Two Irrational Numbers That Give the Last Non-Zero Digits of $n!$ and $n^n$

TL;DR

This paper constructs irrational numbers from sequences of last non-zero digits of factorials and powers, revealing new properties of these digit patterns and their relation to irrationality.

Contribution

It introduces a novel approach to generating irrational numbers using last non-zero digits of factorials and powers, connecting digit patterns to irrationality.

Findings

The constructed numbers are proven to be irrational.

Patterns in last non-zero digits influence the irrationality.

New connections between digit sequences and number theory are established.

Abstract

If we form a decimal where the nth digit is the last non-zero digit of $n!$ (likewise, the last non-zero digit of $n^n$), we obtain an irrational number

Equations

$$N=0.\overline{14765636901636567490}.$$

$$1!$$

$$2!$$

$$3!$$

$$4!$$

$$5!$$

$$6!$$

$$7!$$

$$8!$$

$$9!$$

$$10!$$

$$11!$$

$$12!$$

$$13!$$

$$14!$$

$$F=0.12642242888682\dots$$

$$F=0.1264224288868288682448464484688682224282242866264\dots$$

$$F=0.\dot{1}\dot{2}\dot{8}\dot{8}\dot{4}\dot{4}\dot{8}\dot{2}\dot{2}\dot{6}\dots$$

$$F=0.\dot{1}\dot{2}\dot{8}\dot{8}\dot{4}\dot{4}\dot{8}\dot{2}\dot{2}\dot{6}\dot{2}\dot{4}\dot{6}\dot{6}\dot{8}\dot{4}\dot{8}\dot{2}\dot{2}\dot{6}\dot{4}\dot{8}\dot{2}\dot{2}\dot{6}\dot{8}\dot{6}\dot{4}\dot{4}\dot{2}\dot{2}\dot{4}\dot{6}\dot{6}\dot{8}\dot{6}\dot{2}\dot{8}\dot{8}\dot{4}\dot{2}\dot{4}\dot{6}\dot{6}\dot{8}\dot{2}\dot{4}\dot{6}\dot{6}\dot{8}\dots$$

$$N$$

$$\text{mbox}andP$$

$$\text{mbox}lnzd(12)\cdot \text{mbox}lnzd(53)=2\cdot 3$$

$$\text{mbox}lnzd(12\cdot 53)=\text{mbox}lnzd(636)$$

$$\text{mbox}lnzd(15)\cdot \text{mbox}lnzd(22)=5\cdot 2$$

$$\text{mbox}lnzd(15\cdot 22)=\text{mbox}lnzd(330)$$

$$\text{mbox}lnzd(a\cdot b)$$

$$=$$

$$d$$

$$\text{mbox}andd$$

$$d=\text{mbox}lnzd((A+1)!)=\text{mbox}lnzd(A!)\cdot \text{mbox}lnzd(A+1)=d\cdot K\text{mbox}mod10.$$

$$d=\text{mbox}lnzd((B+1)!)=\text{mbox}lnzd(B!)\cdot \text{mbox}lnzd(B+1)=d\cdot 2K\text{mbox}mod10.$$

$$d=dK\text{mbox}mod10d=2dK\text{mbox}mod10,$$

$$\text{mbox}lnzd(x^x)$$

$$\text{mbox}lnzd(x^x)=(x^{\prime }{)}^{10^i\cdot x^{\prime }}\text{mbox}mod10.$$

$$\text{mbox}lnzd(x^x)$$

$$\text{mbox}lnzd((10^j+100({\lambda }_0+1)!{)}^{10^j+100({\lambda }_0+1)!})=\text{mbox}lnzd((10^j{)}^{10^j}).$$

$$(\text{mbox}lnzd(10^j+100({\lambda }_0+1)!){)}^4\text{mbox}mod10=1,$$

$$(\text{mbox}lnzd({\lambda }_0+1)!{)}^4\text{mbox}mod10=1.$$

Full text

Two Irrational Numbers That Give the Last Non-Zero Digits of $n!$ and $n^{n}$.

Gregory P. Dresden

Washington & Lee University

Lexington, VA 24450

Author’s Note: This is a slightly revised version of the article that appeared in print in Mathematics Magazine in October of 2001. The original proof of Theorem 2 was incorrect; I’ve fixed that mistake here. My thanks to Antonio M. Oller-Marcén and José Mara Grau for pointing out to me the error.

Also, Stan Wagon pointed out in a letter to Mathematics Magazine (February 2002) that the question of the periodicity of the last non-zero digit of $n!$ (our Theorem 1) appeared several times in Crux Mathematicorum in the 1990’s: see v. 18 n. 7 (Sep 1992) page 196 for the statement of the problem, v. 19 n. 8 (Oct 1993) page 228 for an incorrect solution, v. 19 n. 9 (Nov 1993) page 260 for Stan Wagon’s correct solution, and v. 20 n. 2 (Feb 1994) page 44 for another reference.

