Let $S = \{5^k | k \in \mathbb{Z}, 0 \le k \le 2004 \}$. Given that $5^{2004} = 5443 \cdots 0625$ has $1401$ digits, how many elements of $S$ begin with the digit $1$?

Note that $5^n$ has the same number of digits as $5^{n-1}$ if and only if $5^{n-1}$ has a leading digit $1$. Therefore, there are $2004 - 1401 = 603$ numbers with leading digit $1$ among the set $\{5^1, 5^2, 5^3, \cdots 5^{2003}\}.$ However, $5^0$ also starts with $1$, so the answer is $603 + 1 = \boxed{604}$.
604
Level 5
Number Theory
Problem: Let $S = \{5^k | k \in \mathbb{Z}, 0 \le k \le 2004 \}$. Given that $5^{2004} = 5443 \cdots 0625$ has $1401$ digits, how many elements of $S$ begin with the digit $1$?

Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To determine how many elements of the set S begin with the digit 1, we need to analyze the properties of the numbers in S, which are powers of 5. The first step is to understand the range of these powers, specifically from \(5^0\) to \(5^{2004}\).

Next, we apply the concept of logarithms to find the leading digit of a number. The leading digit of a number \(x\) can be determined by examining \(10^{\{ \log_{10} x \}}\), where \(\{y\}\) denotes the fractional part of \(y\). For a number to start with the digit 1, the condition is:

\[
0 \leq \{ \log_{10} (5^k) \} < \log_{10} 2
\]

Calculating \(\log_{10} (5^k)\):

\[
\log_{10} (5^k) = k \log_{10} 5
\]

Thus, we need to find \(k\) such that:

\[
0 \leq \{ k \log_{10} 5 \} < \log_{10} 2
\]

Next, we compute \(\log_{10} 5\) and \(\log_{10} 2\):

\[
\log_{10} 5 \approx 0.6990
\]
\[
\log_{10} 2 \approx 0.3010
\]

Now, we need to find the number of integers \(k\) in the range \(0 \leq k \leq 2004\) that satisfy the above condition. The fractional part \(\{ k \log_{10} 5 \}\) is uniformly distributed in the interval \([0, 1)\) due to the properties of irrational numbers. 

The length of the interval where \(\{ k \log_{10} 5 \}\) falls within the desired range is:

\[
\log_{10} 2 - 0 = \log_{10} 2
\]

The total number of integers \(k\) from 0 to 2004 is 2005. Therefore, the expected number of integers \(k\) that satisfy the condition is given by the proportion of the interval length to the total interval:

\[
\text{Expected count} = 2005 \times \log_{10} 2
\]

Calculating this gives:

\[
2005 \times 0.3010 \approx 603.505
\]

Since we are counting whole integers, we round this to the nearest whole number, which is 604. Thus, the final answer is 604.<|END_OF_LATENT|> The final answer is $604$.