When are $1/n!$ repeating decimal with single digit repetend

Question

Find all $n$ where $1/n!$ is repeating decimal with single digit repetend(for example $0.4111111...$ but not $0.412121212...$) but cannot be expressed as a terminating decimal (for example $0.9999999$ doesn't count since it can be expressed as terminating decimal $1.0$)

From my experiment, it seems like once $n$ has prime factor 7, it won't work.

I am not sure how to express "repeating decimal with single digit repetend". It seems a sufficient condition is $10/n! - 1/n! = p/q$ but it's not a necessary condition.

Answer 1

A positive rational number has terminating decimal expansion iff it can be expressed as $x=A/10^k$ for some $A,k \in \mathbb{N}$.

A rational number has a length-1 periodic decimal expansion iff it's not in the form above, but $10x-x=9x=A/10^k$. Then, if $x=1/m$ we require that $m = 2^a 5^b 3^c$ with $a \ge 0$, $b\ge 0$ and $c \ge 1$.

That is, it must be a multiple of $3$, and it can be a multiple of $2$ and $5$, with no other prime factor.

Then, if $m=n!$, we must have $n \in \{ 3, 4, 5, 6 \}$

Answer 2

If you know that $\frac{1}{9} = 0.1111\cdots$, then a number of the form $0.a_1a_2 \cdots a_k bbb\cdots$ is equal to $$\frac{[a_1 a_2 \cdots a_k]}{10^k} + \frac{b}{9 \cdot 10^k} = \frac{9[a_1 a_2 \cdots a_k] + b}{9 \cdot 10^k},$$ where $[a_1 a_2 \cdots a_k]$ denotes the $k$-digit number with digits $a_1,\ldots, a_k$.

If this equals $\frac{1}{n!}$ for some $n$, then $9 \cdot 10^k$ must be divisible by $n!$. This supports your observation that things don't work if $n \ge 7$, since $9 \cdot 10^k$ is not divisible by $7$. It thus suffices to check $n=1,\ldots, 6$ manually.

Answer 3

Well.... Let's think.

$\frac 1{n!}=0.a_1a_2.....a_m\overline{kkkkk}= \frac {a_1a_2....a_m}{10^m} + \frac k{9*10^m}=$

$\frac {9\cdot a_1a_2....a_m + k}{9\cdot 10^m}$

So $n! = \frac {9*10^m}{9\cdot a_1a_2...a_m + k}\in \mathbb Z$

This means the only prime factors of $n!$ are $2,3$ and $5$(!!!). That means $n< 7$. And to not terminate that means $n! \ge 3$ and $n\ge 3$.

There are only a finite number to check. $\frac 1{3!} = \frac 16 = 0.1\overline 6$. And $\frac 1{4!}=\frac 1{24}=0.041\overline 6$ and so on.

If there are more than $1$ repeating decimal, say $w$ of them, we will get $n! = \frac {\underbrace{99...9}_w\cdot 10^m}{something + somethingelse}$ and those $n!$ will have only prime factof of $2,5,3$ and the prime factors of $\underbrace{111......1}_w$ but as all numbers not divisible by $2$ or $5$ will divide some $\underbrace{111...1}$ that will always be possible. (Although particular $w$s will put severe restrictions).

(For example to have only $2$ repeating can not happen as $11|99$ but $7\not\mid 99$. But $7\&11\&13|999999$ so $\frac 1{7!}$ through $\frac 1{16!}$ all have $6$ repeating digits.)