The Simplest Math Problem No One Can Solve - Collatz Conjecture

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The Collatz Conjecture is the simplest math problem no one can solve - it is easy enough for almost anyone to understand but notoriously difficult to solve. This video is sponsored by Brilliant. The first 200 people to sign up via get 20% off a yearly subscription.

Special thanks to Prof. Alex Kontorovich for introducing us to this topic, filming the interview, and consulting on the script and earlier drafts of this video.

Lagarias, J. C. (2006). The 3x+ 1 problem: An annotated bibliography, II (2000-2009). arXiv preprint math/0608208. -

Lagarias, J. C. (2003). The 3x+ 1 problem: An annotated bibliography (1963-1999). The ultimate challenge: the 3x, 1, 267-341. -

Tao, T (2020). The Notorious Collatz Conjecture -

A. Kontorovich and Y. Sinai, Structure Theorem for (d,g,h)-Maps, Bulletin of the Brazilian Mathematical Society, New Series 33(2), 2002, pp. 213-224.

A. Kontorovich and S. Miller Benford's Law, values of L-functions and the 3x+1 Problem, Acta Arithmetica 120 (2005), 269-297.

A. Kontorovich and J. Lagarias Stochastic Models for the 3x + 1 and 5x + 1 Problems, in "The Ultimate Challenge: The 3x+1 Problem," AMS 2010.

Tao, T. (2019). Almost all orbits of the Collatz map attain almost bounded values. arXiv preprint arXiv:1909.03562. -

Conway, J. H. (1987). Fractran: A simple universal programming language for arithmetic. In Open problems in Communication and Computation (pp. 4-26). Springer, New York, NY. -

Special thanks to Patreon supporters: Alvaro Naranjo, Burt Humburg, Blake Byers, Dumky, Mike Tung, Evgeny Skvortsov, Meekay, Ismail Öncü Usta, Paul Peijzel, Crated Comments, Anna, Mac Malkawi, Michael Schneider, Oleksii Leonov, Jim Osmun, Tyson McDowell, Ludovic Robillard, Jim buckmaster, fanime96, Juan Benet, Ruslan Khroma, Robert Blum, Richard Sundvall, Lee Redden, Vincent, Marinus Kuivenhoven, Alfred Wallace, Arjun Chakroborty, Joar Wandborg, Clayton Greenwell, Pindex, Michael Krugman, Cy 'kkm' K'Nelson, Sam Lutfi, Ron Neal

Written by Derek Muller, Alex Kontorovich and Petr Lebedev
Animation by Iván Tello, Jonny Hyman, Jesús Enrique Rascón and Mike Radjabov
Filmed by Derek Muller and Emily Zhang
Edited by Derek Muller
SFX by Shaun Clifford
Additional video supplied by Getty Images
Produced by Derek Muller, Petr Lebedev and Emily Zhang

3d Coral by Vasilis Triantafyllou and Niklas Rosenstein -
Coral visualisation by Algoritmarte -

