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.
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.
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!
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.
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
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).
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.
