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Comment by IsTom | Hacker News Reader

My favourite prime checking algorithm is that for n < 100 if it looks prime, it is prime.

Up to 100 it is relatively simple, if you remove the even numbers, the multiples of 3 (summing up their digits recursively gives 3 6 or 9), and those that end at 5, you end up with 25 numbers out of which 22 (88%) are primes. If you further exclude 49 which you should remember is 7^2 from the multiplication table, and 77 that is obviously divisible by 7, you are left with 22/23 chance a number that passes these exclusion criteria is not 91 and hence it is a prime.

> (summing up their digits recursively gives 3 6 or 9)

My fun fact is that this type of operation (repeatedly applying a child operation until you reach a fixed point) is called persistence.

https://en.wikipedia.org/wiki/Persistence_of_a_number

The fixed point here being that if you add up a list of 1 digits, you'll always reach the same number (`sum([1]) = 1`). The best known is probably the hailstone sequence.

https://en.wikipedia.org/wiki/Collatz_conjecture

I'm partial to multiplicative persistence.

https://www.youtube.com/watch?v=Wim9WJeDTHQ [15m]

> summing up their digits recursively gives 3 6 or 9

What does this part mean? For example 57.

57 is 3 times 19.

The standard divisibility rule for 3, 6 and 9 in base 10 is to sum the digits until you only have one left and check if it's one of those. Here, 5+7=12, 1+2=3, so 57 is divisible by 3.

Math is not my strong suit at all, so I probably won't grok this, but that kind of blows my mind, so I'm curious... how?! That works for any arbitrarily large number?

Math is crazy!... still don't want to study it though!

Yes. A number like

123456 = 1 * 100000 + 2 * 10000 + 3 * 1000 + 4 * 100 + 5 * 10 + 6 = 1 * (99999+1) + 2 * (9999+1) + 3 * (999+1) + 4 * (99+1) + 5 * (9+1) + 6

When checking whether it is a multiple of some k, you can add/subtract multiples of k without changing the result, and those 99...9 are multiples of both 3 and 9.

So 123456 is a multiple of 3 (or 9) iff

1 * 1 + 2 * 1 + 3 * 1 + 4 * 1 + 5 * 1 + 6 * 1 = 1 + 2 + 3 + 4 + 5 + 6

is. Apply the same rule as often as you want -- that is, until you only have one digit left, because then it won't get simpler anymore.

It is basically because $10 mod 3 == 1$ (as 10 = 3*3 + 1). So if you are in the ring modulo 3, where every number is equal to the remainder of its division by 3, the sum of the digits of the number in its decimal representation equals the number itself (modulo 3), because in that ring 10 is actually 1, so the 10s in the decimal sum become 1s. Ie if n_k is the kth digit of n, you have

    (mod 3) n == n_0 + n_1*10 + n_2*10^2 + ... == n_0 + n_1 + n_2 + ...

Hence, n is divisible by 3 iff $n mod 3 == 0$ iff $(n_0 + n_1 + n_2 + ...) mod 3 == 0$.

Of course, summing up the digits may not give you a 1-digit number, but it gives you a number that you know is divisible by 3 (if the original number is divisible by 3). So you can apply the same idea/process again, summing up the digits of that number, and get another number that is divisible by 3. Repeat until you end up with one digit (hence the recursion mentioned).

Ah I see what is meant by recursively here. Thanks!

5 plus 7 is 12, which of course has a 1 and a 2, the sum of which is 3.

Like the famous Grothendieck prime of course.

Definitely makes me feel better about my own work

91 should be prime, ridiculous that it isn't

Agree. Plus now I now need to release a security patch for the hand-rolled crypto library I built.

The annoying child in me will always remember correcting my freshman math teacher when he needed a prime number and wrote 91 on the chalkboard.

I wonder what the underlying human intuition is for 'prime-ness' and why it might break down with larger numbers. Odd numbers in the rightmost position? The 'shape' of the number (phonaesthesia, the bouba/kiki effect)? Maybe they just sort of feel scary?

For n < 100 to be composite you need a factor < sqrt(100) = 10. Rules for 2, 3, 5 are easy to try quickly. That leaves 7, but up to 7*9 you should remember it from multiplication tables. 77 is quite obviously divisible by 11 too, and then it's 7*13 = 91 as the last boss. But I feel that once you realize how special 91 is in that context, you won't forget it again.

This only works if you know multiplication tables which is not a given in America these days.

"not given" != "not able to learn"

Are there any numbers that don't look prime but are, in fact, prime? [Other than 2, I suppose.]

11, 13, 17 and 19 used to trip me up. And maybe 67

67 has absolutely no right to be prime. Sitting there looking all innocent.

Maybe that’s the real secret behind the 6-7 meme going around

Except 91.

Except 91

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