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30 Jul 2021



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Comentários 52 763
Nathan Mostoller
Nathan Mostoller Hora atrás
I solved it, but I’m not telling you what the answer is…
turtle 56 minutos atrás
Im sure you did
Tako Boutit
Tako Boutit Hora atrás
I feel like we're looking at it the wrong way - Our mathematical definition of an odd number is: n = 2k + 1 or n = 2k - 1 And even numbers are: n= 2k. First of all, if you're using a formula that uses our first two prime/odd numbers to get it to an even-enough of a number to eventually divide it by 2 until it reaches 1, of course it will eventually be divisible by our smallest even number because all even numbers are divisible by two. While, the purpose of the 3x+1 is to just get it to an even number so it can be divided. A) The better question here is, why are we so surprised that it CAN go on to infinity? No matter what number you grab, it will always come back to 1 simply by the nature of how we structured math to work. The foundation of the way we compute numbers will always make it so. Multiplying an odd number by 3 and adding 1 until it equals an even number, will never not give you an even number, and the cycle will continue until eventually it just flows right back to 1. Perhaps the question should instead be: Why is it that we structured math in the way which makes this possible? B) What actually ends up shedding a bit more light on this issue is: Why is it that when we change the formula to 3x - 1, we do not get the same loops - instead, we get the following loops: 2, 1 8, 4, 2, 1 20, 10, 5, 14, 7 50, 25, 74, 37, 110, 55, 164, 82, 41, 122, 61, 182, 91, 272, 136, 68, 34, 17 In theory, as stated above, n = 2k - 1, to get an odd number in the same manner as the above formula, we should be able to arrive to the same number using 3x - 1 but we can't. Hope this was an interesting thought for you all! Thanks! :-)
VagueVahLeh 2 horas atrás
The solution is 4,2,1...
Brandon 2 horas atrás
This is interesting, but you should have included a simple count of x+1 to compare to the results of the Benford's law to better demonstrate the difference since it is obvious that most numbers start with 1.
Szabó Sándor
Szabó Sándor 3 horas atrás
Well, you can pick 0 too.
turtle 52 minutos atrás
0 is not positive
Szabó Sándor
Szabó Sándor 3 horas atrás
So if you get 2 on the "n" it's all over.
grrrrr 3 horas atrás
I was ALMOST bored to death, but I fell to one.
D. von N.
D. von N. 4 horas atrás
22:08: "Why a number couldn't just shoot off to infinity"... infinity is a theoretical concept, it isn't a number. To me it is like an absolute vacuum with nothing in it, which doesn't really exist, in our universe at least. I find this note pointless... if a number shoots somewhere, it will be a number, not infinity. And this is within the Mathematical system operating with numbers 0-9. In a different system it would be different but probably finding the same problem in a similar exercise. Maths describes physics. Maybe there is no infinity and no nothingness in the universe as we know it. But it is good for brain storming.
Putnam 4 horas atrás
"shoot off to infinity" doesn't mean it'll reach infinity, it just means it never stops growing
WanderingKensei92 4 horas atrás
Prime numbers are key here just plug in every prime number till you find the loop you are looking for it’s their I promise it’s just a really long number
sjb cwm
sjb cwm 5 horas atrás
Is brilliant in ather languages like nederlands
The Wizard
The Wizard 5 horas atrás
@15:00 - The animation for perfect squares, is there an interactive(zoom/frame control) copy of that somewhere? I saw an interesting pattern in it that I'd like to see the numbers for.
Kid Amogus
Kid Amogus 5 horas atrás
Its 3 because 3× +1 so plus 1 three times
Bo Bryant
Bo Bryant 5 horas atrás
