nyuuzyou/wikis huggingface dataset:
Numbers such that Divisor Count divides Phi divides Divisor Sum/Examples/3

== Examples of Numbers such that Divisor Count divides Phi divides Divisor Sum ==
The number $3$ has the property that:
:$\map {\sigma_0} 3 \divides \map \phi 3 \divides \map {\sigma_1} 3$
where:
:$\divides$ denotes divisibility
:$\sigma_0$ denotes the divisor count function
:$\phi$ denotes the Euler $\phi$ (phi) function
:$\sigma_1$ denotes the divisor sum function.
== Proof ==
Category:Numbers such that Divisor Count divides Phi divides Divisor Sum
Category:3
Numbers such that Divisor Count divides Phi divides Divisor Sum/Examples/30

== Examples of Numbers such that Divisor Count divides Phi divides Divisor Sum ==
The number $30$ has the property that:
:$\map {\sigma_0} {30} \divides \map \phi {30} \divides \map {\sigma_1} {30}$
where:
:$\divides$ denotes divisibility
:$\sigma_0$ denotes the divisor count function
:$\phi$ denotes the Euler $\phi$ (phi) function
:$\sigma_1$ denotes the divisor sum function.
== Proof ==
Category:Numbers such that Divisor Count divides Phi divides Divisor Sum
Category:30
Numbers such that Divisor Count divides Phi divides Divisor Sum/Examples/35

== Examples of Numbers such that Divisor Count divides Phi divides Divisor Sum ==
The number $35$ has the property that:
:$\map {\sigma_0} {35} \divides \map \phi {35} \divides \map {\sigma_1} {35}$
where:
:$\divides$ denotes divisibility
:$\sigma_0$ denotes the divisor count function
:$\phi$ denotes the Euler $\phi$ (phi) function
:$\sigma_1$ denotes the divisor sum function.
== Proof ==
Category:Numbers such that Divisor Count divides Phi divides Divisor Sum
Category:35
Numbers such that Divisor Count divides Phi divides Divisor Sum/Examples/357

== Examples of Numbers such that Divisor Count divides Phi divides Divisor Sum ==
The number $357$ has the property that:
:$\map {\sigma_0} {357} \divides \map \phi {357} \divides \map {\sigma_1} {357}$
where:
:$\divides$ denotes divisibility
:$\sigma_0$ denotes the divisor count function
:$\phi$ denotes the Euler $\phi$ (phi) function
:$\sigma_1$ denotes the divisor sum function.
== Proof ==
Category:Numbers such that Divisor Count divides Phi divides Divisor Sum
Category:357
Numbers such that Divisor Count divides Phi divides Divisor Sum/Examples/420

== Examples of Numbers such that Divisor Count divides Phi divides Divisor Sum ==
The number $420$ has the property that:
:$\map {\sigma_0} {420} \divides \map \phi {420} \divides \map {\sigma_1} {420}$
where:
:$\divides$ denotes divisibility
:$\sigma_0$ denotes the divisor count function
:$\phi$ denotes the Euler $\phi$ (phi) function
:$\sigma_1$ denotes the divisor sum function.
== Proof ==
Category:Numbers such that Divisor Count divides Phi divides Divisor Sum
Category:420
Numbers such that Divisor Count divides Phi divides Divisor Sum/Examples/56

== Examples of Numbers such that Divisor Count divides Phi divides Divisor Sum ==
The number $56$ has the property that:
:$\map {\sigma_0} {56} \divides \map \phi {56} \divides \map {\sigma_1} {56}$
where:
:$\divides$ denotes divisibility
:$\sigma_0$ denotes the divisor count function
:$\phi$ denotes the Euler $\phi$ (phi) function
:$\sigma_1$ denotes the divisor sum function.
== Proof ==
Category:Numbers such that Divisor Count divides Phi divides Divisor Sum
Category:56
Numbers such that Divisor Count divides Phi divides Divisor Sum/Examples/570

== Examples of Numbers such that Divisor Count divides Phi divides Divisor Sum ==
The number $570$ has the property that:
:$\map {\sigma_0} {570} \divides \map \phi {570} \divides \map {\sigma_1} {570}$
where:
:$\divides$ denotes divisibility
:$\sigma_0$ denotes the divisor count function
:$\phi$ denotes the Euler $\phi$ (phi) function
:$\sigma_1$ denotes the divisor sum function.
== Proof ==
Category:Numbers such that Divisor Count divides Phi divides Divisor Sum
Category:570
Numbers such that Divisor Count divides Phi divides Divisor Sum/Examples/616

