a repunit base $9$ with $n$ digits.
We have:
{{begin-eqn}}
{{eqn | l = m
      | r = \sum_{k \mathop = 0}^{n - 1} 9^k
      | c = Basis Representation Theorem
}}
{{eqn | r = \dfrac {9^n - 1} {9 - 1}
      | c = Sum of Geometric Sequence
}}
{{eqn | r = \dfrac {3^{2 n} - 1} 8
      | c = 
}}
{{eqn | r = \dfrac {\paren {3^n - 1} \paren {3^n + 1} } 8
      | c = Difference of Two Squares
}}
{{eqn | r = \dfrac {\paren {\paren {2 s + 1} - 1} \paren {\paren {2 s + 1} + 1} } 8
      | c = $3^n$ is odd, and so $3^n = 2 s + 1$ for some integer $s$
}}
{{eqn | r = \dfrac {2 s \paren {2 s + 2} } 8
      | c = simplifying
}}
{{eqn | r = \dfrac 1 2 \dfrac {2 s} 2 \dfrac {2 s + 2} 2
      | c = 
}}
{{eqn | r = \dfrac {s \paren {s + 1} } 2
      | c = 
}}
{{end-eqn}}
The result follows from Closed Form for Triangular Numbers.
{{qed}}
Theorem: Square of Small Repunit is Palindromic

The squares of repunits with up to $9$ digits are palindromic.

Proof:
{{begin-eqn}}
{{eqn | l = 1^2
      | r = 1
}}
{{eqn | l = 11^2
      | r = 121
      | c = 
}}
{{eqn | l = 111^2
      | r = 12 \, 321
      | c = 
}}
{{eqn | l = 1111^2
      | r = 1 \, 234 \, 321
      | c = 
}}
{{eqn | l = 11 \, 111^2
      | r = 123 \, 454 \, 321
      | c = 
}}
{{eqn | l = 111 \, 111^2
      | r = 12 \, 345 \, 654 \, 321
      | c = 
}}
{{eqn | l = 1 \, 111 \, 111^2
      | r = 1 \, 234 \, 567 \, 654 \, 321
      | c = 
}}
{{eqn | l = 11 \, 111 \, 111^2
      | r = 123 \, 456 \, 787 \, 654 \, 321
      | c = 
}}
{{eqn | l = 111 \, 111 \, 111^2
      | r = 12 \, 345 \, 678 \, 987 \, 654 \, 321
      | c = 
}}
{{end-eqn}}
but:
:$1 \, 111 \, 111 \, 111^2 = 1 \, 234 \, 567 \, 900 \, 987 \, 654 \, 321$
{{qed}}
Theorem: Largest nth Power which has n Digits

The largest $n$th power which has $n$ digits is $9^{21}$:
:$9^{21} = 109 \, 418 \, 989 \, 131 \, 512 \, 359 \, 209$

Proof:
The $n$th power of $10$ has $n + 1$ digits.
Hence the $n$th power of $m$ such that $m > 10$ has more than $n$ digits.
The $11$th power of $8$ has $10$ digits:
:$8^{10} = 8 \, 589 \, 934 \, 592$
and so $8^n$ where $n > 10$ has fewer than $n$ digits.
Hence the $n$th power of $m$ such that $m < 9$ and $n < 21$ has fewer than $n$ digits.
Finally we note that:
:$9^{22} = 984 \, 770 \, 902 \, 183 \, 611 \, 232 \, 881$
has $21$ digits.
Hence the $n$th power of $9$ such that $n > 21$ has fewer than $n$ digits.
Hence the result.
{{qed}}
Theorem: Smallest Number which is Multiplied by 99 by Appending 1 to Each End

The smallest positive integer which is multiplied by $99$ when $1$ is appended to each end is:
:$112 \, 359 \, 550 \, 561 \, 797  \, 752 \, 809$

