Decimal representation begins with $N$

Question

Let $a \geq 2$ be a positive integer such that $\log_{10}a$ is irrational. Then for any positive integer $N$ there exist infinitely many powers $a^n$ of $a$ whose decimal representation begins with $N$.

I am not sure how to use the fact that $a^n$ begins with $N$. How should I prove this?

Jack D'Aurizio · Answer 1 · 2016-06-24T02:21:57.690

2

If $\log_{10}(a)\not\in\mathbb{Q}$, by setting $r_n=n\cdot\log_{10}(a)$ and considering the sequence $$ r_n\pmod{1}$$ we have that such a sequence is dense in the interval $(0,1)$ by the Dirichlet's box principle.
It follows that for infinite values of $n$ $$ r_n \in M_n+\left(\log_{10} 3,\;\log_{10} 4\right), \qquad M_n\in\mathbb{N}^* $$ holds, so $ 10^{r_n} = a^n $ lies between $3\cdot 10^{M_n}$ and $4\cdot 10^{M_n}$ and the first decimal digit of $a^n$ is $3$.

Obviously you may replace $3$ with any digit $\in\{1,2,3,4,5,6,7,8,9\}$ and the same argument still applies. Moreover, with a minor fix (by working with $\log_{100}$, $\log_{1000}$ and so on) we may also prove that the initial substring of the decimal representation can be any substring not starting with a zero.

edited Jun 24 '16 at 02:21

answered Jun 24 '16 at 01:34

Jack D'Aurizio

353,855

(+1) Nice! I was just thinking about using $\log_{10}(a)\notin \mathbb{Q}$ too, you are fast. – Sungjin Kim Jun 24 '16 at 01:38
Can you explain how $r_n$ is dense in $(0,1)$ using the pigeonhole principle? – Puzzled417 Jun 24 '16 at 01:39
@Puzzled417: http://math.stackexchange.com/questions/843763/a-dense-set-on-0-1 – Jack D'Aurizio Jun 24 '16 at 01:41
@Puzzled417 http://math.stackexchange.com/questions/903142/for-an-irrational-number-a-the-fractional-part-of-na-for-n-in-mathbb-n-is This has been on this site several times. – Sungjin Kim Jun 24 '16 at 01:42
How do you get $r_n \in M_n+\left(\log_{10} 3,;\log_{10} 4\right), \qquad M_n\in\mathbb{N}^* $? – Puzzled417 Jun 24 '16 at 01:48
@Puzzled417 That's the consequence of $r_n$ being dense in $(0,1)$. – Sungjin Kim Jun 24 '16 at 01:50
But that means that $r_n$ could lie in like $(4,5)$, which isn't a subset of $(0,1)$. – Puzzled417 Jun 24 '16 at 01:51
@Puzzled417 $r_n$ is considered modulo $1$. That means you can consider $r_n$ as a fractional part of $n\log_{10}(a)$. – Sungjin Kim Jun 24 '16 at 01:52
@i707107 Right, and how can the fractional part lie outside of $[0,1)$? – Puzzled417 Jun 24 '16 at 01:54
@Puzzled417 No, that does not happen. A fractional part always lies in $(0,1)$. – Sungjin Kim Jun 24 '16 at 01:55
@Puzzled417 I see your point. Maybe I didn't understand $r_n$ correctly. It seems that $r_n$ is exactly equals to $n\log_{10}(a)$, and its fractional parts are dense in $(0,1)$. – Sungjin Kim Jun 24 '16 at 02:06
@JackD'Aurizio Also, I think $n$ is supposed to be a positive integer? So how are you getting $10^{r_n} = a^n$? – Puzzled417 Jun 24 '16 at 02:10
@Puzzled417: $$ 10^{r_n} = 10^{n\cdot \log_{10}(a)} = a^n.$$ – Jack D'Aurizio Jun 24 '16 at 02:22

Decimal representation begins with $N$

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