== Examples of Numbers such that Divisor Count divides Phi divides Divisor Sum ==
The number $616$ has the property that:
:$\map {\sigma_0} {616} \divides \map \phi {616} \divides \map {\sigma_1} {616}$
where:
:$\divides$ denotes divisibility
:$\sigma_0$ denotes the divisor count function
:$\phi$ denotes the Euler $\phi$ (phi) function
:$\sigma_1$ denotes the divisor sum function.
== Proof ==
Category:Numbers such that Divisor Count divides Phi divides Divisor Sum
Category:616
Numbers such that Divisor Count divides Phi divides Divisor Sum/Examples/630

== Examples of Numbers such that Divisor Count divides Phi divides Divisor Sum ==
The number $630$ has the property that:
:$\map {\sigma_0} {630} \divides \map \phi {630} \divides \map {\sigma_1} {630}$
where:
:$\divides$ denotes divisibility
:$\sigma_0$ denotes the divisor count function
:$\phi$ denotes the Euler $\phi$ (phi) function
:$\sigma_1$ denotes the divisor sum function.
== Proof ==
Category:Numbers such that Divisor Count divides Phi divides Divisor Sum
Category:630
Numbers such that Divisor Count divides Phi divides Divisor Sum/Examples/70

== Examples of Numbers such that Divisor Count divides Phi divides Divisor Sum ==
The number $70$ has the property that:
:$\map {\sigma_0} {70} \divides \map \phi {70} \divides \map {\sigma_1} {70}$
where:
:$\divides$ denotes divisibility
:$\sigma_0$ denotes the divisor count function
:$\phi$ denotes the Euler $\phi$ (phi) function
:$\sigma_1$ denotes the divisor sum function.
== Proof ==
Category:Numbers such that Divisor Count divides Phi divides Divisor Sum
Category:70
Numbers such that Divisor Count divides Phi divides Divisor Sum/Examples/78

== Examples of Numbers such that Divisor Count divides Phi divides Divisor Sum ==
The number $78$ has the property that:
:$\map {\sigma_0} {78} \divides \map \phi {78} \divides \map {\sigma_1} {78}$
where:
:$\divides$ denotes divisibility
:$\sigma_0$ denotes the divisor count function
:$\phi$ denotes the Euler $\phi$ (phi) function
:$\sigma_1$ denotes the divisor sum function.
== Proof ==
Category:Numbers such that Divisor Count divides Phi divides Divisor Sum
Category:78
Numbers such that Divisor Count divides Phi divides Divisor Sum/Mistake

== Source Work ==
:
:The Dictionary
::$210$
== Mistake ==
:$\map \phi {210} = 48$ is a factor of $\map \sigma {210} = 576$, and $\map d {210} = 16$ divides both. The sequence of numbers with both these properties starts: $1 \quad 3 \quad 15 \quad 30 \quad 35 \quad 52 \quad 70 \quad 78 \quad 105 \quad 140 \quad 168 \quad 190 \quad 210 \ldots$
The list is incorrect.
$52$ does not have this property:
and neither $6$ nor $24$ divide $98 = 2 \times 7^2$.
Instead it appears that $56$ is the correct number here:
The list should be:
:$1, 3, 15, 30, 35, 56, 70, 78, 105, 140, 168, 190, 210$
== Sources ==
* : $210$
Category:Mistakes/Curious and Interesting Numbers
Numbers that Factorise into Sum of Digits and Reversal

