121–123 of 123 posts
This uses much higher order sieves so that it runs about 32000 times faster than the naive algorithm and was able to search to this point on a single core. It is also possible to thread this algorithm relatively easily.
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Is your algorithm published somewhere?
I literally just invented it, so it's not published yet, although I'll probably throw it up on GitHub at some point. It matches all published results up to an exponent of 3 billion or so, so I'm quite confident it's correct. The short explanation is that 2^x mod 10^k will repeat with a cycle length of 5^(k-1)*4. This is easily obtained from Euler's phi formula on 2^x mod 5^k, plus the fact that 2^x === 0 mod 2^k for…
As mentioned in another comment, the code is at: https://github.com/tdunning/EvenDigits
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10^10 * 36105/39062500 = 9242880, so you're already down to under 10^7 cases to check, which is starting to seem more tractable.
10^7 cases, but almost every case has billions of digits. Even that doesn't seem so bad though, it's on the order of 10^16 total digits to check in the worst case, and far fewer in practice. Maybe someone here can run a program overnight and increase the bound by another few orders of magnitude, or disprove the hypothesis?