Proof:
We have that:
:$112 \, 359 \, 550 \, 561 \, 797 \, 752 \, 809 = 101 \times 1 \, 052 \, 788 \, 969 \times 1 \, 056 \, 689 \, 261$
while:
{{begin-eqn}}
{{eqn | l = 11 \, 123 \, 595 \, 505 \, 617 \, 977 \, 528 \, 091
      | r = 3^2 \times 11 \times 101 \times 1 \, 052 \, 788 \, 969 \times 1 \, 056 \, 689 \, 261
      | c = 
}}
{{eqn | r = 3^2 \times 11 \times 112 \, 359 \, 550 \, 561 \, 797 \, 752 \, 809
      | c = 
}}
{{end-eqn}}
Let $N$ be the smallest integer satisfying $99 N = \sqbrk {1N1}$ when expressed in decimal notation.
Suppose $N$ is $k$ digits long.
Then:
:$\sqbrk {1 N 1} = 10^{k + 1} + 10 N + 1$
Subtracting $10 N$ from $99 N$ gives:
:$89 N = 10^{k + 1} + 1$
One can show, by trial and error, that the smallest $k$ where $10^{k + 1} + 1$ is divisible by $89$ is $21$.
Then $N = \dfrac {10^{22} + 1} {89} = 112 \, 359 \, 550 \, 561 \, 797 \, 752 \, 809$.
{{qed}}
Theorem: Smallest Multiply Perfect Number of Order 6

The number $154 \, 345 \, 556 \, 085 \, 770 \, 649 \, 600$ is multiply perfect of order $6$:
{{begin-eqn}}
{{eqn | l = \map {\sigma_1} {154 \, 345 \, 556 \, 085 \, 770 \, 649 \, 600}
      | r = 926 \, 073 \, 336 \, 514 \, 623 \, 897 \, 600
      | c = 
}}
{{eqn | r = 6 \times 154 \, 345 \, 556 \, 085 \, 770 \, 649 \, 600
      | c = 
}}
{{end-eqn}}
It is the smallest positive integer to be so.

Proof:
From {{DSFLink|154,345,556,085,770,649,600|154 \, 345 \, 556 \, 085 \, 770 \, 649 \, 600}}:
:$\map {\sigma_1} {154 \, 345 \, 556 \, 085 \, 770 \, 649 \, 600} = 926 \, 073 \, 336 \, 514 \, 623 \, 897 \, 600$
{{ProofWanted|That it is the smallest one remains to be proved.}}
Theorem: Sequence of 9 Consecutive Integers each with 48 Divisors

The $9$ integers beginning $17 \, 796 \, 126 \, 877 \, 482 \, 329 \, 126 \, 044$ each has $48$ divisors.