== Theorem ==
The following positive integers can each be expressed as the product of the sum of its digits and the reversal of the sum of its digits:
:$1, 81, 1458, 1729$
== Proof ==
Let $n$ be a positive integer.
Let $S$ the sum of its digits and $S'$ be the reversal of the sum of its digits.
We wish to determine integers $n$ that satisfy:
:$n = S S'$
From this we have:
Suppose $n$ is a $d$-digit integer. 
Suppose $d \le 4$. Then $S \le 9 d \le 36$.
The values of $S S'$ are:
Among these values, only $1, 81, 1458, 1729$ have the desired property.
We claim that there are no integers with more than $4$ digits with this property.
 $d \ge 5$.
Suppose $d$ is a $k$-digit integer.
We have $k = 1 + \floor {\log d}$.
We show that $2 k + 2 \le d - 1$:
:$d = 5,6$ are single-digit integers, so $k = 1$.
:$d - 1 \ge 4 = 2 k + 2$
:For $d \ge 7$, consider the function $\map f d = d - 2 \log d - 5$.
:Then $\map f 7 > 7 - 2 - 5 = 0$.
:We also have $\map {f'} d = 1 - \dfrac 2 {d \ln 10} > 1 - \dfrac 1 d$,
:so $\map {f'} d > 0$ for all $d \ge 7$.
:By Real Function with Strictly Positive Derivative is Strictly Increasing, $f$ is strictly increasing for all $d \ge 7$.
:Then:
:So we have $2 k + 2 \le d - 1$ for all $d \ge 5$.
$9 d$ has not more than $k + 1$ digits.
Since $S \le 9 d$, $S$ cannot have more digits than $9 d$.
We also have that $S'$ cannot have more digits than $S$.
Therefore we have $S, S' < 10^{k + 1}$. 
Then $n = S S' < 10^{2 k + 2} \le 10^{d - 1} \le n$, which is a contradiction.
The result follows by Proof by Contradiction.
== Historical Note ==
== Sources ==
* : $1729$
Category:Fermat Pseudoprimes
Category:1729
Numbers that cannot be made Prime by changing 1 Digit

== Theorem ==
The following positive integers cannot be made into prime numbers by changing just one digit:
:$200, 202, 204, 205, 206, 208, \ldots$
== Proof ==
In order to make any one of these positive integers into a prime number one would have to change the last digit.
Otherwise the number it was changed into would be either even, or divisible by $5$, and so not prime.
But all the other integers between $200$ and $209$ are composite:
In order for there to be a smaller number with this property, a prime gap would need to be found which spans an entire decade.
There are none such.
On the other hand, each of $201, 203, 207, 209$ can themselves be turned into a prime number by changing the initial $2$ into a $1$:
:$101, 103, 107,109$
are all prime.
== Sources ==
* : $101$
* : $200$
Category:Prime Numbers
Category:Recreational Mathematics
Category:Numbers that cannot be made Prime by changing 1 Digit
Numbers the Multiple of whose Reciprocal are Cyclic Permutations

== Theorem ==
Let $m \in \Z_{>0}$.
Consider the reciprocal of $m$.
Let $n \in \Z$ such that $1 \le n < m$.
Then:
:The digits in the decimal expansion of the rational number $\dfrac n m$ form a cyclic permutation of the digits in the decimal expansion of $\dfrac 1 m$
:
:$(1): \quad m$ is the integer power of a prime number $p$
:$(2): \quad$ The period of recurrence of the decimal expansion of $1 / p$ is $p - 1$
:$(3): \quad p$ is not a divisor of $n$.
== Proof ==
== Examples ==
== Sources ==
* : $49$
* : $49$
Category:Reciprocals
Numbers the Multiple of whose Reciprocal are Cyclic Permutations/Examples/49

== Example of Numbers the Multiple of whose Reciprocal are Cyclic Permutations ==
Let $n \in \Z$ such that $1 \le n < 49$ and that $7 \nmid n$, where $\nmid$ denotes that $7$ is not a divisor of $n$.
The digits in the decimal expansion of the rational number $\dfrac n {49}$, for $1 \le n < 49$, form a cyclic permutation of the digits in the decimal expansion of $\dfrac 1 {49}$:
Patterns abound.
== Sources ==
* : $49$
* : $49$
Category:49
Category:Examples of Reciprocals
Numbers which Multiplied by 2 are the Reverse of when Added to 2

== Theorem ==
... and so on:
== Proof ==
We have that:
:$\ds \paren {4 \times 10^n + \sum_{k \mathop = 1}^{n - 1} 9 \times 10^k + 7} + 2 = 4 \times 10^n + \sum_{k \mathop = 0}^{n - 1} 9 \times 10^k$
using the Basis Representation Theorem.
It remains to be demonstrated that:
:$\ds 2 \times \paren {4 \times 10^n + \sum_{k \mathop = 1}^{n - 1} 9 \times 10^k + 7} = \sum_{k \mathop = 1}^n 9 \times 10^k + 4$
again using the Basis Representation Theorem.
Thus:
== Sources ==
* : $47$
* : $499$
* : $47$
* : $499$
Category:Recreational Mathematics
Category:Reversals
Numbers which are Sum of Increasing Powers of Digits

