Any positive number can be written as a sum of three palindromes

Post by pmiller2 » Wed, Sep 19, 2018, 6:00 AM UTC

industriousthou wrote: post #110

shaded-enmity wrote: post #79

Earlier quoted context omitted.

Yeah, for example to check if a number is divisible by 3, we can use the digit sum and check if the result is divisible by 3, e.g. 153=(1)+(5)+(3)=9. This works in base-10, but we can generalize this by converting the dividend X into (3* X+1)-base then doing a digit sum. Let's say we want to check whether 238 is divisible by 17, so our base is 3* 17+1=52, 238 in base-52 is (4)+(30)=34=2* 17. If the convert-to-arbitra…

Is it unreasonable to think that some operations might be cheaper to compute by the converting the base and applying these types of tests?

Post by taejo » Wed, Sep 19, 2018, 12:26 PM UTC

dmurray wrote: post #73

schoen wrote: post #72

Earlier quoted context omitted.

There's a stronger generalization: it's true of any divisor of the base (not just prime divisors and not just proper divisors). For example, it works for divisibility by 6 in base 12, or divisibility by 10 in base 10. (Also, the digital root test, where you add up the digits, works for any divisor that is also a divisor of the base minus 1, so for example hexadecimal has a digital root test for divisibility by 3, 5,…

And the less well known "add and subtract alternative digits" works for dividing by 11, that is, the base plus one, in any base. And for any divisors of 11.