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The most difficult - yet so easy to understand - problem in the world!

Take any natural number \(n\). If \(n\) is even, divide it by \(2\) to get \(n / 2\). If \(n\) is odd, multiply it by \(3\) and add \(1\) to obtain \(3n + 1\). Repeat the process indefinitely. The problem is to prove that no matter what number you start with, you will always eventually reach \(1\).

Don't be fooled by the simplicity of this problem, great mathematicians like Erdős had great trouble proving it. The latter even said: "Mathematics may not be ready for such problems."

Note by حكيم الفيلسوف الضائع
2 years, 10 months ago

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When do you predict that the Collatz conjecture will be solved? @Sharky Kesa @Daniel Liu @Daniel Chiu @Trevor B. Elliott Macneil · 2 years, 10 months ago

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@Elliott Macneil Possibly in the next decade given the number of young mathematicians rising up and are being seen on Brilliant. Sharky Kesa · 2 years, 10 months ago

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I also thought this ....one day....Thanks for posting it Archiet Dev · 2 years, 10 months ago

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The Collatz Conjecture? or The Ulam's Conjecture? Mu'amar Musa Nurwigantara · 2 years, 10 months ago

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@Mu'amar Musa Nurwigantara It's commonly named after Collatz name but it is also named Ulam's conjecture because he used to talk about it in the lectures he gave. حكيم الفيلسوف الضائع · 2 years, 10 months ago

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@حكيم الفيلسوف الضائع I did not understand what were you saying about odd numbers. Even one is true but can you explain me the odd process ? Kushagra Sahni · 2 years, 10 months ago

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@Kushagra Sahni Start with any natural number. If the number is even, divide it by two. If it is odd, triple it and add one. Repeat the process with the new number so formed. Ultimately the number 1 is reached. Shourya Pandey · 2 years, 10 months ago

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@Shourya Pandey Even one is true. But what about odd ? Let the no. be 5. if we triple it and add one, we will get 16 and then 49 then 148 ......... Which I don't think would approach 1. Or am I not getting the process. Can u explain the process with examples. Kushagra Sahni · 2 years, 10 months ago

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@Kushagra Sahni When you got 16, you'll have to use the operation we restricted to even numbers, that is: \(n/2\). And so you will get 8. Repeat it again you'll have 4, then 2, then 1! حكيم الفيلسوف الضائع · 2 years, 10 months ago

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@حكيم الفيلسوف الضائع OK. Now I got it very well. Thanks Kushagra Sahni · 2 years, 10 months ago

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@Kushagra Sahni You're welcome! حكيم الفيلسوف الضائع · 2 years, 10 months ago

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@Kushagra Sahni You have done the operation wrong. Lets call the operation C(x) where x is your starting number.

\(C(5) = 16, C(16) = 8, C(8) = 4, C(4) = 2, C(2) = 1\). And thus the operation ends. Sharky Kesa · 2 years, 10 months ago

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couldn't we just consider it true until proven wrong? =) Gmasha Hussain · 2 years, 9 months ago

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@Gmasha Hussain That wouldn't be helpful in any way. حكيم الفيلسوف الضائع · 2 years, 9 months ago

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@حكيم الفيلسوف الضائع Yep, just taking the lazy way out. Gmasha Hussain · 2 years, 9 months ago

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Le problème de Syracuse ;) Antony Diaz · 2 years, 10 months ago

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Wouldn't it be correct if I say it will also reach every time reach 2 . Utkarsh Dwivedi · 2 years, 10 months ago

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@Utkarsh Dwivedi No. If you start on 2 itself, then you will have to carry out the operation n/2 anyway. You have to COMPLETE the sequence and hence end with 1. Yuxuan Seah · 2 years, 9 months ago

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@Yuxuan Seah But by starting with 2 , 2 comes in our sequence. How will you write the sequence - only 1 or 2 , 1. Consider the following example - We take n = 5 .Then how would you write the sequence like this - 5 , 16 , 8 , 4 , 2 , 1 or simply ignoring 5, by - 16, 8 , 4 , 2 , 1. Of course first one 's correct so we could never ever find a sequence except n = 1 in which two does not come. Utkarsh Dwivedi · 2 years, 9 months ago

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