Aiden Koh
aidenkoh.bsky.social
Aiden Koh
@aidenkoh.bsky.social
I’m not here for any particular reason.
I know no one takes the videos as serious advice but it’s painful to see how they reach that conclusion.
January 18, 2025 at 6:30 AM
I can't find any that satisfy the first, second, and fourth conditions other than 2025 and 11025, let alone all four conditions. I'm not really sure if there are any others. If there was a way to prove an upper bound the problem would actually be solvable but right now it seems impossible.
January 10, 2025 at 3:24 PM
Actually as I wrote this I just realized my solution was clearly wrong, since it has only one zero at the end. I think it might be because loss in floating point precision or something. I’ll get back to you when I get it fixed.
January 1, 2025 at 4:31 AM
I used a simple python script to look through squares of triangular numbers, then checking if the last 2 conditions are met. I’m fairly certain there are infinitely many solutions but I’m not sure if there’s an easily describable pattern. Would be cool if you could find it!
January 1, 2025 at 4:15 AM
278738540624551614024253440 is the next number with no 9s that satisfies these conditions (it’s a bit ambiguous what it means to add 1 to each digit when there is a 9).
January 1, 2025 at 1:38 AM