== Theorem ==
The following integers are the sum of the increasing powers of their digits taken in order begin as follows:
:$0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 89, 135, 175, 518, 598, 1306, 1676, 2427, 2 \, 646 \, 798, 12 \, 157 \, 692 \, 622 \, 039 \, 623 \, 539$
== Proof ==
Single digit integers are trivial:
:$d^1 = d$
for all $d \in \Z$.
Then we have:
Then:
and finally:
== Sources ==
* : $135$
* : $135$
Category:Powers
Category:Recreational Mathematics
Numbers whose Cube equals Sum of Sequence of that many Squares/Examples

== Examples of Numbers whose Cube equals Sum of Sequence of that many Squares ==
=== $47$ as Sum of Sequence of $47$ Squares ===
=== $2161$ as Sum of Sequence of $2161$ Squares ===
=== $99 \, 359$ as Sum of Sequence of $99 \, 359$ Squares ===
Category:Numbers whose Cube equals Sum of Sequence of that many Squares/Examples
Numbers whose Cube equals Sum of Sequence of that many Squares/Examples/2161

== Example of Numbers whose Cube equals Sum of Sequence of that many Squares ==
:$\ds 2161^3 = \sum_{k \mathop = 1}^{2161} \paren {988 + k}^2$
== Proof ==
From Numbers whose Cube equals Sum of Sequence of that many Squares:
:$\ds m^3 = \sum_{k \mathop = 1}^m \paren {n + k}^2$
for $m = 2161$ and for $n$ given by:
:$n = \dfrac {m + 1} 2 \pm \dfrac 1 6 \sqrt {33 m^2 + 3}$
So in this instance:
This gives:
:$n = 988, n = -3150$
which leads to the two solutions:
== Sources ==
* 
Category:Numbers whose Cube equals Sum of Sequence of that many Squares/Examples
Category:2161
Numbers whose Cube equals Sum of Sequence of that many Squares/Examples/47

== Example of Numbers whose Cube equals Sum of Sequence of that many Squares ==
:$\ds 47^3 = \sum_{k \mathop = 1}^{47} \paren {21 + k}^2$
== Proof ==
From Numbers whose Cube equals Sum of Sequence of that many Squares:
:$\ds m^3 = \sum_{k \mathop = 1}^m \paren {n + k}^2$
for $m = 47$ and for $n$ given by:
:$n = \dfrac {m + 1} 2 \pm \dfrac 1 6 \sqrt {33 m^2 + 3}$
So in this instance:
This gives:
:$n = 21, n = -69$
which leads to the two solutions:
Hence we have:
== Sources ==
* 
Category:Numbers whose Cube equals Sum of Sequence of that many Squares/Examples
Category:47
Numbers whose Cube equals Sum of Sequence of that many Squares/Examples/99,359

== Example of Numbers whose Cube equals Sum of Sequence of that many Squares ==
:$\ds 99 \, 359^3 = \sum_{k \mathop = 1}^{99 \, 359} \paren {45 \, 449 + k}^2$
== Proof ==
From Numbers whose Cube equals Sum of Sequence of that many Squares:
:$\ds m^3 = \sum_{k \mathop = 1}^m \paren {n + k}^2$
for $m = 99 \, 359$ and for $n$ given by:
:$n = \dfrac {m + 1} 2 \pm \dfrac 1 6 \sqrt {33 m^2 + 3}$
So in this instance:
This gives:
:$n = 45 \, 449, n = -144 \, 809$
which leads to the two solutions:
Category:Numbers whose Cube equals Sum of Sequence of that many Squares/Examples
Category:99,359
Numbers whose Cube equals Sum of Sequence of that many Squares/Mistake

