nyuuzyou/wikis huggingface dataset:
2,890,625

== Number ==
$2 \, 890 \, 625$ (two million, eight hundred and ninety thousand, six hundred and twenty-four) is:
:$5^7 \times 37$
:The $12$th automorphic number after $1$, $5$, $6$, $25$, $76$, $376$, $625$, $9376$, $90 \, 625$, $109 \, 376$, $890 \, 625$:
::$2 \, 890 \, 625^2 = 8 \, 355 \, 71 \mathbf {2 \, 890 \, 625}$
== Also see ==
* 
Category:Specific Numbers
Category:2,890,625
2,985,984

== Number ==
$2 \, 985 \, 984$ (two million, nine hundred and eighty-five thousand, nine hundred and eighty-four) is:
:$2^{12} \times 3^6$
:The $6$th power of $12$ after $(1)$, $12$, $144$, $1728$, $20 \, 736$, $248 \, 832$:
::$2 \, 985 \, 984 = 12^6$
:The $12$th sixth power after $1$, $64$, $729$, $4096$, $15 \, 625$, $46 \, 656$, $117 \, 649$, $262 \, 144$, $531 \, 441$, $1 \, 000 \, 000$, $1 \, 771 \, 561$:
::$2 \, 985 \, 984 = 12 \times 12 \times 12 \times 12 \times 12 \times 12$
:The $144$th cube number:
::$2 \, 985 \, 984 = 144 \times 144 \times 144$
:The $1728$th square number:
::$2 \, 985 \, 984 = 1728 \times 1728$
== Also see ==
* 
* 
* 
* 
Category:Specific Numbers
Category:2,985,984
2-Digit Numbers divisible by both Product and Sum of Digits

== Theorem ==
The $2$-digit positive integers which are divisible by both the sum and product of their digits are:
:$12, 24, 36$
== Proof ==
We have:
It remains to be demonstrated that these are the only ones.
Let $z$ be a $2$-digit positive integer which is divisible by both the sum and product of its digits.
Neither of its digits may be $0$, because the product of the digits is a positive integer.
So, for $a \in \set {1, \dots, 9}$, $b \in \set {1, \dots, 9}$, $m \in \N$, $n \in \N$ we have:
Consider $(1) = (2)$:
In $(4)$ $a$, $b$, and $m - 1$ are all strictly positive; hence $10 - m$ must also be strictly positive.
Hence $1 < m < 10$.
Consider $(1) = (3)$:
Since $n \in \N$, we must have:
:$\paren {m - 1} \divides 9 m$
Since $m$ and $m - 1$ are coprime, by Euclid's Lemma:
:$\paren {m - 1} \divides 9$
This gives $m \in \set {2, 4}$.
Consider the case $m = 2$:
From $(4)$, $\dfrac b a = 8$.
Hence the only potential value for $b$ is $8$.
From $(5)$, $n = \dfrac {18} 8 \notin \N$.
Hence there are no potential candidates for $z$ for the case $m = 2$.
Consider the case $m = 4$:
From $(4)$, $\dfrac b a = 2$.
Hence the only potential values are $b \in \set {2, 4, 6, 8}$.
From $(5)$, $n = \dfrac {12} b$.
However, $8$ does not divide $12$.
This leaves $b \in \set {2, 4, 6}$.
By virtue of $\dfrac b a = 2$, we have $z \in \set {12, 24, 36}$.
We have now considered all the potential values of $m$.
Hence we conclude that the only $2$-digit positive integers which are divisible by both the sum and product of their digits are $12$, $24$, and $36$.
== Sources ==
* : $36$
* : $36$
Category:Recreational Mathematics
2-Digit Numbers forming Longest Reverse-and-Add Sequence

