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

Take any natural number nn. If nn is even, divide it by 22 to get n/2n / 2. If nn is odd, multiply it by 33 and add 11 to obtain 3n+13n + 1. Repeat the process indefinitely. The problem is to prove that no matter what number you start with, you will always eventually reach 11.

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
5 years, 5 months ago

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Comments

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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 - 5 years, 5 months ago

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Possibly in the next decade given the number of young mathematicians rising up and are being seen on Brilliant.

Sharky Kesa - 5 years, 5 months ago

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The Collatz Conjecture? or The Ulam's Conjecture?

Mu'amar Musa Nurwigantara - 5 years, 5 months ago

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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.

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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 - 5 years, 5 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 - 5 years, 5 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 - 5 years, 5 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/2n/2. And so you will get 8. Repeat it again you'll have 4, then 2, then 1!

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@حكيم الفيلسوف الضائع OK. Now I got it very well. Thanks

Kushagra Sahni - 5 years, 5 months ago

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@Kushagra Sahni You're welcome!

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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)=1C(5) = 16, C(16) = 8, C(8) = 4, C(4) = 2, C(2) = 1. And thus the operation ends.

Sharky Kesa - 5 years, 5 months ago

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I also thought this ....one day....Thanks for posting it

Archiet Dev - 5 years, 5 months ago

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couldn't we just consider it true until proven wrong? =)

Gmasha Hussain - 5 years, 4 months ago

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That wouldn't be helpful in any way.

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Yep, just taking the lazy way out.

Gmasha Hussain - 5 years, 4 months ago

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Wouldn't it be correct if I say it will also reach every time reach 2 .

Utkarsh Dwivedi - 5 years, 5 months ago

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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 - 5 years, 4 months ago

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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 - 5 years, 4 months ago

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Le problème de Syracuse ;)

Antony Diaz - 5 years, 5 months ago

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