Proof:
In the below, $\sigma_0$ denotes the divisor count function.
{{begin-eqn}}
{{eqn | l = \map {\sigma_0} {17 \, 796 \, 126 \, 877 \, 482 \, 329 \, 126 \, 044}
      | r = 48
      | c = {{DCFLink|17,796,126,877,482,329,126,044|17 \, 796 \, 126 \, 877 \, 482 \, 329 \, 126 \, 044}}
}}
{{eqn | l = \map {\sigma_0} {17 \, 796 \, 126 \, 877 \, 482 \, 329 \, 126 \, 045}
      | r = 48
      | c = {{DCFLink|17,796,126,877,482,329,126,045|17 \, 796 \, 126 \, 877 \, 482 \, 329 \, 126 \, 045}}
}}
{{eqn | l = \map {\sigma_0} {17 \, 796 \, 126 \, 877 \, 482 \, 329 \, 126 \, 046}
      | r = 48
      | c = {{DCFLink|17,796,126,877,482,329,126,046|17 \, 796 \, 126 \, 877 \, 482 \, 329 \, 126 \, 046}}
}}
{{eqn | l = \map {\sigma_0} {17 \, 796 \, 126 \, 877 \, 482 \, 329 \, 126 \, 047}
      | r = 48
      | c = {{DCFLink|17,796,126,877,482,329,126,047|17 \, 796 \, 126 \, 877 \, 482 \, 329 \, 126 \, 047}}
}}
{{eqn | l = \map {\sigma_0} {17 \, 796 \, 126 \, 877 \, 482 \, 329 \, 126 \, 048}
      | r = 48
      | c = {{DCFLink|17,796,126,877,482,329,126,048|17 \, 796 \, 126 \, 877 \, 482 \, 329 \, 126 \, 048}}
}}
{{eqn | l = \map {\sigma_0} {17 \, 796 \, 126 \, 877 \, 482 \, 329 \, 126 \, 049}
      | r = 48
      | c = {{DCFLink|17,796,126,877,482,329,126,049|17 \, 796 \, 126 \, 877 \, 482 \, 329 \, 126 \, 049}}
}}
{{eqn | l = \map {\sigma_0} {17 \, 796 \, 126 \, 877 \, 482 \, 329 \, 126 \, 050}
      | r = 48
      | c = {{DCFLink|17,796,126,877,482,329,126,050|17 \, 796 \, 126 \, 877 \, 482 \, 329 \, 126 \, 050}}
}}
{{eqn | l = \map {\sigma_0} {17 \, 796 \, 126 \, 877 \, 482 \, 329 \, 126 \, 051}
      | r = 48
      | c = {{DCFLink|17,796,126,877,482,329,126,051|17 \, 796 \, 126 \, 877 \, 482 \, 329 \, 126 \, 051}}
}}
{{eqn | l = \map {\sigma_0} {17 \, 796 \, 126 \, 877 \, 482 \, 329 \, 126 \, 052}
      | r = 48
      | c = {{DCFLink|17,796,126,877,482,329,126,052|17 \, 796 \, 126 \, 877 \, 482 \, 329 \, 126 \, 052}}
}}
{{end-eqn}}
but then:
{{begin-eqn}}
{{eqn | l = \map {\sigma_0} {17 \, 796 \, 126 \, 877 \, 482 \, 329 \, 126 \, 043}
      | r = 32
      | c = {{DCFLink|17,796,126,877,482,329,126,043|17 \, 796 \, 126 \, 877 \, 482 \, 329 \, 126 \, 043}}
}}
{{eqn | l = \map {\sigma_0} {17 \, 796 \, 126 \, 877 \, 482 \, 329 \, 126 \, 053}
      | r = 8
      | c = {{DCFLink|17,796,126,877,482,329,126,053|17 \, 796 \, 126 \, 877 \, 482 \, 329 \, 126 \, 053}}
}}
{{end-eqn}}
{{qed}}
Theorem: Left-Truncatable Prime/Examples/357,686,312,646,216,567,629,137

The largest left-truncatable prime is $357 \, 686 \, 312 \, 646 \, 216 \, 567 \, 629 \, 137$.

Proof:
First it is demonstrated that $357 \, 686 \, 312 \, 646 \, 216 \, 567 \, 629 \, 137$ is indeed a left-truncatable prime:
{{begin-eqn}}
{{eqn | o = 
      | l = 357 \, 686 \, 312 \, 646 \, 216 \, 567 \, 629 \, 137
      | c = is prime
}}
{{eqn | o = 
      | l = 57 \, 686 \, 312 \, 646 \, 216 \, 567 \, 629 \, 137
      | c = is prime
}}
{{eqn | o = 
      | l = 7 \, 686 \, 312 \, 646 \, 216 \, 567 \, 629 \, 137
      | c = is prime
}}
{{eqn | o = 
      | l = 686 \, 312 \, 646 \, 216 \, 567 \, 629 \, 137
      | c = is prime
}}
{{eqn | o = 
      | l = 86 \, 312 \, 646 \, 216 \, 567 \, 629 \, 137
      | c = is prime
}}
{{eqn | o = 
      | l = 6 \, 312 \, 646 \, 216 \, 567 \, 629 \, 137
      | c = is prime
}}
{{eqn | o = 
      | l = 312 \, 646 \, 216 \, 567 \, 629 \, 137
      | c = is prime
}}
{{eqn | o = 
      | l = 12 \, 646 \, 216 \, 567 \, 629 \, 137
      | c = is prime
}}
{{eqn | o = 
      | l = 2 \, 646 \, 216 \, 567 \, 629 \, 137
      | c = is prime
}}
{{eqn | o = 
      | l = 646 \, 216 \, 567 \, 629 \, 137
      | c = is prime
}}
{{eqn | o = 
      | l = 46 \, 216 \, 567 \, 629 \, 137
      | c = is prime
}}
{{eqn | o = 
      | l = 6 \, 216 \, 567 \, 629 \, 137
      | c = is prime
}}
{{eqn | o = 
      | l = 216 \, 567 \, 629 \, 137
      | c = is prime
}}
{{eqn | o = 
      | l = 16 \, 567 \, 629 \, 137
      | c = is prime
}}
{{eqn | o = 
      | l = 6 \, 567 \, 629 \, 137
      | c = is prime
}}
{{eqn | o = 
      | l = 567 \, 629 \, 137
      | c = is prime
}}
{{eqn | o = 
      | l = 67 \, 629 \, 137
      | c = is the $3 \, 986 \, 726$th prime
}}
{{eqn | o = 
      | l = 7 \, 629 \, 137
      | c = is the $516 \, 434$th prime
}}
{{eqn | o = 
      | l = 629 \, 137
      | c = is the $51 \, 275$th prime
}}
{{eqn | o = 
      | l = 29 \, 137
      | c = is the $3167$th prime
}}
{{eqn | o = 
      | l = 9 \, 137
      | c = is the $1133$rd prime
}}
{{eqn | o = 
      | l = 137
      | c = is the $33$rd prime
}}
{{eqn | o = 
      | l = 37
      | c = is the $12$th prime
}}
{{eqn | o = 
      | l = 7
      | c = is the $4$th prime
}}
{{end-eqn}}
{{ProofWanted|It remains to be shown that it is the largest such.}}
Theorem: Polydivisible Number/Examples/3,608,528,850,368,400,786,036,725