== Theorem ==
Let $m \in \Z_{>0}$ be a positive integer expressed in decimal notation.
Let $\map r m$ be the reverse-and-add process on $m$.
Let $r$ be applied iteratively to $m$.
The $2$-digit integers $m$ which need the largest number of iterations before reaching a palindromic number are $89$ and $98$, both needing $24$ iterations.
== Proof ==
The sequence obtained by iterating $r$ on $89$ is:
:$89, 187, 968, 1837, 9218, 17347, 91718, 173437, 907808, 1716517, 8872688,$
:$17735476, 85189247, 159487405, 664272356, 1317544822, 3602001953, 7193004016, 13297007933,$
:$47267087164, 93445163438, 176881317877, 955594506548, 170120002107, 8713200023178$
Note that $\map r {89} = \map r {98} = 187$, so the sequence obtained by iterating $r$ on $98$ is the same.
To show that no other $2$-digit integers have longer sequences, we define an equivalence relation on the $2$-digit integers as follows:
:$\sqbrk {a_1 a_2} \sim \sqbrk {b_1 b_2} \iff a_1 + a_2 = b_1 + b_2$
For each $2$-digit integer $\sqbrk {a_1 a_2}$:
:$\map r {\sqbrk {a_1 a_2} } = \sqbrk {a_1 a_2} + \sqbrk {a_2 a_1} = 11 \paren {a_1 + a_2}$
Notice that if $\sqbrk {a_1 a_2} \sim \sqbrk {b_1 b_2}$:
:$\map r {\sqbrk {a_1 a_2} } = 11 \paren {a_1 + a_2} = 11 \paren {b_1 + b_2} = \map r {\sqbrk {b_1 b_2} }$
So the value of $r$ is equal in each equivalence class.
For each $2$-digit integer $\sqbrk {a_1 a_2}$, we have:
:$18 = 9 + 9 \ge a_1 + a_2 \ge 1 + 0 = 1$
For $2$-digit integers where $a_1 + a_2 \le 9$:
:$\map r {\sqbrk {a_1 a_2} } = 11 \paren {a_1 + a_2}$ is a $2$-digit multiple of $11$, which is a palindrome.
So we consider each equivalence class of $2$-digit integers with $a_1 + a_2 \ge 10$.
We have the sequences:
:$19, 110, 121$
:$29, 121$
:$39, 132, 363$
:$49, 143, 484$
:$59, 154, 605, 1111$
:$69, 165, 726, 1353, 4884$
:$79, 176, 847, 1595, 7546, 14003, 44044$
:$89, \cdots, 8713200023178$, with $24$ iterations
:$99$
So:
:every $2$-digit integer not in the equivalence class of $89$ has a shorter iteration
and:
:$\eqclass {89} \sim = \set {89, 98}$
== Sources ==
* : $89$
* : $89$
Category:Reverse-and-Add
Category:89
Category:98
2-Digit Permutable Primes

== Theorem ==
The $2$-digit permutable primes are:
:$11, 13, 17, 37, 79$
and their anagrams, and no other.
== Proof ==
It is confirmed that:
:$13$ and $31$ are both prime
:$17$ and $71$ are both prime
:$37$ and $73$ are both prime.
:$79$ and $97$ are both prime.
From Digits of Permutable Prime, all permutable primes contain digits in the set:
:$\left\{ {1, 3, 7, 9}\right\}$
By Prime Repdigit Number is Repunit, all $2$-digit repdigit numbers are composite except for $11$ which is prime.
From Repunit Prime is Permutable Prime, it follows that $11$ is a permutable prime.
Apart from $11$ whose status is known, the possibly permutable primes are:
:$13, 17, 19, 37, 39, 79$
and their anagrams.
We eliminate $13, 17, 37, 79$ from this list, as it has been established that they are permutable primes.
Of those remaining, the following is composite:
It remains to demonstrate that the anagram of $19$, that is, $91$, composite.
We find that:
All contenders are eliminated except for the established permutable primes $11, 13, 17, 37, 79$ and their anagrams.
== Sources ==
* : $113$
* : $113$
Category:Permutable Primes
2-Digit Positive Integer equals Product plus Sum of Digits iff ends in 9