Here is why will be tempted. This problem is simply stated, easily understood, and all too inviting. Just pick a number, any number: If the number is even, cut it in half; if it’s odd, triple it and add 1. Take that new number and repeat the process, again and again. If you keep this up, you’ll eventually get stuck in a loop. At least, that’s what we think will happen. Take 10 for example: 10 is even, so we cut it in half to get 5. Since 5 is odd, we triple it and add 1. Now we have 16, which is even, so we halve it to get 8, then halve that to get 4, then halve it again to get 2, and once more to get 1. Since 1 is odd, we triple it and add 1. Now we’re back at 4, and we know where this goes: 4 goes to 2 which goes to 1 which goes to 4, and so on. We’re stuck in a loop. Or try 11: It’s odd, so we triple it and add 1. Now we have 34, which is even, so we halve it to get 17, triple that and add 1 to get 52, halve that to get 26 and again to get 13, triple that and add 1 to get 40, halve that to get 20, then 10, then 5, triple that and add 1 to get 16, and halve that to get 8, then 4, 2 and 1. And we’re stuck in the loop again. Quantized Academy Patrick Honner, a nationally recognized high school teacher from Brooklyn, New York, introduces basic concepts from the latest mathematical research. See all Quantized Academy Columns The infamous Collatz conjecture says that if you start with any positive integer, you’ll always end up in this loop. And you’ll probably ignore my warning about trying to solve it: It just seems too simple and too orderly to resist understanding. In fact, it would be hard to find a mathematician who hasn’t played around with this problem. I couldn’t ignore it when I first learned of it in school. My friends and I spent days trading thrilling insights that never seemed to get us any closer to an answer. But the Collatz conjecture is infamous for a reason: Even though every number that’s ever been tried ends up in that loop, we’re still not sure it’s always true. Despite all the attention, it’s still just a conjecture. Yet progress has been made. One of the world’s greatest living mathematicians ignored all the warnings and took a crack at it, making the biggest strides on the problem in decades. Let’s take a look at what makes this simple problem so very complicated. To understand the Collatz conjecture, we’ll start with the following function: f(n)={n/23n+1if n is even if n is odd You might remember “piecewise” functions from school: The above function takes an input n and applies one of two rules to it, depending on whether the input is odd or even. This function f enacts the rules of the procedure we described above: For example, f (10) = 10/2 = 5 since 10 is even, and f (5) = 3 × 5 + 1 = 16 since 5 is odd. Because of the rule for odd inputs, the Collatz conjecture is also known as the 3n + 1 conjecture. The Collatz conjecture deals with “orbits” of this function f. An orbit is what you get if you start with a number and apply a function repeatedly, taking each output and feeding it back into the function as a new input. We call this “iterating” the function. We’ve already started computing the orbit of 10 under f, so let’s find the next few terms: f (10) = 10/2 = 5 f (5) = 3 × 5 + 1 = 16 f (16) = 16/2 = 8 f (8) = 8/2 = 4 A convenient way to represent an orbit is as a sequence with arrows. Here’s the orbit of 10 under f: 10 → 5 → 16 → 8 → 4 → 2 → 1 → 4 → 2 → 1 → … At the end we see we are stuck in the loop 1 → 4 → 2 → 1 → …. Similarly, the orbit for 11 under f can be represented as 11 → 34 → 17 → 52 → 26 → 13 → 40 → 20 → 10 → 5 → 16 → 8 → 4 → 2 → 1 → 4 → …. Again we end up in that same loop. Try a few more examples and you’ll see that the orbit always seems to stabilize in that 4 → 2 → 1 → … loop. The starting values of 9 and 19 are fun, and if you’ve got a few minutes to spare, try 27. If your arithmetic is right, you’ll get there after 111 steps. The Collatz conjecture states that the orbit of every number under f eventually reaches 1. And while no one has proved the conjecture, it has been verified for every number less than 268. So if you’re looking for a counterexample, you can start around 300 quintillion. (You were warned!) It’s easy to verify that the Collatz conjecture is true for any particular number: Just compute the orbit until you arrive at 1. But to see why it’s hard to prove for every number, let’s explore a slightly simpler function,ℊ. g(n)={n/2n+1if n is even if n is odd The functionℊ is similar to f, but for odd numbers it just adds 1 instead of tripling them first. Since ℊ and f are different functions, numbers have different orbits under ℊ than under f. For example, here are the orbits of 10 and 11 under ℊ: 10 → 5 → 6 → 3 → 4 → 2 → 1 → 2 → 1 → 2 → … 11 → 12 → 6 → 3 → 4 → 2 → 1 → 2 → 1→ 2 → … Notice that the orbit of 11 reaches 1 faster under ℊ than under f. The orbit of 27 also reaches 1 much faster under ℊ. 27 → 28 → 14 → 7 → 8 → 4 → 2 → 1 → 2 → … In these examples, orbits under ℊ appear to stabilize, just like orbits under f, but in a slightly simpler loop: → 2 → 1 → 2 → 1 → …. We might conjecture that orbits underℊ always get to 1. I’ll call this the “Nollatz” conjecture, but we could also call it the n + 1 conjecture. We could explore this by testing more orbits, but knowing something is true for a bunch of numbers - even 268 of them - isn’t a proof that it’s true for every number. Fortunately, the Nollatz conjecture can actually be proved. Here’s how. First, we know that half of a positive integer is always less than the integer itself. So if n is even and positive, then ℊ(n) = n/2 Now, if n is odd, then ℊ(n) = n + 1 which is bigger than n. But since n is odd, n + 1 is even, and so we know where the orbit goes next: ℊ will cut n + 1 in half. For an odd n the orbit will look like this: … → n → n + 1 → n+12 → … Notice that n+12 = n2 + 12. Since n2 1, then it’s always true that n2 + 12 This tells us that when an orbit under ℊ reaches an odd number greater than 1, we’ll always be at a smaller number two steps later. And now we can outline a proof of the Nollatz conjecture: Anywhere in our orbit, whether at an even or an odd number, we’ll trend downward. The only exception is when we hit 1 at the bottom of our descent. But once we hit 1 we’re trapped the loop, just as we conjectured. Can a similar argument work for the Collatz conjecture? Let’s go back to the original function. f(n)={n/23n+1if n is even if n is odd As with ℊ, applying f to an even number makes it smaller. And as with ℊ, applying f to an odd number produces an even number, which means we know what happens next: f will cut the new number in half. Here’s what the orbit under f looks like when n is odd: … → n → 3n + 1 → 3n+12 → … But here’s where our argument falls apart. Unlike our example above, this number is bigger than n: 3n+12 = 3n2 + 12, and 3n2 = 1.5n, which is always bigger than n. The key to our proof of the Nollatz conjecture was that an odd number must end up smaller two steps later, but this isn’t true in the Collatz case. Our argument won’t work. If you’re like me and my friends back in school, you might now be excited about proving that the Collatz conjecture is false: After all, if the orbit keeps getting bigger, then how can it get down to 1? But that argument requires thinking about what happens next, and what happens next illuminates why the Collatz conjecture is so slippery: We can’t be sure whether 3n+12 is even or odd. We know that 3n + 1 is even. If 3n + 1 is also divisible by 4, then 3n+12 is also even, and the orbit will fall. But if 3n + 1 is not divisible by 4, then 3n+12 is odd, and the orbit will rise. In general we can’t predict which will be true, so our argument stalls out. But this approach isn’t completely useless. Since half of all positive integers are even, there’s a 50% chance that 3n+12 is even, which makes the next step in the orbit 3n+14. For n > 1 this is less than n , so half the time an odd number should get lower after two steps. There’s also a 50% chance that 3n+14 is even, which means there’s a 25% chance that an odd number will be reduced to less than half of where it started after three steps. And so on. The net result is that, in some average way, Collatz orbits decrease when they encounter an odd number. And since Collatz orbits always decrease at even numbers, this suggests that all Collatz sequences must decrease in the long run. This probabilistic argument is widely known, but no one has been able to extend it to a complete proof of the conjecture. Yet several mathematicians have proved that the Collatz conjecture is “almost always” true. This means they’ve proved that, relative to the amount of numbers they know lead to 1, the amount of numbers they aren’t sure about is negligible. In 1976 the Estonian American mathematician Riho Terras proved that, after repeated application of the Collatz function, almost all numbers eventually wind up lower than where they started. As we saw above, showing that the numbers in the orbit consistently get smaller is one path to showing that they eventually get to 1.