I wrote a sequel to this paper, called “Three Transcendental Numbers From the Last Non-Zero Digits of $n^{n}$, $F_{n}$, and $n!$”. It appeared in Mathematics Magazine, April 2008.

We begin by looking at the pattern formed from the last (i.e. unit) digit of $n^{n}$. Since $1^{1}=1$, $2^{2}=4$, $3^{3}=27$, $4^{4}=256$, and so on, we can easily calculate the first few numbers in our pattern to be $1,4,7,6,5,6,3,6\dots$. We construct a decimal number $N=0.d_{1}d_{2}d_{3}\dots d_{n}\dots$ such that the $n^{\mbox{th}}$ digit $d_{n}$ of $N$ is the last (i.e. unit) digit of $n^{n}$; that is, $N=0.14765636\dots$. In a recent paper [1], R. Euler and J. Sadek showed that this $N$ is a rational number with a period of twenty digits:

[TABLE]

This is a nice result, and we might well wonder if it can be extended. Indeed, Euler and Sadek in [1] recommend looking at the last non-zero digit of $n!$ (If we just looked at the last digit of $n!$, we would get a very dull pattern of all [math]’s, as $n!$ ends in [math] for every $n\geq 5$.)

With this is mind, let’s define $\mbox{lnzd}(A)$ to be the last nonzero digit of the positive integer $A$; it is easy to see that $\mbox{lnzd}(A)=A/10^{i}\mbox{ mod }10$, where $10^{i}$ is the largest power of $10$ that divides $A$. We wish to investigate not only the pattern formed by $\mbox{lnzd}(n!)$, but also the pattern formed by $\mbox{lnzd}(n^{n})$. In accordance with [1], we define the “factorial” number $F=0.d_{1}d_{2}d_{3}\dots d_{n}\dots$ to be the infinite decimal such that each digit $d_{n}=\mbox{lnzd}(n!)$, and we define the “power” number $P=0.d_{1}d_{2}d_{3}\dots d_{n}\dots$ to be the infinite decimal such that each digit $d_{n}=\mbox{lnzd}(n^{n})$, and we ask whether these numbers are rational (i.e. are eventually-repeating decimals) or irrational.

Although the title of this article gives away the secret, we’d like to point out that at first glance, our “factorial”  number $F$ exhibits a suprisingly high degree of regularity, and a fascinating pattern occurs. The first few digits of $F$ are easy to calculate:

[TABLE]

[TABLE]

[TABLE]

…

Reading the underlined digits, we have:

[TABLE]

Continuing along this path, we have (to forty-nine decimal places):

[TABLE]

It is not hard to show that (after the first four digits) $F$ breaks up into five-digit blocks of the form $x\ x\ 2x\ x\ 4x$, where $x\in\{2,4,6,8\}$, and the $2x$ and $4x$ are taken mod $10$. Furthermore, if we represent these five-digit blocks by symbols ($\dot{2}$ for $22428$, $\dot{4}$ for $44846$, $\dot{6}$ for $66264$, $\dot{8}$ for $88682$, and $\dot{1}$ for the initial four-digit block of $1264$), we have:

[TABLE]

Grouping these symbols into blocks of five and then performing more calculations (with the aid of Maple) give us $F$ to $249$ decimal places:

[TABLE]

The reader will notice additional patterns in these blocks of five symbols (twenty-five digits). In fact, such patterns exist for any block of size $5^{i}$. However, a pattern is different from a period, and doesn’t imply that our decimal $F$ is rational. Consider the classic example of $0.1\ 01\ 001\ 0001\ 00001\ 000001\ \dots$, which has an obvious pattern but is obviously irrational. It turns out that our decimal $F$ is also irrational, as the following theorem indicates:

Theorem 1. *Let $F=0.d_{1}d_{2}d_{3}\dots d_{n}\dots$ be the infinite decimal such that each digit $d_{n}=\mbox{\rm lnzd}(n!)$. Then, $F$ is irrational. *

As for our “power”  number $P$, it too might seem to be rational at first glance. P is only slightly different from Euler and Sadek’s rational number N, as seen here:

[TABLE]

(Again, calculations were performed by Maple.) Despite this striking similarity between $P$ and $N$, it turns out that $P$, like $F$, is irrational:

Theorem 2. *Let $P=0.d_{1}d_{2}d_{3}\dots d_{n}\dots$ be the infinite decimal such that each digit $d_{n}=\mbox{\rm lnzd}(n^{n})$. Then, $P$ is irrational. *