== Theorem ==
Let $n$ be a $2$-digit positive integer.
Then:
:$n$ equals the sum added to the product of its digits
:
:the last digit of $n$ is $9$.
== Proof ==
Let $n = 10 x + y$ where $0 < x \le 9, 0 \le y \le 9$.
Then:
== Sources ==
* : $39$
Category:Recreational Mathematics
2/Historical Note

== Historical Note on $2$ (two) ==
The number $2$ (two) is understood to have been treated as a special case of a number from the earliest historical records.
Many early languages have specific forms of nouns for when two of an object are under consideration, as well as different forms for singular and plural.
To the ancient Greeks, in addition to having problems with the idea that $1$ is a number, it was questionable whether or not two ($2$) was actually a number either:
:While it has a beginning and an end, it has no middle
:Multiplication by $2$ consists merely of adding a number to itself, and multiplication was expected to do more than just add.
Thus $2$ was an exceptional case.
To the Pythagoreans, odd and even numbers were considered to be either male or female, but sources differ on which was which.
Some suggest that $2$ was considered to be the first male number, and said to personify the principle of diversity.
Such sources state that in contrast, the odd numbers were considered to be female.
However, other sources suggest that it was the even numbers which were female, while the odd numbers were male.
The dyad, as such, is never specifically defined in , but introduced without definition in Power of Two is Even-Times Even Only.
== Sources ==
* : Chapter $1$: Some Preliminary Considerations: $1.3$ Early Number Theory
* : $2$
* : Chapter $\text {A}.2$: Pythagoras (ca. B.C.)
* : $2$
* : Chapter $2$: The Logic of Shape: Pythagoras
Category:Historical Notes
2/Linguistic Note

== Linguistic Note on $2$ (two) ==
There are many words in natural language which indicate a twofold collection:
:dual, duel, brace, couple, pair, double and twin
for example.
Words derived from or associated with the number $2$ include:
:bicycle: a vehicle with $2$ wheels
An archaic term for an instance of the number two is dyad.
This can be found in 's $1908$ translation of , in Power of Two is Even-Times Even Only, but never specifically defined there formally.
== Sources ==
* : $2$
* : Chapter $1$: The Romance of Numbers: Number Words
* : $2$
Category:Linguistic Notes
20

