Too lazy to check, but is this the Rivest from the RSA algorithm?
goddard_guryon
Since you're seeking for answers within physics buzzwords, you're missing a lot of nuance which is causing you to come up with nonsensical theories. For one, the wavefunction doesn't exist 'in' the particle, it is the particle. A wavefunction collapsing is what causes a particle to show up in a specific location (going by the most prominent interpretation of the wavefunction); as long as the wavefunction is non-zero at more than one location, the particle exists in all those locations.
Equipped with this knowledge, phrases like "a wavefunction collapsing in such a way that it absorbs light" or "change the composition of atoms" make no sense, so I'd suggest you to rethink your assumptions, and, if possible, not look at quantum mechanics as a magic black box that can explain everything a mind can conjure up.
As someone who is often unable to even enjoy soda because it immediately starts bubbling up and only remains as a sugary liquid by the time it reaches the esophagus, I guess I can't replicate your observations. Also, I guess I need more water in my mouth before I can let Jesus into it (for a more pleasant experience for both of us)
If only it worked; the mentos would start disintegrating by your saliva and the coke would already form bubbles as soon as it enters your mouth thanks to the rough surface inside 🤓
That won't work tho, you need to make it sys.maxsize//2 to coerce the output into int form
I bought mine through them, it's very slightly more expensive (less expensive than I'd consider a dealbreaker), but I don't see anything that'd make it worse than other registrars. At the same time, I'm not very familiar with how other domain registrars work, so I'm probably not the best person to give a review
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
I tried it once a few years ago and it seemed to work fine enough for me. I can't say how exactly it works now, but there is (or at least used to be) a free plan with limited functionality, so I figured it might be interesting to you
Yup it seems crazy to me how deep insights one needs to have to be able to, say, connect the dots between compression and machine learning. And now it looks to me like he has done a lot of the foundational work in these fields. Super cool stuff