OldMartin 6 horas atrás
i just wrote a python program because i got isnpired, i am currently checking a lot of numbers (to see if they get close to infinity), but i don't think i have enough RAM to run it in the background 24/7, also, i don't think i will find something spectacular soon, cause it does around 50 numbers a second, and i started from that 1 big number u showed at 10:52, i think i have checked 10000 now, most of them take 76 times to get down to 2 update, currently at 295147905179352847778, pretty sure i just saw 1 took 625 steps, but i am not sure, gonna have to read back, but they climbed from 76 steps to 425 steps i stopped it at 295147905179352854200 because i am writing every step to my disk, don't want to kill my hard drive xD
Релёкс84 3 horas atrás
Everything was already checked up to 2^68 (and definitely some more) so your python program will only go through numbers thathave been known to go 1 for a long time already.
Human Overpopulation
Human Overpopulation 6 horas atrás
Math problems like these are useful algorithms for computer games designing (e.g. Casino)
WinterNox - Brawl Stars
13:51 look at Derek
Martin Söderström
Martin Söderström 7 horas atrás
Is it a simple problem if no one can solve it?
bace cern
bace cern 7 horas atrás
its how life works. And lifetime works
countryboycansurvive 7 horas atrás
You never stated why. Why is this being done? Some guy came up with this, but why? He came up with these rules WHY. What's the end game? The goal? To simply prove him wrong? You are doing math, but this is NOT a math problem. The over all thing you are doing, no matter what you call it is completely meaningless as hop scotch or jacks. It's just simply an exercise in mathematical gymnastics. But it's not a "problem" that needs solved or not. Problem with being really smart, you think everything is a problem that needs solved. Just like a hammer, everything looks like a nail.
paul straymond
paul straymond 7 horas atrás
What happens if the number equal to x is infinity or 0?
Релёкс84 3 horas atrás
0 is a trivial loop that would have been worthy of at least a mention in this video, and infinity is not a number.
S H 8 horas atrás
I understand the issue here, clearly explained. Nice job. I'm just confused to where this (3n+1) comes from. Also, why (:2) when even... where did that come from. Anyone maybe have an awnser?
Brodythegameboy 9 horas atrás
I'm 10 in fourth grade and figured out the answer. 3x+1=4 math rules: multiply first, add last. 3x1=3 3+1=4. Lol
Godzilla 11 horas atrás
No one can solve this math problem Me with a calculator: I have no such weakness
REMUS Remus 11 horas atrás
Dude it’s 4 so am I the smartest person in the universe?
Chef R
Chef R 11 horas atrás
I am lost, so are you implying this proves God exists? Or the nature conjecture? Yeah I'm lost
Sjoer van der Ploeg
Sjoer van der Ploeg 11 horas atrás
You will always end up on the 4-2-1 loop as you will never have a division that results in zero and are always adding one to make round numbers! Odd number times odd number always makes an odd number.
Sjoer van der Ploeg
Sjoer van der Ploeg 11 horas atrás
Will you make a video about L-systems some day?
thinknthis 12 horas atrás
Cool, a mathematical problem explaining socialism (it will always fail).
listen2meokidoki 12 horas atrás
Yet it's a trivial question
Ryan Djeddar
Ryan Djeddar 12 horas atrás
I like your funny words, magic man.
Doggy Boy
Doggy Boy 12 horas atrás
Him: Pick a number, any number Me: *Picks 7* Him: 7? Good choice. Me: *Ah yes, and this floor here is made of floor, just like this math problem is hard as impossible.*
listen2meokidoki 13 horas atrás
What about 3 x infinity + 1
d0uch3b4gz0r 13 horas atrás
how about this: pick any 3 digit number (100-999), make it into a six digit number by writing it down twice… then divide it by 7,11 then 13 (you may switch order) and then you get your original number back :) for example: ((712712 / 7)/11)/13= 712
M T 13 horas atrás