== Number ==
$20$ (twenty) is:
:$2^2 \times 5$
:The $1$st positive integer $n$ such that $6 n + 1$ and $6 n - 1$ are both composite:
::$6 \times 20 - 1 = 119 = 7 \times 17$, $6 \times 20 + 1 = 121 = 11^2$
:The $1$st tetrahedral number which is the sum of two tetrahedral numbers:
::$20 = 10 + 10$
:The $1$st primitive abundant number:
:The $3$rd abundant number after $12$, $18$:
::$1 + 2 + 4 + 5 + 10 = 21 > 20$
:The $3$rd central binomial coefficient after $2$, $6$:
::$20 = \dbinom {2 \times 3} 3 := \dfrac {6!} {\paren {3!}^2}$
:The $3$rd number after $1$, $9$ whose square has a divisor sum which is itself square:
:The $4$th tetrahedral number, after $1$, $4$, $10$:
::$20 = 1 + 3 + 6 + 10 = \dfrac {4 \paren {4 + 1} \paren {4 + 2} } 6$
:The $4$th semiperfect number after $6$, $12$, $18$:
:The $2$nd primitive semiperfect number after $6$:
::$20 = 1 + 4 + 5 + 10$
:The $6$th positive integer after $6$, $9$, $12$, $18$, $19$ whose cube can be expressed as the sum of $3$ positive cube numbers:
::$20^3 = 7^3 + 14^3 + 17^3$
:The $9$th even number after $2$, $4$, $6$, $8$, $10$, $12$, $14$, $16$ which cannot be expressed as the sum of $2$ composite odd numbers.
:The $11$th highly abundant number after $1$, $2$, $3$, $4$, $6$, $8$, $10$, $12$, $16$, $18$:
::$\map {\sigma_1} {20} = 42$
:The $13$th after $1$, $2$, $4$, $5$, $6$, $8$, $9$, $12$, $13$, $15$, $16$, $17$ of the $24$ positive integers which cannot be expressed as the sum of distinct non-pythagorean primes.
:The $13$th positive integer after $2$, $3$, $4$, $7$, $8$, $9$, $10$, $11$, $14$, $15$, $16$, $19$ which cannot be expressed as the sum of distinct pentagonal numbers.
:The $13$th harshad number after $1$, $2$, $3$, $4$, $5$, $6$, $7$, $8$, $9$, $10$, $12$, $18$:
::$20 = 10 \times 2 = 10 \times \paren {2 + 0}$
:The $16$th (strictly) positive integer after $1$, $2$, $3$, $4$, $6$, $7$, $8$, $9$, $10$, $12$, $13$, $14$, $18$, $19$ which cannot be expressed as the sum of distinct primes of the form $6 n - 1$
:$20^3 = 11^3 + 12^3 + 13^3 + 14^3$
:The number of faces on an icosahedron
:The number of vertices on a regular dodecahedron
:The number of different ways of playing the first move in chess
== Also see ==
* Definition:Vigesimal System
* Smallest n such that 6 n + 1 and 6 n - 1 are both Composite
* Cube of 20 is Sum of Sequence of 4 Consecutive Cubes
* 
* 
* 
* 
* 
* 
* 
* 
* 
* 
* 
* 
* 
* 
* 
* 
== Historical Note ==
== Sources ==
* : $20$
* : $20$
Category:Specific Numbers
Category:20
20,161

== Number ==
$20 \, 161$ (twenty thousand, one hundred and sixty-one) is:
:The $2281$st prime number
:The largest integer which is not the sum of $2$ abundant numbers
== Also see ==
* Largest Integer not Sum of Two Abundant Numbers
* 
== Sources ==
* : $20,161$
* : $20,161$
Category:Specific Numbers
Category:20,161
20,615,673

== Number ==
$20 \, 615 \, 673$ is:
:$3 \times 13 \times 331 \times 1597$
:The $2$nd known positive integer whose fourth power can be expressed as the sum of $3$ fourth powers, thus refuting the Euler Quartic Conjecture:
::$20 \, 615 \, 673^4 = 2 \, 682 \, 440^4 + 15 \, 365 \, 639^4 + 18 \, 796 \, 760^4$
== Also see ==
* 
== Sources ==
* : $20,615,673$
Category:Specific Numbers
Category:20,615,673
20,736

== Number ==
$20 \, 736$ (twenty thousand, seven hundred and thirty-six) is:
:$2^8 \times 3^4$
:$10 \, 000$ in duodecimal
:The $4$th power of $12$ after $(1)$, $12$, $144$, $1728$:
::$20 \, 736 = 12^4$
:The $12$th fourth power after $1$, $16$, $81$, $256$, $625$, $1296$, $2401$, $4096$, $6561$, $10 \, 000$, $14 \, 641$:
::$20 \, 736 = 12 \times 12 \times 12 \times 12$
:The $144$th square number:
::$20 \, 736 = 144 \times 144$
== Also see ==
* 
* 
* 
== Sources ==
* : $20,736$
* : $20,736$
Category:Specific Numbers
Category:20,736
20/Historical Note

== Historical Note on $20$ (Twenty) ==
Occurrences of the number $20$ in various cultures in history:
:There were $20$ shillings in $1$ pound Sterling in pre-decimal British coinage.
:There are $20$ fluid ounces in the imperial pint.
The word score means a set of $20$.
== Also see ==
* Definition:Vigesimal System
== Sources ==
* : $20$
* : $20$
Category:Historical Notes
200

