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Using Bertrand's postulate which states:

For every integer $n \geq 1$ there is a prime number p such that $n<p\leq 2n$

Prove that there exists infinitely many primes whose decimal expansion starts with $1$.

I'm guessing I need to use something equal to $10$ in this as it is a decimal expansion, but I'm not seeing where to start?

Any guidance would be great, thank you

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Just take the sequence $a_n = 10^n$. Since for $n > 1$, we have $n < p < 2n$, so there is a prime between $a_n, 2a_n$ but all numbers in between begin with $1$.

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