The largest polydivisible number has $25$ digits:
:$3 \, 608 \, 528 \, 850 \, 368 \, 400 \, 786 \, 036 \, 725$

Proof:
{{begin-eqn}}
{{eqn | l = 3
      | r = 1 \times 3
}}
{{eqn | l = 36
      | r = 2 \times 18
}}
{{eqn | l = 360
      | r = 3 \times 120
}}
{{eqn | l = 3608
      | r = 4 \times 902
}}
{{eqn | l = 36 \, 085
      | r = 5 \times 7217
}}
{{eqn | l = 360 \, 852
      | r = 6 \times 60 \, 142
}}
{{eqn | l = 3 \, 608 \, 528
      | r = 7 \times 515 \, 504
}}
{{eqn | l = 36 \, 085 \, 288
      | r = 8 \times 4 \, 510 \, 661
}}
{{eqn | l = 360 \, 852 \, 885
      | r = 9 \times 40 \, 094 \, 765
}}
{{eqn | l = 3 \, 608 \, 528 \, 850
      | r = 10 \times 360 \, 852 \, 885
}}
{{eqn | l = 36 \, 085 \, 288 \, 503
      | r = 11 \times 3 \, 280 \, 480 \, 773
}}
{{eqn | l = 360 \, 852 \, 885 \, 036
      | r = 12 \times 30 \, 071 \, 073 \, 753
}}
{{eqn | l = 3 \, 608 \, 528 \, 850 \, 368
      | r = 13 \times 277 \, 579 \, 142 \, 336
}}
{{eqn | l = 36 \, 085 \, 288 \, 503 \, 684
      | r = 14 \times 2 \, 577 \, 520 \, 607 \, 406
}}
{{eqn | l = 360 \, 852 \, 885 \, 036 \, 840
      | r = 15 \times 24 \, 056 \, 859 \, 002 \, 456
}}
{{eqn | l = 3 \, 608 \, 528 \, 850 \, 368 \, 400
      | r = 16 \times 225 \, 533 \, 053 \, 148 \, 025
}}
{{eqn | l = 36 \, 085 \, 288 \, 503 \, 684 \, 007
      | r = 17 \times 2 \, 122 \, 664 \, 029 \, 628 \, 471
}}
{{eqn | l = 360 \, 852 \, 885 \, 036 \, 840 \, 078
      | r = 18 \times 20 \, 047 \, 382 \, 502 \, 046 \, 671
}}
{{eqn | l = 3 \, 608 \, 528 \, 850 \, 368 \, 400 \, 786
      | r = 19 \times 189 \, 922 \, 571 \, 072 \, 021 \, 094
}}
{{eqn | l = 36 \, 085 \, 288 \, 503 \, 684 \, 007 \, 860
      | r = 20 \times 1 \, 804 \, 264 \, 425 \, 184 \, 200 \, 393
}}
{{eqn | l = 360 \, 852 \, 885 \, 036 \, 840 \, 078 \, 603
      | r = 21 \times 17 \, 183 \, 470 \, 716 \, 040 \, 003 \, 743
}}
{{eqn | l = 3 \, 608 \, 528 \, 850 \, 368 \, 400 \, 786 \, 036
      | r = 22 \times 164 \, 024 \, 038 \, 653 \, 109 \, 126 \, 638
}}
{{eqn | l = 36 \, 085 \, 288 \, 503 \, 684 \, 007 \, 860 \, 367
      | r = 23 \times 1 \, 568 \, 925 \, 587 \, 116 \, 695 \, 993 \, 929
}}
{{eqn | l = 360 \, 852 \, 885 \, 036 \, 840 \, 078 \, 603 \, 672
      | r = 24 \times 15 \, 035 \, 536 \, 876 \, 535 \, 003 \, 275 \, 153
}}
{{eqn | l = 3 \, 608 \, 528 \, 850 \, 368 \, 400 \, 786 \, 036 \, 725
      | r = 25 \times 144 \, 341 \, 154 \, 014 \, 736 \, 031 \, 441 \, 469
}}
{{end-eqn}}
From No Polydivisible Number with 26 Digits Exists, there are no polydivisible numbers with more digits.
{{ProofWanted|It remains to be shown there is no larger one with $25$ digits.}}
Theorem: Probability of All Players receiving Complete Suit at Bridge

