# Some Theorems on the divisibility of N! Factor

Paul Francisco Marrero Romero, Samuel Flores, Eduardo J. Acuña T

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Distributed under a Creative Commons Attribution 4.0 International License
# SOME THEOREMS ON THE DIVISIBILITY OF $N!$ FACTOR

A PREPRINT

Paul Marrero

Samuel J. Flores M.

Eduardo J. Acuña T.

Paulqed0@protonmail.com

sflores4@uc.edu.ve

eduardonumberst9@gmail.com

January 27, 2022

## ABSTRACT

This document is a compilation of theorems on factorials, a brief proof of the relationship of factorials with multiples of 9 from $n! \ge 6$, and general.

## 1 Introduction

A factorial function is the operation where all positive integers from 1 to $n!$ must be multiplied. In search of the properties of module 9, during the investigation [1] Tarazona, E. J. A., Flores, S., and Marrero, P. F. (2021). When we realized that every factorial greater than or equal to 6 was multiples of 9 up to infinity, it was from this that we started to look for the rest of the properties that could appear.

## 2 Theorem 1

If $n \in \mathbb{N}/n \ge 6$, then $9|n$

### 2.1 Proof

As we can see the base case implies that $n = 6$.

1. If $n = 6$, this preposition is $9|6$!; If we use the recurrence ratio of the factorial, which is $n! = n(n-1)!$, then we verify that:

$$6! = 6(6 - 1)! = 6(5)! = 720$$

From the above result we can say that:

$$9|6!$$

2. Let $k \ge 6$. We have to prove that: If

$$9|k! \rightarrow 9|(k + 1)!$$

For the above we will use the direct method. Suppose $9|k!$, therefore $k! = 9c$. For any $c \in \mathbb{N}$
Now we notice that $k! = k(k - 1)! = 9c$. So we have the following:

$$ \begin{aligned} (k + 1)[(k + 1) - 1]! &= (k + 1)k! \\ &= (k + 1)9c \\ &= 9c(k + 1) \\ &= 9[c(k + 1)] \end{aligned} $$

As $[c(k+1)] \in \mathbb{N}$, this shows that $(k+1)[(k+1)-1]!$ It's a whole multiple of 9, then $9|(k+1)[(k+1)-1]! = 9|(k+1)!$
Now we have proven that $9|k!$ Implies that $9|(k+1)!$.

Finally by induction method we have shown that $9|n!$ For all natural $n \ge 6$.

☐

## 2.2 Corollary 1

If $n \in \mathbb{N}$, $n \ge 6$ then $k|n!$ with $k = 3, 4, 5$.

## 2.3. Proof.

By Theorem 1 for $n \ge 6$, then

$$n! = \prod_{i=1}^{n} i = 3 \prod_{i \in [n]_3}^{n} i = 4 \prod_{i \in [n]_4}^{n} i = 5 \prod_{i \in [n]_5}^{n} i \quad (1)$$

where $[n]_s \equiv \{1, \dots, n\} - \{s\}$ for $n, s \in \mathbb{N}$.

### AXIOM

Being commutative multiplication in $\mathbb{N}$. It is known that:

* if $n \ge 2$, then $2|n$

☐

# 3 Theorem 2.

If $n, q \in \mathbb{N}$, $n \ge q$ then $q|n$

## 3.1 Proof

The base case determines that $n = q$. From this, we have to:

1. By recurrence ratio, we know that $n! = n(n-1)!$ so if $n = q$, we've got $n! = q(q-1)!$ and as we know $(q-1)! \in \mathbb{N}$.

Then if we do $(q-1)! = c$, where $c \in \mathbb{N}$ we can fundamentally verify that $n! = qc$, and form here we conclude that $n!$ it's a multiple of $q$. $q|n!$ for $n = q$.

2. Let $n \ge q$. Now suppose that $q|k!$ i.e. $k! = qc$ for any $c \in \mathbb{N}$.

By recurrence ratio we have to $k! = k(k-1)!$. We can see that:

$$\begin{aligned} (k+1)[(k+1)-1]! &= (k+1)k! \\ &= (k+1)qc \\ &= qc(k+1) \\ &= q[c(k+1)]. \end{aligned}$$

As $[c(k+1)] \in \mathbb{N}$, it is shown that $(k+1)!$ is a natural multiple of $q$. Then we have to $q|(k+1)!$. Now we've shown that $q|k! \rightarrow q|(k+1)!$.

☐

## 3.2 corollary

If $n \in \mathbb{N}$, $n \ge q$ if $q|n!$ then:

(i) $q^n|(n!)^n$.

(ii) $q(n!)^m|q^{1-n}(n!)^{n+m}$.
## 3.3 Proof

### 3.3.1 (i)

if $n, q \in \mathbb{N}$, $n \ge q$ with $q|n!$, then $q^n|(n!)^n$.

