Відео

This is just playing semantics. The word "end" was the wrong word to use here, unless it's being used as a definition that we stop when reaching one (which isn't unrealistic in this framing since the framing is pretty scarce in this video). But anyway, a better phrasing would be to say that it always reaches 1. This is of course true of 4 and 2 as well, but 1 sounds better.

same 💀💀

LOL me

are you talking about the fraction 2/2?

Positive numbers. 0 is not considered positive.

Notation is about communication. Communicate properly and all sorts of notation is used for all sorts of things.

3n+1. You add 1 making it even. Divide and from there it can either be even or odd. The base cases for these 2 types of behavior 1>1*3+1>4>2 2>2/2>1

1/2 numbers are reduced to about 1/2 (2n) 1/4 are reduced to about 3/4 (4n+1) There are infinite numbers of these equations with all ratios of 3^a/2^b. Some of course meaning having decreased and others not.

@@piercexlr878 first number decrease is based on 5 and last max number is based on 7, this formula is for a tesseract, 4 d formula..

@Marc.2.0 It's an even or odd game. 5s and 7s don't have a major impact. Your big players are powers of 2 and multiples of 3, depending on your approach.

nope u didn't

It's a race. Look at the numbers in the answers, look at the numbers in the info, and go with your gut.

They can’t make it too specialized either or it would just become unfair so reading graphs is really the only fair way they can do this.

it’s 421 loop dude..

Wdym solve gravity

Do you really think 5th grade you is smarter than some of the worlds greatest mathematicians? your "solution" is literally you just saying that "yeah you'll reach 2 eventually i guess"

Fuck are you yapping about lil bro

The problem is the Loop itself,The negative has 3 loop and positive has only one

"you'll eventually reach 2 which halves into one" doesn't make any sense in mathematics. What we need is a proof not a belief

That doesn’t prove there isn’t another loop with very large numbers or that some starting number can’t spin off to infinity. If you wanted a sequence that is guaranteed to have at least N steps I can quickly give you such a number no matter how big N is. But without testing the whole sequence I can’t promise that it doesn’t just keep growing forever.

Pretty sure positive whole numbers

no becuase its not even or odd because of the decimal

Only positive integers. If you allow fractions then EVERY number is divisible by two.

Still didn't solve it yet however what you just said is absolutely correct

this is so dumb fr but we still have to give it to apply to forgein unis

The most interesting aspect of this problem is that anyone with a basic knowledge of arithmetic can easily understand the problem. The problem is simply: Let's take any natural number x. ie x; Any of the numbers 1, 2, 3, 4, ... If x is an even number, let's divide this number by 2, so we get the number x/2; If x is an odd number, let's multiply this number by 3 and add 1, so we get 3x + 1. Let's apply the same process again to this new number we obtained, and similarly, let's continue by applying this process to the number we obtained later. If we continue in this way, the problem says that we will eventually reach the number 1.

Let's take the number 12 as an example. if we apply this operation to 12: 12 -> 6 -> 3 -> 10 -> 5 -> 16 -> 8 -> 4 -> 2 -> 1 and we get the number 1. If attention is paid, if we continue to apply this process after reaching the number 1, it becomes 1 -> 4 -> 2 -> 1 and we enter an endless loop. Therefore, the important thing here is to reach the number 1 at least once. World-renowned Hungarian mathematician Paul Erdos for this problem said, "Mathematics is not yet ready for such problems." and the american mathematician jeff lagarias, "this is a very difficult problem that we will never be able to solve with today's mathematics." he said.

As of 2020, all numbers up to 2^68 (2 to the 68th) have been tested using computers, and no number has been found to falsify the conjecture. This is one of the most important arguments for many mathematicians that the conjecture is highly likely to be true, even if it does not provide a mathematical proof for the conjecture. This problem remains unresolved to date, despite hundreds of published and unpublished articles (including numerous erroneous evidence) written about it.

@@huseyinkagantoy5469It’s not really an argument at all. You can easily construct a polynomial that has a thousand zeroes beyond 2^68, so checking that it has no zeroes below that means jack. The question is, reformulated, does the sequence inevitably reach a power of 2 even though those are increasingly rare as the numbers in the sequence increase.

⁠@@magicmulder 3n+1 tends to infinity so you will always reach a power of 2 no matter the density

Just one power close

Actually 2^100000 -1.

Who is “we?” Is that you and your lame friends or something? I solved this problem all the way up to 2^92nd when I was in solitary confinement, just to have something to do to pass the time. It gets pretty boring being in the hole, after a while. I don’t wish that torture on anyone. I did nearly 2 full days in there, all alone, and it was awful. I just about lost my mind.

​@@chriswebster24uh

​@@chriswebster24 um,1; there are 6-7 months gap bettwen your reply and the comments. 2, did you publish the problems you solved anywhere?

And then look for patterns

It comes down to probability - does the increasing sparseness of powers of 2 “win” or does the infinite sequence eventually hit a power of 2.

I guess a follow up question is "Is there a number that forms a closed loop that does not contain 2^n?" which has been no until 2^68 so far.

@@Fahad-21 Indeed, if the sequence doesn't end up at 1, it can either grow infinitely or enter a loop above 1. Which of the two, if any, or both, apply is the question.

Actually, not 2^n. 3x+1=2^n, so ×=(2^n-1)/3. If n=even, there is a natural number x that satisfies the equation, but not for odd. so, does every number reach (4^n-1)/3? Then, let's think about later one, written as x' = 2(4^n-1)/3 (which is 2x) or (x-1)/3. This gives us an infinite form. x(a) = 2(x(a-1)) or x(a-1)/3. But, there are some problems. f(a) has tooooo many cases. (2^a) (This is way too big on large numbers)

I am so glad to hear you learned so much!!

No, it’s how easy it is to plow through your mother on a Sunday night

No. The ACT is a standardized test like the MCAT, SAT, LSAT, CSAT, etc. If you know how these tests work, they have multiple subjects and a time limit and are made so everyone has some sort of chance. The ACT in particular for a section like math is organized from easy, medium, to hard level questions. This is just the science section though. The ACT also has really tough time limits compared to the SAT.

Lol the main problem ain't about the difficulty level of papers but actually the time management which makes it ten times harder..u can't even give a minute to each mcq esp for english u can max give 30 seconds to each mcq

I SERIOUSLY need to make a 20 on my act tomorrow

yeah, because they don't breed like pest thus having less competition.

Thanks Kyle!!

Absolutely! Send me an email at contact@jjtestprep.com, and I can guide you :)