== Number ==
$200$ (two hundred) is:
:$2^3 \times 5^2$
:The $1$st positive integer that cannot be made into a prime number by changing just $1$ digit
:The $9$th positive integer after $50$, $65$, $85$, $125$, $130$, $145$, $170$, $185$ which can be expressed as the sum of two square numbers in two or more different ways:
::$200 = 10^2 + 10^2 = 14^2 + 2^2$
:The $23$rd powerful number after $1$, $4$, $8$, $9$, $16$, $25$, $\ldots$, $108$, $121$, $125$, $128$, $144$, $169$, $196$
== Also see ==
* 
* 
* 
== Sources ==
* : $200$
Category:Specific Numbers
Category:200
2002

== Number ==
$2002$ (two thousand and two) is:
:$2 \times 7 \times 11 \times 13$
:The $14$th palindromic integer after $0$, $1$, $2$, $3$, $11$, $22$, $101$, $111$, $121$, $202$, $212$, $1001$, $1111$ whose square is also palindromic integer
::$2002^2 = 4 \, 008 \, 004$
== Also see ==
* 
Category:Specific Numbers
Category:2002
201

== Number ==
$201$ (two hundred and one) is:
:$3 \times 67$
:The $39$th lucky number:
::$1$, $3$, $7$, $9$, $13$, $15$, $21$, $\ldots$, $133$, $135$, $141$, $151$, $159$, $163$, $169$, $171$, $189$, $193$, $195$, $201$, $\ldots$
== Also see ==
* 
Category:Specific Numbers
Category:201
2013

== Number ==
$2013$ (two thousand and thirteen) is:
:$3 \times 11 \times 61$
:The $1$st term of the $3$rd triplet of consecutive positive integers all of which are sphenic:
::$2013 = 3 \times 11 \times 61$, $2014 = 2 \times 19 \times 53$, $2015 = 5 \times 13 \times 31$
== Also see ==
* 
Category:Specific Numbers
Category:2013
2014

== Number ==
$2014$ (two thousand and fourteen) is:
:$2 \times 19 \times 53$
:The $2$nd term of the $3$rd triplet of consecutive positive integers all of which are sphenic:
::$2013 = 3 \times 11 \times 61$, $2014 = 2 \times 19 \times 53$, $2015 = 5 \times 13 \times 31$
== Also see ==
* 
Category:Specific Numbers
Category:2014
2015

== Number ==
$2015$ (two thousand and fifteen) is:
:$5 \times 13 \times 31$
:The $3$rd Lucas-Carmichael number after $399$, $935$:
::$\paren {5 + 1} \divides \paren {2015 + 1}$, $\paren {13 + 1} \divides \paren {2015 + 1}$, $\paren {31 + 1} \divides \paren {2015 + 1}$
:The $3$rd term of the $3$rd triplet of consecutive positive integers all of which are sphenic:
::$2013 = 3 \times 11 \times 61$, $2014 = 2 \times 19 \times 53$, $2015 = 5 \times 13 \times 31$
== Also see ==
* 
* 
Category:Specific Numbers
Category:2015
2016