== Source Work ==
:
:The Dictionary
::$103,823$
== Mistake ==
:$103,823 = 47^3 = 22^2 + 23^2 + \ldots + 67^2 + 68^2$, the smallest representation of a cube as the sum of consecutive squares. The next smallest is $2161^3$.
What was omitted from here is the fact that the number of squares is $47$.
That is, this is the smallest $m$ such that $m^3$ equals the sum of $m$ consecutive squares, not just any number of them.
Whether there are indeed any other cubes less than $47^3$ or between $47^3$ and $2161^3$ which are the sum of any number of consecutive squares (apart from the trivial $6$th powers) needs to be established.
== Sources ==
* : $103,823$
Category:Mistakes/Curious and Interesting Numbers
Numbers whose Cyclic Permutations of 3-Digit Multiples are Multiples

== Theorem ==
Let $n$ be a two-digit positive integer with the following property:
:Let $m$ be a $3$-digit multiple of $n$.
:Then any cyclic permutation of the digits of $m$ is also a multiple of $n$.
Then $n$ is either $27$ or $37$.
== Proof ==
Let $m$ be a multiple of $n$ with $3$ digits.
Then we have:
Let us now cyclically permute the digits of $m$ by multiplying by $10$.
Then we have:
From the above, we see that:
:$n$ is a divisor of a cyclic permutation of $m$
:
:$n \divides \paren {10^3 - 1 }$
We now note that:
:$10^3 - 1 = 37 \times 27 = 37 \times 3^3$
Upon inspection, we see that the only $2$-digit factors are $27$ and $37$.
== Also see ==
* Cyclic Permutations of 5-Digit Multiples of 41
== Sources ==
* : $27$
* : $27$
Category:Number Theory
Category:27
Category:37
Numbers whose Difference equals Difference between Cube and Seventh Power/Mistake

== Source Work ==
:
:The Dictionary
::$125$
== Mistake ==
:$5^3 - 2^7 = 5 - 2$. The only known matching pattern is $13^3 - 3^7 = 13 - 3$.
In fact $5^3 - 2^7 = -3$, not $3$.
== Sources ==
* 
* : $125$
Category:Mistakes/Curious and Interesting Numbers
Numbers whose Digits are Unchanged when Subtracting Reversal

== Theorem ==
The following sequence consists of the integers which have the property that subtraction of their reversals results in anagrams of them:
:$954, 1980, 2961, 3870, 5823, 7641, 9108, 19980, 29880, 29961, 32760, \ldots$
== Proof ==
== Sources ==
* : $1980$
* : $1980$
Category:Reversals
Category:Anagrams
Numbers whose Divisor Sum is Square/Examples

== Examples of Numbers whose Divisor Sum is Square ==
=== $\sigma_1$ of $3$ is Square ===
=== $\sigma_1$ of $22$ is Square ===
=== $\sigma_1$ of $66$ is Square ===
=== $\sigma_1$ of $70$ is Square ===
=== $\sigma_1$ of $81$ is Square ===
=== $\sigma_1$ of $94$ is Square ===
=== $\sigma_1$ of $115$ is Square ===
=== $\sigma_1$ of $119$ is Square ===
=== $\sigma_1$ of $400$ is Square ===
Category:Numbers whose Divisor Sum is Square
Numbers whose Fourth Root equals Number of Divisors

