# Does this number exist?

Does there exist $$x\in\mathbb{R}$$ such that $$\lfloor 10^nx\rfloor$$ is a prime number for all $$n\in\mathbb{N}$$?

• Like this ... $2$ is prime, $29$ is prime, $293$ is prime, $2939$ is prime; can we continue this indefinitely? For this start, no, at $29399999$ we get stuck. But what about other starting values? Jul 30, 2023 at 16:40
• All the integer parts will be right-truncatable primes of which there are finitely many, so the answer is no. Jul 30, 2023 at 16:58
• No, we can choose $E(x)$ a prime number of more of one digit Jul 30, 2023 at 17:19
• @GHfromMO Wojowu's comment doesn't technically answer the question, since the starting point of the sequence needn't be a single digit. Jul 31, 2023 at 0:29
• Euristically the answer seems a clear no. Say that the first prime in the sequence has length $m$ (so about $10^m/(m\log(10))$ possibilities for it) and we try to add a digit, we have a probability of $\frac{10}{\log(10^m)}$ to successfully land on a prime again, so we see that once $m$ is larger than $2$ the expected number of live branches decreases exponentially and soon will come to an end with probability $1$. Jul 31, 2023 at 12:37

(This is an extended comment.) There couldn't be anything special about base 10, could there?

Notation: Given two positive integers $$m,n$$, let $$m\oplus n$$ be the integer that results from prepending the digits of $$m$$ to the left of the digits of $$n$$ (in whatever base we are considering).

Base 2: Here we can only append digits $$0$$ and $$1$$. We can't append a $$0$$ digit to a prime and have it remain prime, since the result would be divisible by $$2$$. So, we can only repeatedly append $$1$$. But such a process cannot always result in a prime. Indeed, $$p\oplus \underbrace{111\cdots 1}_{p-1\text{ times}}$$ is divisible by $$p$$, when $$p$$ is prime.

Base 3: Again, we can ignore the digit $$0$$, due to 3-adic considerations. We can also ignore the digit $$1$$, due to 2-adic considerations. This only leaves us with the ability to append the digit $$2$$ repeatedly, which leads to a similar contradiction as at the very end of the base 2 case.

Base 4: We can ignore the digits $$0$$ and $$2$$, due to 2-adic considerations. Looking 3-adically, we see that $$n\oplus 1\equiv n+1\pmod{3}$$ and $$n\oplus 3\equiv n\pmod{3}$$. Thus, the digit $$1$$ can be appended only finitely many times (before we reach a number that is $$0\pmod{3}$$), and hence eventually we must only see the digit $$3$$ used. That leads to a contradiction as before.

Base 5: We can ignore the digit $$0$$. Also, due to 2-adic considerations, we can ignore the digits $$1$$ and $$3$$. This leaves the digits $$2$$ and $$4$$. If $$n\equiv 1\pmod{3}$$, then $$n\oplus 2\equiv 1\pmod{3}$$ while $$n\oplus 4\equiv 0\pmod{3}$$. Thus, for numbers that are $$1\pmod{3}$$, we can only append the digits $$2$$ repeatedly, and this leads to a contradiction. Similar considerations apply to numbers that are $$2\pmod{3}$$.

Base 6: We can ignore the digits $$0,2,3,4$$, leaving only $$1,5$$. Now, looking 5-adically, we see that $$n\oplus 1\equiv n+1\pmod{5}$$ while $$n\oplus 5\equiv n\pmod{5}$$. So, the digit $$1$$ can be appended only finitely many times, leaving us to eventually repeat only the digit $$5$$, which leads to a contradiction.

Base 7: We can ignore the digit $$0$$. Due to 2-adic considerations, we can ignore $$1,3,5$$. This leaves digits $$2,4,6$$. Suppose that we have reached a prime $$p\equiv 1\pmod{3}$$. Then $$p\oplus 2\equiv 0\pmod{3}$$, $$p\oplus 4\equiv 2\pmod{3}$$, and $$p\oplus 6\equiv 1\pmod{3}$$. Similar considerations apply when $$p\equiv 2\pmod{3}$$. So, what has to happen is that we append some number of $$6$$'s, then $$4$$, then some number of $$6$$'s, then $$2$$, then some number of $$6$$'s, then $$4$$ again, etc...

Looking 5-adically, we can get some additional restrictions on the sequence of appended digits. At this point, additional $$p$$-adic considerations, for primes $$p\geq 11$$, give additional restrictions, but they don't seem to solve the problem.

Base 10: A similar analysis, using 2-adic, 3-adic, and 5-adic considerations, leads to eventually only having use of the digits $$3$$ and $$9$$. Moreover, $$p$$-adic considerations, for primes $$p\geq 7$$, give limitations on the allowable sequences of $$3$$'s and $$9$$'s. Ultimately, I believe there is no local obstruction (but I could be wrong). This base seems more tractable than for bases 7,8,9.