@maralorn Reminds me of https://www.smbc-comics.com/comic/math-translations
@maralorn I did!
Cool, isn't it ...
@maralorn Here is the proof: A number in the form abcabc (where each letter represents a digit in the number) is equal to:
100000a + 10000b + 1000c + 100a + 10b + c
because of the place value of each digit. This simplifies to: 100100a + 10010b + 1001c
Now it can be factorised into: 77(1300a + 130b + 13), which means it's divisible by 77. However, it is also divisible by 1, 7, 11, 13, 77, 91, 143, and 1001, because it can be factorised by those numbers too. #maths #math
@amazingadam999 Yeah, I actually derived it from realizing that every number of the form abcabc is divisible by 1001. Thus all the divisors you list are possible.
chaos.social – a Fediverse instance for & by the Chaos community