• $\pi(n) = \text{ number of prime number smaller than or equal to n}$ .
  • $rub(n) = \text{ number of palindromic number smaller than or equal to n}$ .

For a given $p$ and $q$ find maximum such $n$ so that, $\pi(n)\le \frac{p}{q}\times rub(n)$

Inside Math:⌗

  • $\pi(n) \approx \frac{n}{ln(n)}$ [Prime number approximation]
  • $rub(n) \approx 2\sqrt{n}$
  • maximum value of $\frac{p}{q} = 42$

Hence , \begin{align} \pi(n) &\le \frac{p}{q}\times rub(n) \newline \Longrightarrow \frac{n}{ln(n)} &\le 42 \times 2\sqrt{n} \newline \Longrightarrow \frac{\sqrt{n}}{ln(n)} &\le 84 \newline \Longrightarrow n_{max} &\approx \boxed{1415344} \newline \end{align}

Now we can just precalculate all prime,and the palindrome in the range $[0,n_{max}]$, then find out maximum such $n$ for which $\pi(n) \le \frac{p}{q}\times rub(n)$.

Reference:⌗