== Number ==
$2016$ (two thousand and sixteen) is:
:$2^5 \times 3^2 \times 7$
:The $2$nd number after $360$ whose ratio to aliquot sum is $4 : 9$:
::$\dfrac {\map {\sigma_1} {2016} - 2016} {2016} = \dfrac 9 4$
:where $\sigma_1$ denotes the divisor sum function.
:The $9$th positive integer after $504$, $756$, $806$, $1008$, $1148$, $1209$, $1472$, $1512$ which can be expressed as the product of $2$ two-digit numbers in $2$ ways such that the factors in one of those pairs is the reversal of each of the factors in the other:
::$2016 = 24 \times 84 = 48 \times 42$
:The $32$nd hexagonal number after $1$, $6$, $15$, $28$, $45$, $66$, $91$, $\ldots$, $703$, $780$, $861$, $946$, $1035$, $1225$, $1326$, $1431$, $1540$, $1653$, $1770$, $1891$:
::$2016 = \ds \sum_{k \mathop = 1}^{32} \paren {4 k - 3} = 32 \paren {2 \times 32 - 1}$
:The $63$rd triangular number after $1$, $3$, $6$, $10$, $15$, $\ldots$, $1326$, $1378$, $1431$, $1485$, $1540$, $1596$, $1653$, $1711$, $1770$, $1770$, $1830$, $1891$, $1953$:
::$2016 = \ds \sum_{k \mathop = 1}^{63} k = \dfrac {63 \times \paren {63 + 1} } 2$
:$2016 = 3^3 + 4^3 + 5^3 + 6^3 + 7^3 + 8^3 + 9^3$
:The divisor sum of $672$:
::$\map {\sigma_1} {672} = 2016 = 3 \times 672$
:The middle term of the amicable triplet $\tuple {1980, 2016, 2556}$:
::$\map {\sigma_1} {1980} = \map {\sigma_1} {2016} = \map {\sigma_1} {2556} = 6552 = 1980 + 2016 + 2556$
== Also see ==
* Ratio of 2016 to Aliquot Sum
* 
* 
* 
* 
* 
== Historical Note ==
Category:Specific Numbers
Category:2016
202

== Number ==
$202$ (two hundred and two) is:
:$2 \times 101$
:The $2$nd positive integer after $200$ that cannot be made into a prime number by changing just $1$ digit
:The $9$th Smith number after $4$, $22$, $27$, $58$, $85$, $94$, $121$, $166$:
::$2 + 0 + 2 = 2 + 1 + 0 + 1 = 4$
:The $10$th palindromic integer after $0$, $1$, $2$, $3$, $11$, $22$, $101$, $111$, $121$ whose square is also palindromic integer
::$202^2 = 40 \, 804$
:The smallest positive integer which can be expressed as the sum of $2$ distinct lucky numbers in $13$ different ways
:The $18$th noncototient after $10$, $26$, $34$, $50$, $52$, $58$, $86$, $100$, $116$, $122$, $130$, $134$, $146$, $154$, $170$, $172$, $186$:
::$\nexists m \in \Z_{>0}: m - \map \phi m = 202$
:where $\map \phi m$ denotes the Euler $\phi$ function
:The $29$th nontotient:
::$\nexists m \in \Z_{>0}: \map \phi m = 202$
:where $\map \phi m$ denotes the Euler $\phi$ function
== Also see ==
* 
* 
* 
* 
* 
* 
Category:Specific Numbers
Category:202
2024

== Number ==
$2024$ (two thousand and twenty-four) is:
:$2^3 \times 11 \times 23$
:With $2295$, an element of the $5$th quasiamicable pair:
::$\map {\sigma_1} {2024} = \map {\sigma_1} {2295} = 4320 = 2024 + 2295 + 1$
:The $22$nd tetrahedral number, after $1$, $4$, $10$, $20$, $35$, $\ldots$, $816$, $969$, $1140$, $1330$, $1540$, $1771$:
::$2024 = \ds \sum_{k \mathop = 1}^{22} \frac {k \paren {k + 1} } 2 = \dfrac {22 \paren {22 + 1} \paren {22 + 2} } 6$
== Also see ==
* 
* 
Category:Specific Numbers
Category:2024
2025