Before we begin with the (slightly technical) proofs, let us pause and see if we can get a feel for why these two numbers must be irrational. There is no doubt that both $F$ and $P$ are highly “regular”, in that both exhibit a lot of repetition. The problem is that there are too many patterns in the digits, acting on different scales. Taking $P$, for example, we note that there is an obvious pattern (as shown by Euler and Sadek in [1]) repeating every $20$ digits with $1^{1}$, $2^{2}$, $3^{3},\dots,9^{9}$ and $11^{11},12^{12},\dots,19^{19}$, but this is broken by a similar pattern for $10^{10},20^{20},\dots,90^{90}$ and $110^{110}\dots 190^{190}$, which repeats every $200$ digits. This, in turn, is broken by another pattern repeating every $2000$, and so on. A similar behaviour is found for $F$, but in blocks of $5$, $25$, $125$, and so on, as mentioned above. So, in vague terms, there are always “new patterns”  starting up in the digits of $P$ and of $F$, and this is what makes them irrational.

Are there some simple observations that we can make about $P$ and $F$ which might help us to prove our theorems? To start with, we might notice that every digit of $F$ (except for the first one) is even. Can we prove this? Yes, and without much difficulty:

Lemma 1. *For $n\geq 2$, then $\mbox{\rm lnzd}(n!)$ is in $\{2,4,6,8\}$. *

Proof: The lemma is certainly true for $n=2,3,4$. For $n\geq 5$, we note that the prime factorization of $n!$ contains more $2$’s than $5$’s, and thus even after taking out all the $10$’s in $n!$, the quotient will still be even. To be precise, the number of $5$’s in $n!$ (and thus the number of trailing zeros in its base-$10$ representation) is $\displaystyle e_{5}=\sum_{i=1}^{\infty}\left[n/5^{i}\right]$, which is strictly less than the number of $2$’s, $\displaystyle e_{2}=\sum_{i=1}^{\infty}\left[n/2^{i}\right]$ (here, $[\cdot]$ represents the greatest integer function). Hence, $n!/10^{e_{5}}$ is an even integer not divisible by $10$, and so $\mbox{lnzd}(n!)=n!/10^{e_{5}}\mbox{ mod }10$, which must be in $\{2,4,6,8\}$. This completes the proof.

Another helpful observation is to note that the lnzd function appears to be multiplicative. For example,

[TABLE]

However, we note that at times this “rule”  fails:

[TABLE]

So, we can only prove a limited form of multiplicativity, but it is useful none the less:

Lemma 2. *Suppose $a,b$ are integers such that $\mbox{\rm lnzd}(a)\not=5$, $\mbox{\rm lnzd}(b)\not=5$. Then, lnzd is multiplicative; that is, $\mbox{\rm lnzd}(a\cdot b)=\mbox{\rm lnzd}(a)\cdot\mbox{\rm lnzd}(b)$ mod $10$. *

Proof: Let $x^{\prime}$ denote the integer $x$ without its trailing zeros; that is, $x^{\prime}=x/10^{i}$, where $10^{i}$ is the largest power of $10$ dividing $x$. (Note that $\mbox{lnzd}(x)=x^{\prime}$ mod $10$.) By hypothesis, $a^{\prime}$ and $b^{\prime}$ are both $\not=0$ mod $5$, and so $(a\cdot b)^{\prime}\not=0$ mod $5$ and so $(a\cdot b)^{\prime}=a^{\prime}\cdot b^{\prime}$. Thus,

[TABLE]

This completes the proof.

We are now ready to supply the proof of Theorem 1, in which we show that $F$ is irrational. The proof is a little technical, but it relies first on assuming that $F$ has a repeating decimal expansion, then on choosing an appropriate multiple of the period $\lambda_{0}$ and choosing an appropriate digit $d$, in order to arrive at a contradiction.

Proof of Theorem 1: We argue by contradiction. Suppose $F$ is rational. Then $F$ is eventually periodic; let $\lambda_{0}$ be the period (i.e. for every $n$ sufficiently large, then $d_{n}=d_{n+\lambda_{0}}$). Write $\lambda_{0}=5^{i}\cdot K$ such that $5\not|\,K$ (we acknowledge that $K$ could be $1$) and let $\lambda=2^{i}\cdot\lambda_{0}=10^{i}\cdot K$. Then, $\mbox{lnzd}(\lambda)=\mbox{lnzd}(K)$, and since $5\not|\,K$, then $10\not|\,K$ and so $\mbox{lnzd}(K)=K$ mod $10$. Note also that $\mbox{lnzd}(2\lambda)=2K$ mod $10$. Choose $M$ sufficiently large so that both of the following are true: $\mbox{lnzd}(10^{M}+\lambda)=\mbox{lnzd}(\lambda)$ (this can easily be done by demanding that $10^{M}>\lambda$), and for all $n\geq M$, then $d_{n}=d_{n+\lambda_{0}}$, which of course would then equal $d_{n+\lambda}$. Finally, let $d=\mbox{lnzd}((10^{M}-1)!)$. By Lemma 1, $d\in\{2,4,6,8\}$, and since $10^{M}!=(10^{M}-1)!\cdot 10^{M}$, then $d$ also equals $\mbox{lnzd}(10^{M}!)$.

