Interesting math fact of the day #527:
There is no integer value of n such that 2ⁿ, 3ⁿ, 4ⁿ, …, and 9ⁿ all start with the same digit.
Is this true in arbitrary base?
I do not have any sources for that, so i cannot say… sorry!
it can’t be true in arbitrary base because positive integers always start with the same digit in binary, that digit being 1.
I think the original statement needs to be modified somehow to exclude n=0? like, as written I think it is false. anyway this sounds cool! I’m interested in what the proof is like.
I think the right generalization to arbitrary base is something like "if n > 0, then for any b > 2, 2^n, 3^n, ..., and (b-1)^n cannot all have the same leading digit." I'm not sure if this is true in the general case because I think the proof of the base 10 version uses something about 2^n * 5^n = 10^n to only have to worry about 3 being the leading digit (but not sure if that's the right approach I haven't actually worked it out yet)
wait, what? this statement doesn’t make sense for the base 3 case, where it would be saying 2^n cannot have the same leading digit as itself for all n > 0. you also have a counterexample very quickly in base 4, as 2^4 and 3^4 give you 100_4 and 1101_4, respectively.
it's only true in bases that aren't prime powers
