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DFA, lower bound on number of states, language with primes and remainders

This is an exercise from old exam on formal languages that I don't know how to solve:

Let $$p \ge 5$$ be a prime number and $$L_p$$ be a language of words over $$\{0,1\}$$ that read in binary from right (i.e. from least significant bit) give a number that gives remainder modulo $$p$$ from the set $$\{1,2, \ldots, \frac{p-1}{2}\}$$.

How to show that:

Every DFA recognizing $$L_p$$ has at least $$2p$$ states.

?

One fact that I know of and is somehow related (has DFA and primes in the statement) is:

Any DFA recognizing language $$\{0^n : n \text{ is not divisible by } p\}$$ has at least $$p$$ states.

This can be seen by observing that the language is infinite, hence any DFA must have a reachable cycle, from which some accepting state is reachable. And if that cycle had less than $$p$$ states, then because any number smaller than $$p$$ is coprime with $$p$$, we could loop sufficiently many times in that cycle and arrive at the aforementioned accepted state with a word $$0^{kp}$$ for some natural $$k$$ - a contradiction.

Maybe it's possible to use this fact, or alter this proof somehow to make it fit for the theorem with $$L_p$$?