So what is the problem posed in this video? These mathematicians can't earn a decent living, can only work as professors at colleges and universities. Comes up with some BS challenge claiming no one can solve, not even themselves. Then comes up with all these grandiose labels of the step they took in attempt to solve. Basic point is this... Any odd number times 3 is still an odd number, then add 1 to make it an even number. Even numbers can always be divided by two. And repeat until you get down to 4-2-1 pattern. Sarcastically wow!!! Their intellectual can explain why and how the world exists and would eventually destroy itself. Now pay me $200K per year.
Galileo Chiu
Galileo Chiu 14 horas atrás
Of course it is a Soviet invention
SAGE OF MUGEN 16 horas atrás
3:17 white space omori theme
Lachcorrupted 17 horas atrás
Me: picks random number… try 14000391
Beta Prime
Beta Prime 21 hora atrás
humans add 1 to 1 over and over and are amazed to find patterns its as simple as this.
Samedi 21 hora atrás
What happends if, for example 5x+1 instead of 3x+1 ? Another final loop ? Another conjecture ?
Ice Kiln
Ice Kiln 22 horas atrás
Can it be solved in an octa decimal numeric system?
Stanley Lubarski
Stanley Lubarski 22 horas atrás
This is easy. It's one more than 3X, duh! (just joking)
Dark Carnival
Dark Carnival 23 horas atrás
OK, I know nothing about mathematics other than day to day usage. This is all very interesting, to some more than others. I've always understood a problem as being something needing to be solved for a specific reason. What is the point of this so-called "problem"? What is the "problem" that is trying to be solved? Once solved, if solved, what "problem" is no longer a "problem"? If the "problem" to be solved is to identify a number that will not end in the 4-2-1 loop, and that number (or numbers) is/are identified, what exactly have you accomplished? I believe I've just wasted 22 minutes and 8 seconds of my life. Now THAT"S a problem......
x Miuna x
x Miuna x 23 horas atrás
Scariest thing about this is that he knew i picked 7 👀
Ken Berthiaume
Ken Berthiaume Dia atrás
So what is the question? Write a function for the number of steps it takes?
Alek Stuper
Alek Stuper Dia atrás
i didn't watch the video, also i suck at maths. still i'm going to give my answer... is it +3?
Ege Ayvala
Ege Ayvala Dia atrás
I dont see a problem the universe is infinite you bozos lmao
Gary GW Hicks
Gary GW Hicks Dia atrás
I am not a mathematician, so I am OK with leaving the Collatz Conjecture unproven.
Griffin Dia atrás
Very, very irritating that you do not define the problem. Wtf is the problem or the question?
AussieOtterYT Dia atrás
7:30 4*3=12+1=13
MergezMan2000 Dia atrás
bruh I chose 4
Tuftycamel Dia atrás
I don't really understand what the question actually is?
John Atonic
John Atonic Dia atrás
why is this a "problem"? and what would be considered a "solution"? If the "solution" is that this should reiterate infinitely, then wouldn't that be considered a "problem" too? maybe this is just a "law" of mathematics that when you invent a rule based equation there is going to be a certain conclusion. What if it's 9/x + 1, if even add 1, if odd +1, if it's an even negative then -1, odd negative +1, anyone can make up rule based equation but what's the point? And why would a loop outcome be considered a problem when it's just the natural outcome of a contrived equation that really has no purpose but waste every ones time? Now I'm curious to see how 9/x + 1 works out or how about a/x + y? I kinda gets ridiculous when you don't define the answer that you are looking for.
Irfan Banning
Irfan Banning Dia atrás
One approach may be to work backwards from arbitrary even numbers. Each even number has at most two precedents, and at least one, that number times 2. The only numbers that have two precedents are one 1 more than a multiple of 3. But this is not a "simple" problem, as the title implies, because of recursion, which adds complexity at each instance of recursion. Moreover, it is *branched* recursion.
I wonder how one could apply chaos theory here. This is the general idea of periodic trends in complex systems, maybe it could shed light as to why this occurs?
KayleePlays Dia atrás
Merrýden Ghosttech Mage☆
Intersynchmenthres shiftmetrics
Merrýden Ghosttech Mage☆
Merrýden Ghosttech Mage☆
Same like researched in 1922
Merrýden Ghosttech Mage☆
Use language systems ignoring by interbridgework variables
Merrýden Ghosttech Mage☆
Symbolica, not numbers
Not going to infinity makes perfect sense. 📉 ☺ Not having infinite loops makes no sense. 🔃?? 😠
Pablo Dia atrás
If no one can solve why is it called simple