The probability of all $4$ players in a game of Bridge being dealt a complete suit is $1$ in $2 \, 235 \, 197 \, 406 \, 895 \, 366 \, 368 \, 301 \, 560 \, 000$.

Proof:
{{ProofWanted|Straightforward but boring exercise in combinatorics}}
Theorem: Powers of 10 Expressible as Product of 2 Zero-Free Factors

The powers of $10$ which can be expressed as the product of $2$ factors neither of which has a zero in its decimal representation are:
{{begin-eqn}}
{{eqn | l = 10^1
      | r = 2 \times 5
      | c = 
}}
{{eqn | l = 10^2
      | r = 4 \times 25
}}
{{eqn | l = 10^3
      | r = 8 \times 125
}}
{{eqn | l = 10^4
      | r = 16 \times 625
}}
{{eqn | l = 10^5
      | r = 32 \times 3125
}}
{{eqn | l = 10^6
      | r = 64 \times 15 \, 625
}}
{{eqn | l = 10^7
      | r = 128 \times 78 \, 125
}}
{{eqn | l = 10^9
      | r = 512 \times 1 \, 953 \, 125
}}
{{eqn | l = 10^{18}
      | r = 262 \, 144 \times 3 \, 814 \, 697 \, 265 \, 625
}}
{{eqn | l = 10^{33}
      | r = 8 \, 589 \, 934 \, 592 \times 116 \, 415 \, 321 \, 826 \, 934 \, 814 \, 453 \, 125
}}
{{end-eqn}}

Proof:
Let $p q = 10^n$ for some $n \in \Z_{>0}$.
Then $p q = 2^n 5^n$.
{{WLOG}}, suppose $p = 2^r 5^s$ for $r, s \ge 1$.
Then $2 \times 5 = 10$ is a divisor of $p$ and so $p$ ends with a zero.
Thus for $p$ and $q$ to be zero-free, it must be the case that $p = 2^n$ and $q = 5^n$ (or the other way around).
The result follows from Powers of 2 and 5 without Zeroes.
{{qed}}
Theorem: Sequence of Palindromic Sophie Germain Primes

The number $N = 191 \, 918 \, 080 \, 818 \, 091 \, 909 \, 090 \, 909 \, 190 \, 818 \, 080 \, 819 \, 191$ has the property that:
:$N$ is a palindromic Sophie Germain prime
:$2 N + 1$ is also a palindromic Sophie Germain prime
:$2 \left({2 N + 1}\right) + 1$ is also a palindromic prime, but not a Sophie Germain prime.

