I disagree with you. Firstly, puzzle #67 has a range width of 66 bits.
The second is that 1BY8GQbnueY has a length of 56 bits and accordingly in the entire range of 66 bits there are approximately 1024 such combinations.
1BY8GQbnueY How did you know it was 56 bit?
You lack proper understanding of things, every char in base58 encodes, well, 58 bits.
So 10 base58 chars can encode at most 58**10 values. This is a rough estimate without even needing to bother going deeper to binary or even hex.
2**66 / 58**10 is around 171, and Etar was even more precise and correct with his calculus, whether you like it or not. Nowhere near your "correct answer is 4", which is an illusion and correct for you only.
Sorry to smash your nice dreams of finding a hash collision 250x faster than mathematically possible.