# What is the factorial of 1 3

**What is a faculty?**

...... | Every natural number n has a factorial. It is the product of the natural numbers that are less than or equal to the number n. You write it as n! = 1 * 2 * 3 * ... * (n-1) * n and read it n faculty. |

One can n! as a function with the domain D = | N, i.e. as a sequence.

Then a_{n}= n! or written recursively a_{i}= i * a_{i-1}. (i = 1, 2, 3, ..., n).

These are the first 15 numbers, such as those provided by the TI-30 pocket calculator.

1!=12!=23!=64!=245!=120 | 6!=7207!=50408!=403209!=36288010!=3628800 | 11!=3991680012!=47900160013!=622702080014!=8.717829120*10^{10}15!=1.307674368*10^{12} |

All numbers are composite numbers that have ever increasing prime numbers as divisors. Among them are always pairs of 2 and 5, which lead to a growing number of ending zeros.

On this page you will find mostly entertaining faculties.

**Great faculties **Top

calculator

Faculties are growing rapidly. Up is 15! already 13 digits.

The largest number the TI-30 calculator can display is 69! = 1.711224524 * 10^{98}.

In between there is a number that is easy to remember, namely 25! is about 10^{25}, more precisely 1.6 * 10^{25}.

Google calculator

Larger faculties up to 170! can be found by typing them into the Google search box and pressing Enter For example, the Google calculator finds 100! = 9.33262154 × 10

^{157}.

Tables

...... | Since stochastics found its way into schools, faculties have gained in importance there. Tables with faculties and binomial coefficients are available today. |

Numerical representations

Even larger faculties can be found online at

*Wolfram Demonstration Project*or at

*Nitrxgen*(URLs below) and then also in the exact number spelling.

100!=9332621544394415268169923885626670049071596826438162146859296389521759999322

9915608941463976156518286253697920827223758251185210916864000000000000000000000000

The number has 158 digits, including 24 ending zeros.

Stirling formula

Large faculties can also be determined with the help of the Stirling formula (and other approximation formulas).

Obviously the quotient does not converge strongly, as the following numerical example shows.

50! = 50 ^ 50 * e ^ (- 50) sqrt (2 * pi * 50) = 4.2 * 10

^{64}. - It should be 50! = 3.0 * 10

^{64}.

It is noteworthy that the constants e and pi appear. In the background is the gamma function, which can be used to prove the Stirling formula. See also (2), page 97f.

**Number of ending zeros **Top

The top turns 100! mentioned with 158 digits and 24 ending zeros.

The question arises as to how one can determine the number of ending zeros.

The following rule applies.

It is enough to examine the last number, here 100. In a first step you divide 100 by 5, which is 20. Then you divide 20 again by 5 and get 4. The sum of the quotients is 20 + 4 = 24, and that is the number of final zeros.

This example is not typical because there are no remainders when dividing.

A second example is 81! .

In a first step 81 is divided by 5, which is 16. The rest is of no interest.

Then divide 16 by 5 again and get 3. Again, leave out the remainder.

The sum of the quotients is 16 + 3 = 19, and that is the number of ending zeros.

**Number of digits **Top

A game with faculties is to determine the number of their digits. (1)

It then allows the numbers to be written as figures.

2909005041013 0580032964972 0646107774902 5791441766365 7322653190990 5153326984536 5268082403397 7639893487202 9657993872907 8134368160972 8000000000000 0000000000000 | 081 39675 8240290 900504101 30580032964 9720646107774 902579144176636 57322653190990515 3326984536526808240 339776398934872029657 99387290781343681609728 000000 0000000000000000000 | The number 105! has 169 digits. So the digits form a 13 * 13 square. Because of 169 = 1 + 3 + 5 + ... + 25 = 13² |

7 9 7 1 2 6 0 2 0 7 4 7 3 6 7 9 8 5 8 7 9 7 3 4 2 3 1 5 7 8 1 0 9 1 0 5 4 1 2 3 5 7 2 4 4 7 3 1 6 2 5 9 5 8 7 4 5 8 6 5 0 4 9 7 1 6 3 9 0 1 7 9 6 9 3 8 9 2 0 5 6 2 5 6 1 8 4 5 3 4 2 4 9 7 4 5 9 4 0 4 8 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 | The number 81! has 121 digits. This number is the sum of the triangular numbers d Therefore you can build a figure from two triangles. |

0 5 9 2 0 8 2 4 7 0 6 6 6 7 2 3 1 7 0 3 0 6 7 8 5 4 9 6 2 5 2 1 8 6 2 5 8 5 5 1 3 4 5 4 3 7 4 9 2 9 2 2 1 2 3 1 3 4 3 8 8 9 5 5 7 7 4 9 7 6 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 | The hexagon is given by the number 91 of the faculty 65! certainly. It applies (i.e. |

**Prime numbers **Top

The powers of two 2^{n} with many factors lead to prime numbers if they are decreased by 1 or increased by 1. These are then the well-known Mersenne numbers 2^{n}-1 or Fermat numbers 2^{n}+1. The question is whether one also gets prime numbers if one forms n! -1 and n! +1 with the help of the faculties.

An answer can be found at N. J. A. Sloane.

n! -1 is a prime number for 3, 4, 6, 7, 12, 14, 30, 32, 33, 38, 94, 166, ...

n! +1 is a prime number for 1, 2, 3, 11, 27, 37, 41, 73, 77, 116, ...