Proof:
By direct calculation:
{{begin-eqn}}
{{eqn | o =
      | r = 191 \, 918 \, 080 \, 818 \, 091 \, 909 \, 090 \, 909 \, 190 \, 818 \, 080 \, 819 \, 191
      | c = is palindromic and prime
}}
{{eqn | o = 
      | r = 2 \times 191 \, 918 \, 080 \, 818 \, 091 \, 909 \, 090 \, 909 \, 190 \, 818 \, 080 \, 819 \, 191 + 1
      | c = 
}}
{{eqn | r = 383 \, 836 \, 161 \, 636 \, 183 \, 818 \, 181 \, 818 \, 381 \, 636 \, 161 \, 638 \, 383
      | c = is palindromic and prime
}}
{{eqn | o = 
      | r = 2 \times 383 \, 836 \, 161 \, 636 \, 183 \, 818 \, 181 \, 818 \, 381 \, 636 \, 161 \, 638 \, 383 + 1
      | c = 
}}
{{eqn | r = 767 \, 672 \, 323 \, 272 \, 367 \, 636 \, 363 \, 636 \, 763 \, 272 \, 323 \, 276 \, 767
      | c = is palindromic and prime
}}
{{eqn | o = 
      | r = 2 \times 767 \, 672 \, 323 \, 272 \, 367 \, 636 \, 363 \, 636 \, 763 \, 272 \, 323 \, 276 \, 767 + 1
      | c = 
}}
{{eqn | r = 1 \, 535 \, 344 \, 646 \, 544 \, 735 \, 272 \, 727 \, 273 \, 526 \, 544 \, 646 \, 553 \, 535
      | c = is neither prime nor palindromic
}}
{{end-eqn}}
{{qed}}
Theorem: Smallest Integer Divisible by All Numbers from 1 to 100

The smallest positive integer which is divisible by each of the integers from $1$ to $100$ is:
:$69 \, 720 \, 375 \, 229 \, 712 \, 477 \, 164 \, 533 \, 808 \, 935 \, 312 \, 303 \, 556 \, 800$

Proof:
Let $N$ be divisible by each of the integers from $1$ to $100$.
Each prime number between $2$ and $97$ must be a divisor of $N$.
Also:
:$2^6 = 64 \divides N$
:$3^4 = 81 \divides N$
:$5^2 = 25 \divides N$
:$7^2 = 49 \divides N$
Every other integer between $1$ and $100$ is the product of a subset of all of these.
Hence by Euclid's Lemma:
{{begin-eqn}}
{{eqn | l = N
      | o = \ge
      | r = 2^6 \times 3^4 \times 5^2 \times 7^2 \times 11 \times 13 \times 17 \times 19 \times 23 \times 29 \times 31 \times 37 \times 41 \times 43 \times 47 \times 53 \times 59 \times 61 \times 67 \times 71 \times 73 \times 79 \times 83 \times 89 \times 97
      | c = 
}}
{{eqn | r = 69 \, 720 \, 375 \, 229 \, 712 \, 477 \, 164 \, 533 \, 808 \, 935 \, 312 \, 303 \, 556 \, 800
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Integer which is Multiplied by 9 when moving Last Digit to First

Let $N$ be the positive integer:
:$N = 10 \, 112 \, 359 \, 550 \, 561 \, 797 \, 752 \, 808 \, 988 \, 764 \, 044 \, 943 \, 820 \, 224 \, 719$
$N$ is the smallest positive integer $N$ such that if you move the last digit to the front, the result is the positive integer $9 N$.

Proof:
From Integer which is Multiplied by Last Digit when moving Last Digit to First, $N$ is equal to the recurring part of the fraction:
:$q = \dfrac {a_1} {10 a_1 - 1}$
where $a_1 = 9$.
Thus:
:$q = \dfrac 9 {10 \times 9 - 1} = \dfrac 9 {89}$
Hence:
=== Decimal Expansion ===
{{:Integer which is Multiplied by 9 when moving Last Digit to First/Decimal Expansion}}{{qed|lemma}}
The result follows.
{{qed}}
Theorem: Power of 2 containing no Digit 2

$2^{168}$ contains no $2$ anywhere in its decimal representation.