== Number ==
$2025$ (two thousand and twenty-five) is:
:$3^4 \times 5^2$
:$2025$ is $5$-smooth:
::$\max \set {3, 5} = 5$
:The sum of the first $9$ cubes:
::$2025 = 1^3 + 2^3 + 3^3 + 4^3 + 5^3 + 6^3 + 7^3 + 8^3 + 9^3$
:The $45$th square number after $1$, $4$, $9$, $16$, $25$, $36$, $\ldots$, $1296$, $1369$, $1444$, $1521$, $1600$, $1681$, $1764$, $1849$, $1936$:
::$2025 = 45 \times 45$
:Adding $1$ to each of its digits yields another square:
::$2025 + 1111 = 3136 = 56^2$
:The roots of those squares also differ by a repunit:
::$45 + 11 = 56$
:Not quite a Friedman number, but this is amusing:
::$\sqrt {2025} = \paren {2 + 0!}^2 \times 5$
:... and so is this:
::$\sqrt {2025} = 20 + 25$
== Also see ==
* Difference between Two Squares equal to Repunit
* Smooth Number/Examples/2025
* 
* 
== Sources ==
* : $2025$
* : $2025$
Category:Specific Numbers
Category:2025
203

== Number ==
$203$ (two hundred and three) is:
:$7 \times 29$
:The $6$th Bell number after $(1)$, $1$, $2$, $5$, $15$, $52$
:The $33$rd happy number after $1$, $7$, $10$, $13$, $19$, $23$, $\ldots$, $130$, $133$, $139$, $167$, $176$, $188$, $190$, $192$, $193$:
::$203 \to 2^2 + 0^2 + 3^2 = 4 + 0 + 9 = 13 \to 1^2 + 3^2 = 1 + 9 = 10 \to 1^2 + 0^2 = 1$
== Also see ==
* 
* 
Category:Specific Numbers
Category:203
2035

== Number ==
$2035$ (two thousand and thirty-five) is:
:$5 \times 11 \times 37$
:The $37$th pentagonal number after $1$, $5$, $12$, $22$, $35$, $\ldots$, $925$, $1001$, $1080$, $1162$, $1247$, $1335$, $1426$, $1520$, $1617$, $1717$, $1820$, $1926$:
::$2035 = \ds \sum_{k \mathop = 1}^{37} \paren {3 k - 2} = \dfrac {37 \paren {3 \times 37 - 1} } 2$
:The $73$rd generalized pentagonal number after $1$, $2$, $5$, $7$, $12$, $15$, $\ldots$, $1617$, $1650$, $1717$, $1751$, $1820$, $1855$, $1926$, $1962$:
::$2035 = \ds \sum_{k \mathop = 1}^{37} \paren {3 k - 2} = \dfrac {37 \paren {3 \times 37 - 1} } 2$
== Also see ==
* 
* 
Category:Specific Numbers
Category:2035
204

== Number ==
$204$ (two hundred and four) is:
:$2^2 \times 3 \times 17$
:The $3$rd positive integer after $200$, $202$ that cannot be made into a prime number by changing just $1$ digit
:The $8$th square pyramidal number after $1$, $5$, $14$, $30$, $55$, $91$, $140$:
::$204 = 1 + 4 + 9 + 16 + 25 + 36 + 49 + 64 = \ds \sum_{k \mathop = 1}^8 k^2 = \dfrac {8 \paren {8 + 1} \paren {2 \times 8 + 1} } 6$
:The $40$th positive integer $n$ such that no factorial of an integer can end with $n$ zeroes.
== Also see ==
* Sum of Cubes of 3 Consecutive Integers which is Square
* 
* 
* 
== Sources ==
* : $204$
* : $204$
Category:Specific Numbers
Category:204
2047

== Number ==
$2047$ (two thousand and forty-seven) is:
:$23 \times 89$
:The $1$st Mersenne number which is composite:
::$2047 = 2^{11} - 1 = 23 \times 89$
:The $5$th Mersenne number after $3$, $7$, $31$, $127$:
::$2047 = 2^{11} - 1$
:The $8$th Woodall number after $1$, $7$, $23$, $63$, $159$, $383$, $895$:
::$2047 = 8 \times 2^8 - 1$
:The $8$th Poulet number after $341$, $561$, $645$, $1105$, $1387$, $1729$, $1905$:
::$2^{2047} \equiv 2 \pmod {2047}$: $2047 = 23 \times 89$
== Historical Note ==
== Also see ==
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== Sources ==
* : $2047$
* : $2047$
Category:Specific Numbers
Category:2047
2048

