The cattle easily have sufficient mass to collapse into a black hole.

@@snowgods2195 pedantry warning: So do protons, but yeah I get what you mean.

That is a solution to another problem; it isn't turtles all the way down, it's cows.

Not cows but "pieces of cattle"!

@@X_Baron The only non-gendered word I can think of is oxen

surely he is referring to some academy person, which is available for consultation even today :)

Friendly sure is an assumption 😜

Me too! 😖

As a kid, I sent an email to Wolfram, asking if I had discovered something. About a decade later, they emailed back, "Nope". But, I do suspect that there's a math teacher somewhere in your neck of the woods.

@Suki Desu - Dustin's comment was a joke

That would be more interesting than the solution. The technique used to solve the problem with the last two requirements involved Pell equations. Pell equations are reported as having been first studied by Brahmagupta 400 years after Archimedes lived. It is unlikely that Archimedes solved the problem, but he might have determined that a solution existed.

Firstly you have a solvable system of equations. Enough formulas compared to variables that a solution should exist. For example: x=1 - 1 equation, 1 variable. x=y, y=2 - 2 and 2 ... This isn’t a proof, but the system of equations only uses rational numbers. There is an infinite number of these (far fewer than the real numbers). There is no reason for a solution not to exist.

@@MusicBent essentially, this is an underdetermined system, in integers. Some of them are unsolvable, so 🤷‍♂️

Yeah if it's solvable by rational #'s, then it stands to reason that there should be infinite solutions. Dat triangular kicker tho..

@@MusicBent Yes, but we've got eight variables and seven "normal" equations, which gives infinitely many solutions and then two extra, that are different. They point out that variables are in certain sets. If from the first we know that solutions are in form of x=q*x_0, where q is a rational vector, we don't really know if there is any x_0 for which both (or only one of them) of the specified x'es belongs to those sets.

I love this community lol

Dang, you beat me to it! :)

Out. 😂

I legit didn't realise they meant his name was Gotthold until they said he found the letter. I thought the sentence was finishing "Gotthold of the letter" or words to that effect.

Anybody else conceive 'Lessing' as a verb? To rephrase Bucky, "By Lessing we can Moring".

@@epsi you mean imaginary numbers? Yeah they're fun.

@@epsi Get real!

If each cow is an atom, fill the Universe, shrink to the size of an atom, repeat twice.. 3 of these should do it.

@@tim40gabby25 so let's do that. We've got a universe containing 10^80 particles, each of which contains 10^80 of the below universes' particles. That's (10^80)^2=10^160. Now each of those 10^160 particles contains 10^80 giving 10^160*10^80=10^240. Still incomparable. You don't cube the exponent (80^3 = 512000 giving 10^512000), you cube the whole number ((10^80)^3=10^240). You'd have to do this 2575 times to get (10^80)^2575=10^206000=~the number of cattle

What andrew meant is not a universe of 10^80 atoms, but a universe of 93 billion light-years diameter, packed solid with atoms (as he said, "fill the universe"). So a much larger number of atoms in each notional universe.

4 years later, and I could see this in an asdfmovie.

Ok, but thats COWS(1) and its way bigger than TREE(1)

@@Wemdiculous What would COWS(2) be?

Can't see the forest for them

@@Lexivor COWS(2) is the second smallest number that solves the equation, I guess.

@@Shadow81989 That makes sense. I think the second smallest solution is roughly the size of the square of the first solution.

That was the most baffling part to me! Where's the Extras video?

I agree with you, but I also have a simple guess. I think it's got to do with the fact that each digit is another order of magnitude. It's way easier to say when a bowl of rice has between 10,000 and 100,000 grains than to say when it has between 1,000 and 10,000 grains. This is still true on the most massive scale.

Why so low? I had to use the search function to find this ONE comment about 42 being (included in) the answer!

Patrick Ryan too low

Archimedes 2K years later "I've earned this $50 joint 😁"

ima flexing on those who couldnt read that in ancient greek!