Proof:
:$2^{168} = 374 \, 144 \, 419 \, 156 \, 711 \, 147 \, 060 \, 143 \, 317 \, 175 \, 368 \, 453 \, 031 \, 918 \, 731 \, 001 \, 856$
{{qed}}
Theorem: Smallest Cube whose Sum of Divisors is Cube

The smallest cube $N$ such that $\map {\sigma_1} N$ is also a cube is:
:$27 \, 418 \, 521 \, 963 \, 671 \, 501 \, 273 \, 905 \, 190 \, 135 \, 082 \, 692 \, 041 \, 730 \, 405 \, 303 \, 870 \, 249 \, 023 \, 209$
where $\map {\sigma_1} N$ denotes the divisor sum of $N$.

Proof:
We have that:
{{begin-eqn}}
{{eqn | l = N
      | r = 27 \, 418 \, 521 \, 963 \, 671 \, 501 \, 273 \, 905 \, 190 \, 135 \, 082 \, 692 \, 041 \, 730 \, 405 \, 303 \, 870 \, 249 \, 023 \, 209
      | c = 
}}
{{eqn | r = 3^9 \times 7^3 \times 11^3 \times 13^3 \times 17^3 \times 41^3 \times 43^3 \times 47^3 \times 443^3 \times 499^3 \times 3583^3
      | c = 
}}
{{eqn | r = 30 \, 154 \, 214 \, 043 \, 975 \, 990 \, 969^3
      | c = 
}}
{{end-eqn}}
Then from {{DSFLink|27,418, 521,963, 671,501, 273,905, 190,135, 082,692, 041,730, 405,303, 870,249, 023,209|27 \, 418 \, 521 \, 963 \, 671 \, 501 \, 273 \, 905 \, 190 \, 135 \, 082 \, 692 \, 041 \, 730 \, 405 \, 303 \, 870 \, 249 \, 023 \, 209}}:
{{begin-eqn}}
{{eqn | l = \map {\sigma_1} N
      | r = 65 \, 400 \, 948 \, 817 \, 364 \, 742 \, 403 \, 487 \, 616 \, 930 \, 512 \, 213 \, 536 \, 407 \, 552 \, 000 \, 000 \, 000 \, 000 \, 000
      | c = 
}}
{{eqn | r = 2^{39} \times 3^6 \times 5^{15} \times 7^3 \times 11^3 \times 13^3 \times 17^3 \times 29^3 \times 37^3 \times 61^3 \times 157^3
      | c = 
}}
{{eqn | r = 40 \, 289 \, 760 \, 243 \, 532 \, 800 \, 000^3
      | c = 
}}
{{end-eqn}}
{{ProofWanted|It remains to be shown that this is the smallest such number.}}
Theorem: Uniqueness of Polynomial Ring in One Variable

Let $R$ be a commutative ring with unity.
Let $\struct {R \sqbrk X, \iota, X}$ and $\struct {R \sqbrk Y, \kappa, Y}$ be polynomial rings in one variable over $R$.
Then there exists a unique ring homomorphism $f: R \sqbrk X \to R \sqbrk Y$ such that:
:$f \circ \iota = \kappa$
:$\map f X = Y$
and it is an isomorphism.

Proof:
The existence and uniqueness of $f$ follows from the universal property.
Likewise, there is a unique ring homomorphism $g: R \sqbrk Y \to R \sqbrk X$ such that:
:$g \circ \kappa = \iota$
:$\map g Y = X$
and a unique ring homomorphism $h: R \sqbrk X \to R \sqbrk X$ such that:
:$h \circ \iota = \iota$
:$\map X = X$
By uniqueness and Identity Mapping is Ring Homomorphism, $h = \operatorname{id}$ is the identity mapping on $R \sqbrk X$.
Again by uniqueness and Composition of Ring Homomorphisms is Ring Homomorphism, $g\circ f = \operatorname{id}_{R \sqbrk X}$.
By symmetry, $f \circ g = \operatorname{id}_{R \sqbrk Y}$.
Thus $f$ is an isomorphism.
{{qed}}
Category:Polynomial Theory
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