== Number ==
$1024$ (one thousand and twenty-four) is:
:$2^{11}$
:The $11$th power of $2$ after $(1)$, $2$, $4$, $8$, $16$, $32$, $64$, $128$, $256$, $512$, $1024$:
::$2048 = 2^{11}$
:The $12$th almost perfect number after $1$, $2$, $4$, $8$, $16$, $32$, $64$, $128$, $256$, $512$, $1024$:
::$\map {\sigma_1} {2048} = 4095 = 2 \times 2048 - 1$
== Also see ==
* 
* 
Category:Specific Numbers
Category:2048
2049

== Number ==
$2049$ (two thousand and forty-nine) is:
:$3 \times 683$
:The $9$th Cullen number after $1$, $3$, $9$, $25$, $65$, $161$, $385$, $897$:
::$2049 = 8 \times 2^8 + 1$
== Also see ==
* 
Category:Specific Numbers
Category:2049
205

== Number ==
$205$ (two hundred and five) is:
:$5 \times 41$
:The $4$th positive integer after $200$, $202$, $204$ that cannot be made into a prime number by changing just $1$ digit
:The number of pairs of twin primes less than $10^4$
:The $10$th positive integer after $50$, $65$, $85$, $125$, $130$, $145$, $170$, $185$, $200$ which can be expressed as the sum of two square numbers in two or more different ways:
::$205 = 14^2 + 3^2 = 13^2 + 6^2$
:The $13$th integer after $7$, $13$, $19$, $35$, $38$, $41$, $57$, $65$, $70$, $125$, $130$, $190$ the decimal representation of whose square can be split into two parts which are each themselves square:
::$205^2 = 42 \, 025$; $4 = 2^2$, $2025 = 45^2$
:The $40$th lucky number:
::$1$, $3$, $7$, $9$, $13$, $15$, $21$, $\ldots$, $163$, $169$, $171$, $189$, $193$, $195$, $201$, $205$, $\ldots$
== Also see ==
* Integer Greater than 205 is Sum of Distinct Primes of form 6 n + 1
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== Sources ==
* : $205$
* : $205$
Category:Specific Numbers
Category:205
2053

== Number ==
$2053$ (two thousand and fifty-three) is:
:The $310$th prime number
:The $11$th prime $p$ such that $p \# - 1$, where $p \#$ denotes primorial (product of all primes up to $p$) of $p$, is prime:
::$3$, $5$, $11$, $13$, $41$, $89$, $317$, $337$, $991$, $1873$, $2053$
== Also see ==
* 
* 
Category:Specific Numbers
Category:2053
2056

== Number ==
$2056$ (two thousand and fifty-six) is:
:$2^3 \times 257$
:The magic constant of a magic square of order $16$, after $1$, $(5)$, $15$, $34$, $\ldots$, $870$, $1105$, $1379$, $1695$:
::$2056 = \ds \dfrac 1 {16} \sum_{k \mathop = 1}^{16^2} k = \dfrac {16 \paren {16^2 + 1} } 2$
== Also see ==
* 
Category:Specific Numbers
Category:2056
2059

== Number ==
$2059$ (two thousand and fifty-nine) is:
:$29 \times 71$
:The $29$th heptagonal number after $1$, $7$, $18$, $34$, $55$, $81$, $112$, $\ldots$, $874$, $970$, $1071$, $1177$, $1288$, $1404$, $1525$, $1651$, $1782$, $1918$:
::$2059 = \ds \sum_{k \mathop = 1}^{29} \paren {5 k - 4} = \dfrac {29 \paren {5 \times 29 - 3} } 2$
== Also see ==
* 
Category:Specific Numbers
Category:2059