As someone who tried to self-host it like a month ago (and seemingly still hasn’t got it fully working), I’ll just write out the overview of what I’ve done and let you (and others) comment on how correct and feasible it is.

Since my ISP doesn’t allow me to get a static IP address, I rented a VPS connection and made a wireguard tunnel from the VPS to my computer. This tunnel forwards traffic at all the necessary ports between the two machines. I really wasn’t familiar with all the necessary components for an entire mail server, so I chose mailcow since it packages everything into one single software (well, more like a bunch of docker containers). Another reason I went with mailcow was that I could easily find a github tutorial for how to set up mailcow with wireguard tunneling (it’s a bit outdated IMO, but the changes are minor). Mailcow also gives a nice portal interface listing out all the DNS entries you need to put in place to get it working perfectly.

In the end, I still see a few incoming emails getting dropped and reception time being an hour or so, and I’m not sure if it’s a problem with my tunnel or DNS or something else. But overall, I’d say it was much easier than setting up all the individual services myself.

But does it make jack jork?

Last I checked (which was some time ago), pandoc-bin doesn’t require the haskell dependencies. I saved quite some installation time (and screen space during installation) by switching.

IIRC it doesn’t; that has caused me pain so many times when trying to generate fractional range

That won’t work tho, you need to make it sys.maxsize//2 to coerce the output into int form

That may be exactly what I need when searching for some highly technical problem lol

Huh I didn’t notice that. That does sound a bit weird :/

Indeed, an integer is divisible by 3 if and only if the sum of its digits is divisible by 3.

For proof, take the polynomial representation of an integer n = a_0 * 10^k + a_1 * 10^{k-1} + … + a_k * 1. Note that 10 mod 3 = 1, which means that 10^i mod 3 = (10 mod 3)^i = 1. This makes all powers of 10 = 1 and you’re left with n = a_0 + a_1 + … + a_k. Thus, n is divisible by 3 iff a_0 + a_1 + … + a_k is. Also note that iff answers your question then; all multiples of 3 have to, by definition, have digits whose sum is a multiple of 3

For Meet, I’d also suggest Jitsi. For Drive, I’d recommend giving cryptpad.fr a shot; that seems to be the closest to Drive’s file editing capabilities.

You are

I see your encourageMint and raise you

Commenting so you see your post one more time

Not exactly what you wanted, but njalla is a privacy-focussed domain registrar that basically buys domains on your behalf under their own name and gives you all the access to it that you need

2. Strairdrac is three crabs in a trenchcoat, now teaching others of his kind how to blend in with humans