== Theorem ==
There are $4$ positive integers whose $4$th root equals the number of its divisors:
== Proof ==
Suppose $N = \map {\sigma_0} {N^4}$.
By Divisor Count Function is Odd Iff Argument is Square, $N$ must be odd.
The case $N = 1$ is trivial.
Suppose $N$ is a prime power.
Write $N = p^n$.
By Divisor Count Function of Power of Prime:
:$N = \map {\sigma_0} {p^{4 n} } = 4 n + 1$
By Bernoulli's Inequality:
:$N = p^n \ge 1 + n \paren {p - 1}$
This gives us the inequality:
:$4 n + 1 \ge 1 + n \paren {p - 1}$
which can be simplified to:
:$4 \ge p - 1$
The only odd primes satisfying the inequality are $3$ and $5$.
We have:
:$\map {\sigma_0} {3^4} = 5 > 3^1$
:$\map {\sigma_0} {3^8} = 9 = 3^2$
:$\map {\sigma_0} {3^{4 n} } = 4 n + 1 < 3^n$ for $n > 2$
:$\map {\sigma_0} {5^4} = 5 = 5^1$
:$\map {\sigma_0} {5^{4 n} } = 4 n + 1 < 5^n$ for $n > 1$
:$\map {\sigma_0} {p^{4 n} } = 4 n + 1 < p^n$ for any $p > 5$
Hence $625$ and $6561$ are the only prime powers satisfying the property.
Note that Divisor Count Function is Multiplicative.
To form an integer $N$ with our property, we must choose and multiply prime powers from the list above.
The product $625 \times 6561 = 4 \, 100 \, 625$ gives equality.
If we chose any $\tuple {p, n}$ with $\map {\sigma_0} {p^{4 n} } < p^n$, we must choose $3^4$ in order for equality to possibly hold.
Then $\map {\sigma_0} {3^4} = 5 \divides N$, so a tuple $\tuple {5, n}$ must be chosen.
If $\tuple {5, 1}$ was chosen, $5^2 \nmid N$.
But $\map {\sigma_0} {3^4 \times 5^4} = 25 \divides N$, which is a contradiction.
Suppose $\tuple {5, n}$ with $n \ge 2$ was chosen.
Then $\map {\sigma_0} {3^4 \times 5^{4 n} } = 20 n + 1 < 3 \times 5^n$, a contradiction.
Thus we have exhausted all cases.
== Sources ==
* : $4,100,625$
Category:Fourth Powers
Category:Divisor Count Function
Numbers whose Product is Square are Similar Plane Numbers

== Theorem ==
== Proof ==
Let $a$ and $b$ be natural numbers such that $a b$ is square.
From :
:$a : b = a^2 : a b$
We have that $a b$ and $a^2$ are both square.
By Square Numbers are Similar Plane Numbers they are similar plane numbers.
From :
:there exists one mean proportional between $a^2$ and $a b$.
From :
:there exists one mean proportional between $a$ and $b$.
The result follows from .
== Sources ==
* : Book $\text{IX}$. Propositions
Category:Euclidean Number Theory
Numbers whose Square is Palindromic with Even Number of Digits

== Theorem ==
The sequence of positive integers whose square is a palindromic number with an even number of digits begins:
:$836, 798 \, 644, 64 \, 030 \, 648, 83 \, 163 \, 115 \, 486, 6 \, 360 \, 832 \, 925 \, 898, \ldots$
== Proof ==
== Sources ==
* : $836$
* : $698,896$
* 
* 
* : $836$
* : $698,896$
Category:Square Numbers
Category:Palindromic Numbers
Numbers whose Squares are Consecutive Odd or Even Integers Juxtaposed

== Theorem ==
Integers whose squares consists of $2$ consecutive odd or even integers juxtaposed include:
:$1127^2 = 01 \, 270 \, 129$
:$8874^2 = 78 \, 747 \, 876$
Such integers come in pairs which add to $1$ more than a power of $10$:
:$1127 + 8874 = 10 \, 001$
== Proof ==
== Historical Note ==
== Sources ==
* : $1127$
* : $1127$
Category:Square Numbers
Category:Recreational Mathematics
Numbers with 6 or more Prime Factors

== Theorem ==
The sequence of positive integers with $6$ or more prime factors (not necessarily distinct) begins:
:$64, 96, 128, 144, 160, 192, 216, 224, 240, 256, \ldots$
== Proof ==
== Sources ==
* : $64$
* : $96$
* : $64$
* : $96$
Category:Prime Numbers
Numbers with All Digits Have a Common Factor are Divisible by This Factor

== Theorem ==
A number expressed in decimal notation is divisible by $d$ if all its digits are divisible by $d$.
That is:
:$N = \sqbrk {a_0 a_1 a_2 \ldots a_n}_{10} = a_0 + a_1 10 + a_2 10^2 + \cdots + a_n 10^n$ is divisible by $d$ 
if
:$\gcd \set {a_0, a_1, \ldots, a_n}$ is divisible by $d$.
This theorem is in fact true in all base $b$.
== Proof ==
We can find integers $k_0, k_1, \dots, k_n$ such that $a_i = k_i d$ for all $0 \le i \le n$.
Then:
Since $k_0 + k_1 b + k_2 b^2 + \cdots + k_n b^n$ is a sum of integers, it is also an integer.
This proves that $a_0 + a_1 10 + a_2 10^2 + \cdots + a_n 10^n$ is divisible by $d$.
Category:Divisibility Tests
Numbers with Euler Phi Value of 72