The base case determines that $n = q$. From this, we have that:

a) As we know $n! = n(n-1)!$, therefore $q! = q(q-1)!$, since $n = q$. Now, how $q \in \mathbb{N}$, We can say that $(q-1)! = b$, such that $b \in \mathbb{N}$. then we have that:

$$ (n!)^n = (qb)^n $$

$$ (n!)^n = q^n b^n. $$

therefore, it is evident that:

$$ q^n|q^n b^n \therefore q^n|(n!)^n \text{ for } n = q. $$

b) let $n \ge q$. For this case we will test if $q^k|(k!)^k$, then $q^{k+1}|(k+1)!^{k+1}$. Now we will use the direct method. Suppose that $q^k|(k!)^k \therefore (k!)^k = q^k = c$, for some $c \in \mathbb{N}$. now we observe that $(k!) = k(k-1)!$, then also we can affirm that $(k!)^k = [k(k-1)!]^{k+1}$ therefore it is valid if we say that:

$$ \begin{aligned} [(k+1)!]^{k+1} &= [(k+1)((k+1)-1)] \\ &= [(k+1)(k)!]^{k+1} \\ &= (k+1)^{k+1}(k!)^{k+1} \\ &= (k+1)^{k+1}(k!)^k(k)! \end{aligned} $$

And as by hypothesis we know that $q|n!$, for $n \ge q$, we have:

$$ \begin{aligned} [(k+1)]^{k+1} &= (k+1)^{k+1}q^kcd, \text{ such that } d \in \mathbb{N} \\ &= (k+1)^{k+1}q^kqcd \\ &= (k+1)^{k+1}q^{k+1}cd \\ &= q^{k+1}[(k+1)^{k+1}cd]. \end{aligned} $$

therefore $q^{k+1}|(k+1)^{k+1}$.

in conclusion, by mathematical induction, we have proved that $q^n|(n!)^n$ for all $n, q \in \mathbb{N}$. $\square$

### 3.3.2 (ii)

if $n, m, q \in \mathbb{N}$, $n \ge q$ with $q|n!$, then $q(n!)^m|q^{1-n}(n!)^m$.

a) As we can see, the base case implies that $n = m = q$. In addition to the above, we know that $(n!)^{m+n} = (n!)^m(n!)^n$, from where also:

$$ (n!)^{m+n} = [n(n-1)!]^m[n(n-1)!]^n. $$

therefore, therefore, as $n = m = q$, it is possible to say that $(n!)^{m+n} = [q(q-1)]^q[q(q-1)]^q$. Based on the above equality, we are finally left with the following:

$$ \begin{aligned} q^{1-n} &= q^{1-q}[q(q-1)]^{2q} \\ &= qq^{-q}q^{2q}[(q-1)!]^{2q} \\ &= qq^q[(q-1)!]^{2q} \\ &= qq^q[(q-1)!]^q[(q-1)!]^q \\ &= q[q(q-1)]^q[(q-1)!]^q. \end{aligned} $$

And as $q \in \mathbb{N}$, then it is valid to say that $[(q-1)]^q = c$ such that $c \in \mathbb{N}$. Therefore:

$$ q^{1-n}(n!)^{m+n} = q[q(q-1)]^q c. $$

In this case we study $n = m = q$, finally we have that

$$ q^{1-n}(n!)^{m+n} = q[n(n-1)]^m c. $$

From the above we conclude that $q(n!)^m|q^{1-n}(n!)^{m+n}$ for $n = m = q$.
b) Case $n \ge m \ge q$. The objective in this case is to prove that

$$q(k!)^m |q^{1-k}(k!)^{m+k} \Rightarrow q[(k+1)!]^m |q^{1-(k+1)}[(k+1)!]^{m+k+1}.$$

Now suppose that $q(k!)^m |q^{1-k}(k!)^{m+k}$. Then, by factorial recurrence relationship, we can say that:

$$q^{1-k}(k!)^{m+k} = q^{1-k}[k(k-1)!]^{m+k}$$

from the above equality we have that

$$\begin{align*}
q^{1-(k+1)}[(k+1)!]^{m+(k+1)} &= q^{1-k}[(k+1)((k+1)-1)!]^{m+(k+1)} \\
&= q^{-k}[(k+1)k!]^{m+k+1} \\
&= q^{-k}(k+1)^{m+k+1}(k!)^{m+k+1} \\
&= q^{-k}(k+1)^{m+k+1}(k!)^{m+k}(k!).
\end{align*}$$

According to **Theorem 2**, because $k, q \in \mathbb{N}/k \ge q \therefore q|k!$ ergo:

$$\begin{align*}
q^{1-(k+1)}[(k+1)!]^{m+(k+1)} &= q^{-k}(k+1)^{m+k+1}(k!)^{m+k}qc, \quad c \in \mathbb{N} \\
&= q^{1-k}(k!)^{m+k}(k+1)^{m+k+1}c \\
&= q(k!)^m b(k+1)^{m+k+1}c \\
&= q(k+1)^m(k!)^m(k+1)^{m+1}bc \qquad = q[(k+1)k!]^m(k+1)^{m+1}bc.
\end{align*}$$

And by the recurrence relation of the factorials, we know that

$$(k+1)! = (k+1)k!/k \in \mathbb{N} \therefore q^{1-(k+1)}[(k+1)!]^{m+(k+1)} = q[(k+1)!]^m(k+1)^{m+1}bc,$$

from where it is evident that $(k+1)^{m+1}bc = d$ such that $d \in \mathbb{N}$. Therefore, we finally have that

$$q^{1-(k+1)}[(k+1)!]^{m+(k+1)} = q[(k+1)!]^m d.$$

From the above we conclude that $q[(k+1)!]^m |q^{1-(k+1)}[(k+1)!]^{m+(k+1)}$ for $n \ge m \ge q$. in this way by mathematical induction we have proved (ii). $\square$

# 4 Special Acknowledgments

We would like to thank Dr. Samuel Flores for his guidance during our research, for his ideas and corrections, for his patience and support.

# Bibliography

[1] E. J. A. Tarazona, S. Flores, and P. F. Marrero. Matrize_φ^9. clasificación de números enteros. algebra primordial. *Revista MATUA* ISSN: 2389-7422, 8(1):10-45, 2021.