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Comment by CydeWeys | Hacker News Reader

For those wondering, this works in every base ≥ 5, according to the original paper: https://arxiv.org/abs/1602.06208v2

Every time I see something like this that is dependent on base-10 representation, I'm always curious to know whether it generalizes, or if it's a quirk specific to base-10 representation. For example, being able to tell if a number is divisible by 2 or 5 by only looking at the last digit.

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And according to the Numberphile video (https://www.youtube.com/watch?v=OKhacWQ2fCs) there is a prior paper showing it for bases 2, 3, and 4

Oh nice, so it works in every base then. Awesome.

Not base 2. All palindromes in base 2 have a 1 in the last position, because there's a 1 in the first position. This means that any addition of 3 palindromes must be odd because the result of the addition is guaranteed a 1 in the ones position, meaning even numbers can't possibly be represented that way. This is discussed in the original paper at the bottom of the second page: https://arxiv.org/pdf/1602.06208.pdf , which I've expanded out a bit since they just mention the evenness without the argument I give. They say base 2 needs at least 4 summands, but their phrasing suggests to me that's intended as a lower bound, not that they have a proof 4 is always sufficient. (Not to mention it would have to be more complicated than that anyhow since it would have to be something like "3 for odd and 4 for even above some threshold", or some other bounds.)

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If you pause at 3:33 in the video and read the abstract, it actually says that numbers in base 2 require "at most 4 palindromes", rather than exactly 3, and "similar results" were found for bases 3 and 4. So it's not quite the same result as for bases of 5 or greater.

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-arbitrary-base operation was particularly cheap to compute, we could convert to base X+1 and check if the last number is 0 (same logic as: is_even = (i & 1) == 0) :)

The convert to base X operation can't be cheaper than just dividing by X, can it?

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

Referring specifically to divisibility tests, converting bases takes more work than just doing the division. What other operations were you thinking of?

> For example, being able to tell if a number is divisible by 2 or 5 by only looking at the last digit.

That still seems like it is generalizable to any base by taking prime factors of any base-n.

It is, but there's nothing special about 2 or 5 except that they're factors of 10.

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, or 15.)

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.

You mean any divisors of [base + 1], right?

Yes.

And much more generally, "numbers divisble by n written base b" is a regular language decidable by an n-state DFA (and thus in linear time, looking at each digit exactly once). And see https://blog.plover.com/math/divisibility-by-7.html for a state machine that's even simpler (you only need edges for +1 and ×b, rather than +d×b for each digit d)

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