Back to initial equation

In the Brute force solution, we found that \( 1 < x < 10 \) and \( 10^{n-1} > x^n \) for \( n \) sufficiently large. Rather than iterating through all \( n \) until \( 10^{n-1} > x^n \), solving the equation \( 10^{n-1} = x^n \) for \( n \) gives the exact limit for \( n \).

\[ \begin{align} &\Leftrightarrow 10^{n-1} = x^n \\ &\Leftrightarrow \log_{10}\left(10^{n-1}\right) = \log_{10}\left(x^n\right) \\ &\Leftrightarrow n - 1 = n\log_{10}\left(x\right) \\ &\Leftrightarrow n(1 - \log_{10}\left(x\right)) = 1 \\ &\Leftrightarrow n = \frac{1}{1 - \log_{10}\left(x\right)} \\ \end{align} \]

Since \( n \) is an integer, the limit is the largest integer for which the inequality holds true.

From solution2.py:

def powerful_digit_counts():
    return sum(floor(1 / (1 - log10(i))) for i in range(1, 10))