Hello again

??

Yes

Even Matt Parker and Curtis Cooper wouldn't be impressed

They printed it though

But TREE(3) is still Way bigger

tomiesz there are not enough atoms to print it

tomiesz exactly!

Who says he did? Maybe he was just poking at Eratosthenes. "Here... If you can't solve this, you must be an idiot."

It's an underdetermined system, so it actually should have an infinite number of solutions.

@@Jodabomb24 It's kind of amazing to think that the 206,545 digit solution is the smallest of an infinite number of solutions.

@@Lexivor Only if you ignore the last two statements.

Joseph McGowan it's not obvious that there are an infinite number of solutions that are square and triangular, and if there are an infinite number of solutions, that would be an interesting proof There are underdetermined systems that don't have any solutions at all, so your assumption doesn't hold up. Edit: removed bad example Edit: better example of a simple underdetermined system with no solutions: {x=0, x=1, a+b+c+x=2} Four variables, only three equations, but still 0 solutions

Dubious existence of a proof. Sounds familiar....

Probably started with the first and last sentence of the problem and worked his way from there.

false.

Collatzes into laughter?

collaughz

Or what do I know... maybe Archimedes Number is the first that computationally brakes the Collatz!!!?

??

Funnily enough, modern understanding of mathematics does allow a smaller solution, namely no cattle whatsoever. I wonder if the phrasing of the original disqualifies it.

Oh a cow pun. I've rarely herd these before.

lemmysfibroma Do you have a beef with my comment?

@@p11111 - Perhaps @lemmysverruca's comment is moot?

Not to mention TREE(Graham's number)...

@@RedwoodRhiadra And Googolplex for instance.

They did Rayo's number though lol

@@carbrickscity tree(g(rayo's number))

Brilliant!

You ... you ... are right, sir.

Yeah, the solution would be that Odysseus’ troup killed them all

Nr.3042 but 0 isn’t a triangular number so it doesn’t work.

@@genessab 0 is regarded as a triangular number by most people.

Isn't 0 a solution?

Adriano Giacomini the FBI wants to know your location

@@adrianogiacomini6541 yes but it's a trivial solution

Cows are a torus as any fule kno.

I love turning anything into something described by spherical cows in a vacuum. I once set up an english literary analysis project for a 10th grade project that seemed normal, just with the occasional slight tangent, until I got to the end, where I spent a single paragraph to show that, given my normal analyses of the individual portions, the argument posed by the book could be applied equally to spherical cows in a vacuum. Then, I logically restated it to fit those cows in a less confusing way while still including every major and minor point of the original argument. And then, once reframed to do so, I showed that it could then be reframed back to human societies, in a way where the transformation makes logical sense, but the end result is diametrically opposed to the original argument. As such, I showed that the book's takeaway, the moral of the story, was utterly (udderly) meaningless. Oh, and this was an oral presentation in front of the entire class. The teacher tried to stop me when I started talking about spherical cows in a vacuum, but I said "This is the last paragraph, and it's relevant. In fact, it's the entire point of my project" before restarting the paragraph. I knew that it was somewhat confusing, so I came prepared with about two dozen printed copies of my speech (each one was just one piece of paper, double sided) to hand to anyone who wanted to read it over. I got a begrudging A+ on the project. TL;DR: Showed a persuasive argument to be meaningless by reframing it to spherical cows in a vacuum and then reframing it back to its original subject.

@@Nilguiri Only when they open their mouths.

@@CaTastrophy427 Please tell me you still have copies of that saved somewhere!

In short, Amthor figured out that the solution involved raising a more than 20 digit number to the power of a four digit number. This was too long to calculate by hand, but using logarithm tables he was able to get an approximate solution.

Glad you liked it.

Wir sind nicht so dumm.

HA

I did what you see there.

??

Exactly

He comes from France.