Релёкс84 Dia atrás
Simple to formulate and understand
Dark Knight
Dark Knight Dia atrás
FJB. Fix the 2020 FRAUD!!!
Flynn Powers
Flynn Powers Dia atrás
12:40 had us in the first half not gonna lie ahaha
Anton Gromek
Anton Gromek Dia atrás
Congrats, you've found the roots of existence - why there is something rather then nothing🥳
Prranjal will Nuke U all
Eric Foreman: I'm quite good at math. Red Foreman: Oh yeah? So what's the value of x?
Youtube Videos
Youtube Videos Dia atrás
amazing math: 9X=YZ where, Y+Z=X...put any number above 0 and see the magic
Abderrahim Imaankaf
Abderrahim Imaankaf
MrWhatwhathuh Lala
3X+1=4X Stop over thinking. You're welcome.
Giorgi Weibolt
Giorgi Weibolt Dia atrás
What if x is 8
Marc Varta18
Marc Varta18 Dia atrás
Thermal Dynamics. What's the problem? Have a nice day. Four two and one. There will be different numbers. Zit zeroith. Law. That's the answer. Have some poultry math thought. It's thanks giving. Wait.wait. don't tell me. Infinity is wrong also. No more plates to count. That's the last number. Show your work? Infinity would require all sounds possible. That can't happen. No Infinite numbers. ARE POSSIBLE 3x+1=Zero Infinity Relative to temp. To measure /\0/\ will cause temperature. Graph a cooling system. Remember dark is the absence of light and motion. Light is the display of motion and the effect of said motion on the COSMOS. lol Let's look at 4. 2. 1. 7 where 6 and 6 Of probability will lead to 7. Die+Die less a number or integer brings us to 11 7+6=13 same buss. On b0th sides. Plus an integer. Less an in integer. Where two die with out motion are static. 52 is enough probability. 2. Snake eyes. How many rolls to roll a 2 again. Less than 52 rolls. All day long.
Im Kiki
Im Kiki Dia atrás
pls add Arabic subs I really enjoy your content
Jyps Ridic
Jyps Ridic Dia atrás
6:48 This is a fancy way of saying that you'll dismiss it if the election results are the ones you prefer.
Arka 0
Arka 0 Dia atrás
Propesyonal Dia atrás
This equation is taboo. its a violation against nature. Adding 5+5 =10 is simple enough but adding +1, forcing or bending the reality will cause serious problems. Like so, being stuck in 4, 2, 1 is the penalty of messing with the laws of existence. The penalty or living in the same moment again and again or to not have an existence at all. However, the problem can still always be resolved once the subject begins the reverse equation of 3×+1 and ÷ 2 at the moment of the loop.
Ishan Parajuli
Ishan Parajuli Dia atrás
World greatest mathematician 😅🤣🤣😅🤣
Paulo Lellis
Paulo Lellis Dia atrás
if you reverse the process: start with 1, apply both inverse control functions - multiply by 2 - multiply by 3 then add one for each newly generated number, repeat the steps... 1-> 2,4 2-> 4,9 4-> 8,17 8-> 16,33 9-> 18,28 17-> 34,52 ... you will end up with all possible numbers to be generated by the conjecture, not the opposite, I mean, not looking for how many times/ steps it decays to the 4->2->1 loop That should be possible for a skilled mathematician (definitely not me, just a computer engineering here) to intersect the original conjecture group with this newly generated set and check if there's any "outliers", using all those advanced-crazy group/ set operations. If the conjecture is true, then the intersect will be equals to both of the sets; otherwise, the data not in the intersection will likely be the circle, the non-convergent or any other combination of them; don't get me wrong, but for me, who handles with algorithms and loves to learn about math, origins, histories, how the genius solved determined problem, etc., this problem lacks some special "shine" (other than drive people nuts) to be worthy to actually be solved - and I know that there are infinite counter-examples of problems people thought that was not worthy and ended up changing the world, just don't think that this is one of them.
Paulo Lellis
Paulo Lellis 3 horas atrás
@Putnam I believe in you..
Putnam 4 horas atrás
this has been done many, many, many times
eric kestner
eric kestner Dia atrás
We brute forced Pi, so we can brute force it. Some day..
doggollamaman Dia atrás
I’m telling you 295,147,905,179,352,824,857 is the one to break the pattern
ONE 1 SEVEN 7 TEN 10 are the answers to everything, in a perfect world.
NoPlan Man
NoPlan Man Dia atrás
