Prime Generating Irrational Number

Does there exist an irrational number such that every time it is multiplied by 100 its integer part gives a prime number?

$$\phi= 0,a_0a_1a_2a_3\cdots$$
$$\lfloor 10^{2n}\phi \rfloor \in \mathbb P,\quad \forall n \in \mathbb N$$

Or in a more general way multiply by $10^{p.n}$, where p is a fixed prime number.

For example, let $\phi = 0,1163\cdots$
$$\lfloor 10^2 \phi \rfloor = 11$$
And $$\lfloor 10^4 \phi \rfloor = 1163$$
While 11 and 1163 are primes, 63 by itself is not. So, $a_{2i}a_{2i+1}$ is not necessarily prime for $i \in \mathbb N$

Answer

This is similar to right-truncatable primes, but removing two decimal digits at a time.

Equivalently, we are looking for right-truncatable primes in base $100$. http://oeis.org/A076586 tells us that there are exactly $9823399067$ such primes. Hence there is no infinite example and no such irrational number.

For the more general question of multiplying by $10^{p\cdot n}$, we can note that the natural density of primes is $0$, and that arbitrarily large prime gaps exists, to conclude that the number of right-truncatable primes in any given base is almost certainly finite. However, I don't believe a proof exists. See for example Proof for finite number of truncatable primes .