He looked like Ice Cube

he got a 5 head :/

Haha, that was the first thought I had about him!

negative Grahams number? Or zero, if you ignore negative numbers

Kaedenn - The initial equations (Archimedes' "no expert" level) are as follows. Unknowns are Bw, Bb, Bd, By, Cw, Cb, Cd and Cy. Bw = 5Bb/6 + By Bb = 9Bd/20 + By Bd = 13Bw/42 + By Cw = 7(Bb + Cb)/12 Cb = 9(Bd + Cd)/20 Cd = 11(By + Cy)/30 Cy = 13(Bw + Cw)/42 Note that we have seven equations and eight unknowns, so the solution to the first part is not unique; this makes the second part solvable.

TREE(3): Amateurs

This may not apply precisely but this will hopefully help. One problem I worked on a while ago, from Project Euler (lots of cool math problems that have big numbers) was to find the lowest number which was a multiple of all integers from 1 to 20. That can be thought of as a simple version of this problem - a handful of equations, some constraints, what's the lowest possible number. Well, let's say we only do one constraint - that the number must be a multiple of 2 (we can drop the multiple-of-1 argument, that is trivially true). That's easy! Just look at the even numbers, the lowest answer is 2. So now, add one more constraint - it must be a multiple of 2, and a multiple of 3. This is equivalent to saying it's a multiple of 6, by the way - thinking back now I could have made the problem easier if I'd thought about that, but anyway. Suddenly the numbers just got a lot more spaced out. Now it's no longer every other number, but only one in every six numbers meets every requirement. Now add that it must be a multiple of 4 as well - that makes the lowest number 12, only one in twelve numbers meets the requirement. And, a number that is a multiple of 2, 3, 4, and 5 (and 6 by default)? The lowest one is 60 - and 60 isn't a multiple of 7, 8, 9, 11, 13, 14, 16, 17, 18, or 19, so better keep looking. Every additional constraint weeds out a ton more numbers and pushes back the first number that satisfies the equation a ton. I don't remember the final answer to this problem, but I do know I wrote a computer program to solve it rather than even attempt to do so by hand, and it was a large number. I hope this helps makes it clear how adding constraints like this leads to large numbers, fast.

@@superlukey3 Answer will be (2^4)(3^2)(5)(7)(11)(13)(17)(19). Think about prime factorisation of each number 1 to 20.

@@additionaddict5524 Ooh, clever! Ultimately my program started at 20, jumped forward 20 numbers at a time, and just checked the modulus of each number in reverse order (so checked for 19, then 18, then 17... down to 2). My immediate thought was just cutting down on ones that wouldn't need to be checked, ie multiples of 18 are multiples of 2, 3, and 6 as well so no need to check those again. Even so, my naive approach solved the problem pretty fast - but your analytic solution is much better!

You are correct.

It makes me feel lucky to live now, in a time where if we want an answer to a fairly straightforward problem, all it takes is a few dozen lines of code to crank it out. We can literally just tell a machine how to find an answer, and receive it not much later. What's even cooler though, is the realization that there are problems today that even our supercomputers cannot solve, but might be trivial in a century or even just a few decades.

In contrast: It only took 15 years and three supercomputers to MacGyver a DHD for the Tau'ri gate.

It doesn't really take a genius to make puzzles, he didn't find the solution himself either

I was thinking of that as well, but did the Greeks have the concept of zero as an integer or natural number already?

@@BaronSamedi1959 This assumes that the problem was indeed designed by the greeks, and not much later. Though 0 fails once you add back in the old condition: and there are more bulls of each color, than cows of the same color.

@@AleaumeAnders Where is that condition? I must have missed it. All the equations (overview shown at 2:50 ) are satisfied when plugging in 0.

@@Shadow81989 They ommited it in the video, and it's not used anymore. But in the original text it's stated.

​@@AleaumeAnders No wonder I didn't see it, if they just left it out! ;-)

Underrated

The captions are written in invisible ink

Lol 🤣

Depends on the class of problem 😋