your evil! why would you show me this! this will bug me for a while
Mr Marvellous
Mr Marvellous Dia atrás
Alex Dia atrás
3x + 1? What’s x though
JaiPod Dia atrás
Well technically if you start with (-1) you get: (-1) x 3 + 1 = (-2), (-2)/2 = (-1) and so you end up in a loop. So that doesn't go to 1.
Joy Wanderi
Joy Wanderi Dia atrás
what if the solution can only be found by a person who is not limited by discovery. for instance, what if you say 1 and 0 are neither odd nor even. what if 1 and 0 are like quantum computing, where they represent the oddness or evenness of a number where necessary. and what if all the answers are in nature, such that each number is represented symbolically? what if the reason we can't solve for infinite numbers is because they cannot be mathematically solved, but solved by symbols? for instance, this guy who is a savant, what if he has been able to tap a higher percentage of his brain compared to the other people, such that he can communicate with nature? he had to go through a specific experience in order to attain the knowledge, right? what if what hinders us from tapping that knowledge is the fact that we are unable to tap from our brain more? i don't know where i am getting at but i am in that level where i believe the answers we are expect to find hinder us from finding the answer itself. maybe after technological singularity, i can find proof of my thoughts that the smallest particle is not a neutron, or electron or a proton, but it is infinite. it is like looking through 2 mirrors placed in parallel. these are just my thoughts but i wish there was someone out there who has the same questions i do
Everfox Marketing
Reach41 Dia atrás
Wish I hadn’t watched this. Now I won’t be able to sleep.
swearing_crumb Dia atrás
i would rather listen to this than listen to math class
Leo Duncan
Leo Duncan Dia atrás
Thought it was just 3(x+1)
Alla Greta
Alla Greta Dia atrás
The utility of it in this world? Zero. Waste of precious time and intelligence.
J Modified
J Modified Dia atrás
Possibly the effort could be better spent, but that was true for a lot of math developed in the past that is now very useful.
Emm1e 889
Emm1e 889 Dia atrás
I did the 3x+1 and if it’s even I divide it. Here’s what I got from 1-14. I accidentally forgot that 4 was even. I changed it to 2. It would be 9 if it was an odd number! Correct me if I’m wrong :) 1: 3 moves || 2: 1 move 3: 7 moves || 4: 2 moves 5: 5 moves || 6: 8 moves 7: 16 moves || 8: 3 moves 9: 19 moves || 10: 6 moves 11: 14 moves || 12: 9 moves 13: 9 moves || 14: 17 moves Edit (15-34): 15: 17 moves || 16: 4 moves 17: 12 moves || 18: 20 moves 19: 18 moves || 20: 6 moves 21: 7 moves || 22: 15 moves 23: 15 moves || 10 moves 25: 21 moves || 26: 10 moves 27: 28 moves || 28: 18 moves 29: 18 moves || 30: 18 moves 31: 25 moves || 32: 5 moves 33: 24 moves || 34: 13 moves
83_RIPTIDE_83 Dia atrás
Who else is looking in the comments for someone who just puts it in here
Emm1e 889
Emm1e 889 Dia atrás
I did the 3x+1 and if it’s even I divide it. Here’s what I got from 1-14. I accidentally forgot that 4 was even. I changed it to 2. It would be 9 if it was an odd number! Correct me if I’m wrong :) 1: 3 moves || 2: 1 move 3: 7 moves || 4: 2 moves 5: 5 moves || 6: 8 moves 7: 16 moves || 8: 3 moves 9: 19 moves || 10: 6 moves 11: 14 moves || 12: 9 moves 13: 9 moves || 14: 17 moves Edit 15-23 15: 17 moves || 16: 4 moves 17: 12 moves || 18: 20 moves 19: 18 moves || 20: 6 moves 21: 7 moves || 22: 15 moves 23: 15 moves
RVM451 Dia atrás
Are there any interesting similar equations? "5X + 1" "3X + 5" or something similar?
Emm1e 889
Emm1e 889 Dia atrás
I did the 3x+1 and if it’s even I divide it. Here’s what I got from 1-14. I accidentally forgot that 4 was even. I changed it to 2. It would be 9 if it was an odd number! Correct me if I’m wrong :) 1: 3 moves || 2: 1 move 3: 7 moves || 4: 2 moves 5: 5 moves || 6: 8 moves 7: 16 moves || 8: 3 moves 9: 19 moves || 10: 6 moves 11: 14 moves || 12: 9 moves 13: 9 moves || 14: 17 moves Edit 15-23 15: 17 moves || 16: 4 moves 17: 12 moves || 18: 20 moves 19: 18 moves || 20: 6 moves 21: 7 moves || 22: 15 moves 23: 15 moves
Myfriendadmin24 Dia atrás
because 3x+1 is 3x1 = 3 or reverse it's 1x3 = 1
Myfriendadmin24 Dia atrás
it's 3
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