Wilson theorem

For every prime number, the following applies: The term (n-1)! + 1 is divisible by n if and only if n is a prime number.

To illustrate the sentence, a table with examples follows. The red numbers are prime numbers. Then the division opens.

n (n-1)! + 1 [(n-1)! + 1] : n | 2 1 | 3 3 | 7 1,8 | 25 5 | 121 20,2 | 721 103 | 3628801 329891 | 479001601 36846277 |

It also applies: If the term t (p) = (p-1)! + 1 is divisible by p², then p is a Wilson prime number.

With N. J. A. Sloane one finds the following Wilson primes 5, 13 and 563 known up to now.

**Derivatives **Top

The nth derivative of the power function f (x) = x^{n }is f ^{(n)}(x) = n!

Faculties appear in the Taylor series.

f (x) = f (a) + [f '(a) / 1!] (x-a) + [f' '(a) / 2!] (x-a) ² + ... + [f

^{(n)}(a) / n!] (x-a)

^{n}

The meaning of this formula lies in the fact that one can use it to lead to series expansions of i.a.

^{x}, ln (1 + x), sin (x), cos (x), tan (x) and arc tan (x). You can find it at Wikipedia

*Taylor series*.

**Permutations **Top

If one gives n different elements of a set, then there are n! different arrangements of the elements.

This is how one encounters the faculties at school.

example

Let the set {a, b, c} be given.

The order of the three elements can be changed in six ways, namely to abc, acb, bac, bca, cab, cba.

That's 3! = 6 arrangements or, as they say, 3!

*Permutations*.

If one wants to prove in general that n elements n! Have permutations, one chooses the method of complete induction.

The initial set is {a

_{1}, a

_{2}, a

_{3}, ..., a

_{n-1}, a

_{n}} or for simplicity {1, 2, 3, ..., n-1, n}.

proof

Prerequisite: The number of permutations of n elements is n !.

The cases n = 1 and n = 2 are trivial.

For n = 3 the following applies (see above): There are 3! = 6 permutations of three elements.

It has to be shown that n + 1 elements (n + 1)! Permutations have, provided that n elements n! Have permutations.

If you add to the n! Permutations add the element n + 1, so you can put it in n + 1 places of each of the n! Set permutations. So there are a total of n! * (N + 1) permutations, and that is (n + 1) !, wzbw.

**Binomial coefficient **Top

Pascal triangle

Binomial coefficients are the numbers of the Pascal triangle. In the 2nd line are the numbers 1, 2, 1. These are the prefixes of (a + b) ² = a² + 2ab + b². The 3rd line contains the numbers 1, 3, 3, 1. These are the prefixes of (a + b) ³ = a³ + 3a²b + 3ab² + b³. In the 4th line are the numbers 1, 4, 6, 4, 1 from (a + b) ^{4}= a^{4}+ 4a³b + 6a²b² + 4ab³ + b^{4}.... The nth line contains the prefixes of the calculated term to (a + b) ^{n}. |

The number that is in the kth position in the nth line can be calculated explicitly as "n over k", as an expression that is formed from factorials.

They are used in several places on my homepage.

Triangular numbers

Numbers lottery

If one sets k = 6 and n = 49, then the following applies: n! / [k! (n-k)!] =49!/(6!43!) =(49*48*47*46*45*44)/(2*3*4*5*6)... =13 983 816. | In the lottery 6 out of 49, the number of possibilities to tick six numbers is calculated as a binomial coefficient. More on my page 13 983 816. There n! / [K! (N-k)!] Is derived. |

Taxi route

...... | In taxi geometry, the problem arises, among other things, of determining the number of routes with a distance of 8 from A to B. |

Soma body

**Curiosities **Top

4!+1=5² | 5!+1=11² | 7!+1=71² | 145=1!+4!+5! | 40585=4!+0!+5!+8!+5! | 10!=6!*7!=3!*5!*7! |

**Faculties on the Internet **Top

German

Martin Steen

Faculty (numbers between 1 and 2500 are allowed as input)

Thomas Peters

The faculties

Wikipedia

Faculty (mathematics), faculty prime, Wilson prime, Stirling formula, gamma function, binomial coefficient, Taylor series

English

Eric W. Weisstein (MathWorld)

Factorial, Factorial Prime

Jiel Beaumadier, Matej Hausenblas

All about factorial notation

Nitrxgen

Factorial Calculator (Allowed range: 0 to 200,000)

N. J. A. Sloane (On-Line Encyclopedia of Integer Sequences)

Factorial numbers: A000142,

Pascal's triangle read by rows: A007318

Numbers n such that n! - 1 is prime. A002982

Numbers n such that n! +1 is prime. A002981

Wilson primes: primes p such that (p-1)! == -1 mod p ^ 2. A007540

The Wolfram Demonstrations Project

Factorial

Wikipedia

Factorial, Factorial prime, Wilson prime, Stirling's approximation, Gamma function, Binomial coefficient, Taylor series

**credentials **Top

(1) Martin Gardner: Geometry with Taxis, the heads of the Hydra and other mathematical gimmicks, Basel 1997

[ISBN 3-7643-5702-9]

(2) Jean-Paul Delahaye: Pi - The Story, Basel, Boston, Berlin 1999 [ISBN 3-7643-6056-9]

**Feedback:**Email address on my main page

URL of my homepage:

http://www.mathematische-basteleien.de/

© 2009 Jürgen Köller

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