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.

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jeffbee 12 hours ago | parent [-]

> summing up their digits recursively gives 3 6 or 9

What does this part mean? For example 57.

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zdimension 12 hours ago | parent | next [-]

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.

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squigz 3 hours ago | parent | next [-]

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!

	▲	moring 2 hours ago \| parent \| next [-]
		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.
	▲	an hour ago \| parent \| prev [-]
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jeffbee 12 hours ago | parent | prev [-]

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

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hollerith 12 hours ago | parent | prev [-]

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