I want 1 to be first

Calculus Level 5

$\displaystyle \{ a_n \}_{n=1}^{\infty}$ 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: $\displaystyle \lfloor 1000L \rfloor$ .

The answer is 301.

1 solution

Maggie Miller
Aug 21, 2015

A number $k$ has 1 as its first digit whenever the fractional part of $\log(k)$ is less than $\log(2)$ , where $\log$ refers to log base 10. Note $\log(2^m)=m\log(2)$ . Since $\log(2)$ is irrational, the set of fractional parts $\{[m\log(2)]\mid m\in\mathbb{Z}\}$ is uniformly distributed in $[0,1)$ . Therefore, given random $m$ , the probability that $[\log(2^m)]<\log(2)$ is given by $\log(2)$ .

Thus, $L=\log(2)$ , so $\lfloor 1000L\rfloor=\boxed{301}$ .

I want 1 to be first

The answer is 301.

1 solution

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