\( \displaystyle \{ a_n \}_{n=1}^{\infty}\) be a sequence such that \(a_n = 2^{n-1} \).

Define \( X(m) = \{ a_p | \text{ The first digit of } a_p \text{is } 1 ; 1 \leq p \leq m \} \).

Let \( \displaystyle \{ b_n \}_{n=1}^{\infty} \) be such that \( b_n \) is the cardinality of \( X(n) \).

The fraction \( \dfrac{b_n}{n} \) is found to converge to \(L\) as the value of \(n\) becomes very large.

Evaluate: \(\displaystyle \lfloor 1000L \rfloor \).

×

Problem Loading...

Note Loading...

Set Loading...