What do stoners & prime numbers have in common?

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kafka_lite@reddit

Not to be that guy but there are prime numbers with over 300,000 digits that are just two apart from one another.

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RoundTwoLife@reddit

You are that guy. True but unnecessary.

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bumscum@reddit

It's not a joke if it isn't true is it?

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Equal-Bobcat204@reddit

It is probably fair to say that most people don't know the prime numbers with over 300,000 digits so we can cut OP some slack. Or just not find it funny if you prefer.

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bumscum@reddit

Right I get it you wanna laugh but you don't know how to.. lol. Just say hahaha.

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Equal-Bobcat204@reddit

Unnecessary but interesting. Well done that guy.

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xXNoeticXx@reddit

Since it’s 4/20, here’s a fun prime number fact!

Both 419 and 421 are prime numbers!

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ExistingBathroom9742@reddit

Since both of those numbers are prime, then 420 must be divisible by 6. (Twin primes always surround a number divisible by six)

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WikiWantsYourPics@reddit

Kind of obvious, though.

n-1 is prime, so not divisible by 2 or by 3
n is in between
n+1 is prime, so not divisible by 2 or 3.

Because every second number is divisible by 2 and every 3rd number is divisible by 3, n has to be divisible by both 2 and 3.

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With the lamentable exceptions of 2 and 3, all primes are one away from a multiple of six, but it could be higher or lower. With twin primes you know it’s between them. Not a huge mindblower, but a fact people on r/Jokes might not know.

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-suspended-@reddit

A more mathematical proof:

Take the set (6n-3, 6n-2, 6n-1, 6n, 6n+1, 6n+2). Notice that 6n+3 = 6(n+1)-3, showing that the set repeats. With n as all integer, every integer exists in this set. 6n-3 and 6n are divisible by 3. 6n-2, 6n, and 6n+2 are divisible by 2. This means that prime integers >3 must be of the form 6n-1 or 6n+1. So any pair of primes >3 is going to have a multiple of 6 between them.

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