Since $\lambda$ is a multiple of the period $\lambda_{0}$, then if we let $A=10^{M}-1+\lambda$ and $B=10^{M}-1+2\lambda$, then:

[TABLE]

Let’s now look at the last two terms in the above equation; it is here we will find our contradiction. Note that since $\mbox{lnzd}(A!)=d$, then $\mbox{lnzd}(A!)\not=5$. Also, since $\mbox{lnzd}(A+1)=\mbox{lnzd}(10^{M}+\lambda)=\mbox{lnzd}(\lambda)=K$ mod $10$, we know that $\mbox{lnzd}(A+1)\not=5$. Thus, we can apply Lemma 2 to $\mbox{lnzd}(A!\cdot(A+1))$ to get:

[TABLE]

Likewise, working with $B$, we find:

[TABLE]

Combining these two equations, we get:

[TABLE]

and this becomes $d(1-K)=0=d(1-2K)$ mod $10$. Since $d$ is even, this implies that $1-K=0$ mod $5$ and $1-2K=0$ mod $5$, which is a contradiction. Thus, there can be no period $\lambda_{0}$ and so $F$ is irrational. This completes the proof.

We now turn our attention to the “power”  number P derived from the last non-zero digits of $n^{n}$. This part was more difficult, but a major step was the discovery that the sequence $\mbox{lnzd}(100^{100})$, $\mbox{lnzd}(200^{200})$, $\mbox{lnzd}(300^{300})\dots$ was the same as the sequence $\mbox{lnzd}(100^{4})$, $\mbox{lnzd}(200^{4})$, $\mbox{lnzd}(300^{4})\dots$. This relies not only on the fact that $4|100$ but also on the fact that $a^{b}=a^{b+4}\mbox{ mod }10$ for $b>0$, as used in the following lemma:

Lemma 3. *Suppose $100\mid x$. Then, $\mbox{\rm lnzd}(x^{x})=(\mbox{\rm lnzd }x)^{4}$ mod $10$. *

Proof: As in Lemma 2, let $x^{\prime}$ denote the integer $x$ without its trailing zeros; that is, $x^{\prime}=x/10^{i}$, where $10^{i}$ is the largest power of $10$ dividing $x$. Now,

[TABLE]

Since $10\not|\,x^{\prime}$, then $10\not|\,(x^{\prime})^{10^{i}\cdot x^{\prime}}$, and so:

[TABLE]

Since $100\mid x$, then $4\mid 10^{i}\cdot x^{\prime}$, and since $(x^{\prime})^{n}=(x^{\prime})^{n+4}$ mod $10$ for every positive $n$, we have:

[TABLE]

This completes the proof.

With Lemma 3 at our disposal, the proof of Theorem 2 is now fairly easy.

Proof of Theorem 2: Again, we argue by contradiction. Suppose $P$ is rational. Let $\lambda_{0}$ be the period, and choose $j$ sufficiently large such that $10^{j}>100(\lambda_{0}+1)!$ and such that $\mbox{lnzd}((10^{j}+n\lambda_{0})^{10^{j}+n\lambda_{0}})=\mbox{lnzd}((10^{j})^{10^{j}})$ for every positive $n$. Choosing $n=100(\lambda_{0}+1)(\lambda_{0}-1)!$, we get:

[TABLE]

We reduce the left side of the above equation by Lemma 3 and the right side is obviously $1$, so we have:

[TABLE]

but since $10^{j}>100(\lambda_{0}+1)!$ and $\mbox{lnzd}(100(\lambda_{0}+1)!)=\mbox{lnzd}((\lambda_{0}+1)!)$, we can rewrite the above equation as:

[TABLE]

Note that by Lemma 1, the only values of $\mbox{lnzd}((\lambda_{0}+1)!)$ are $2,4,6,$ and $8$, and raising these to the fourth power mod $10$ gives us:

[TABLE]

which is a contradiction. Thus, $P$ is irrational. This completes the proof.

We close by asking the obvious (and very difficult) question: Are $F$ and $P$ algebraic or transcendental? I suspect the latter, but it is only a hunch, and I hope some curious reader will continue along this interesting line of study.