== Example of Use of Euler $\phi$ Function ==
There are $17$ positive integers for which the value of the Euler $\phi$ function is $72$:
:$73, 91, 95, 111, 117, 135, 146, 148, 152, 182, 190, 216, 222, 228, 234, 252, 270$
== Proof ==
== Sources ==
* : $72$
Category:Euler Phi Function
Category:72
Numbers with Square-Free Binomial Coefficients

== Theorem ==
For every $n$ greater than $23$, there exists a binomial coefficient $\dbinom n k$ that is not square-free.
More specifically, the list of numbers $n$ such that $\dbinom n k$ are square-free for all $k = 0, \dots, n$ is given by:
:$1, 2, 3, 5, 7, 11, 23$
== Proof ==
=== Lemma ===
=== Case 1: $n \ge 25$ ===
Consider the case $n \ge 25$. 
 for all $0 \le m \le n$, $\dbinom n m$ is square-free.
Then for any prime $p$:
:$p^2 \nmid \dbinom n m$
for all $0 \le m \le n$.
In particular we have:
:$2^2 \nmid \dbinom n m$
and:
:$5^2 \nmid \dbinom n m$
There exists an integer $r$ where $2^r \le n < 2^{r + 1}$.
There also exists an integer $s$ where $5^s \le n < 5^{s + 1}$.
We have $r \ge 4$ and $s \ge 2$ from $n \ge 25 = 5^2 \ge 16 = 2^4$.
By the Lemma, we have:
:$2^{r - 1} \divides \paren {n + 1}$
and:
:$5^{s - 1} \divides \paren {n + 1}$
Because $2^{r - 1}$ and $5^{s - 1}$ are coprime, we have by Product of Coprime Factors:
:$2^{r - 1} 5^{s - 1} \divides \paren {n + 1}$
Thus:
which is a contradiction.
Therefore $\dbinom n m$ is not square-free for any $0 \le m \le n$.
=== Case 2: $9 \le n \le 24$ ===
Consider the case $9 \le n \le 24$.
Suppose for all $0 \le m \le n$, $\dbinom n m$ is square-free.
Then for any prime $p$:
:$p^2 \nmid \dbinom n m$
for all $0 \le m \le n$.
In particular we have:
:$2^2 \nmid \dbinom n m$
and:
:$3^2 \nmid \dbinom n m$
There exists an integer $r$ where $2^r \le n < 2^{r + 1}$.
There also exists an integer $s$ where $3^s \le n < 3^{s + 1}$.
From:
:$n \ge 9 = 3^2 \ge 8 = 2^3$
we have $r \ge 3$ and $s \ge 2$
By the Lemma, we have:
:$2^{r - 1} \divides \paren {n + 1}$
and:
:$3^{s - 1} \divides \paren {n + 1}$
So we have:
:$2^2 \divides \paren {n + 1}$
and:
:$3^1 \divides \paren {n + 1}$
Since $4$ and $3$ are coprime, we have by Product of Coprime Factors:
:$12 \divides \paren {n + 1}$
In the range $9 \le n \le 24$, only $11$ and $23$ satisfy this condition.
=== Case 3: $4 \le n \le 8$ ===
Consider the case $4 \le n \le 8$.
Suppose for all $0 \le m \le n$, $\dbinom n m$ is square-free.
Then for any prime $p$:
:$p^2 \nmid \dbinom n m$
for all $0 \le m \le n$.
In particular we have:
:$2^2 \nmid \dbinom n m$
There exists an integer $r$ where $2^r \le n < 2^{r + 1}$.
From $n \ge 4 = 2^2$:
:$r \ge 2$
By the Lemma, we have:
:$2^{r - 1} \divides \paren {n + 1}$
So we have:
:$2^1 \divides \paren {n + 1}$
In the range $4 \le n \le 8$, only $5$ and $7$ satisfy this condition.
=== Manual Checking ===
We have narrowed our list of possible candidates to:
:$1, 2, 3, 5, 7, 11, 23$
Manually checking each $n$ in the list shows that $\dbinom n m$ is square-free for each $0 \le m \le n$.
== Sources ==
* 
* : $23$
Category:Square-Free Integers
Category:Binomial Coefficients
Category:23