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."

## Comments

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TopNewestWhen do you predict that the Collatz conjecture will be solved? @Sharky Kesa @Daniel Liu @Daniel Chiu @Trevor B. – Elliott Macneil · 3 years, 2 months ago

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– Sharky Kesa · 3 years, 2 months ago

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

I also thought this ....one day....Thanks for posting it – Archiet Dev · 3 years, 2 months ago

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The Collatz Conjecture? or The Ulam's Conjecture? – Mu'amar Musa Nurwigantara · 3 years, 2 months ago

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– حكيم الفيلسوف الضائع · 3 years, 2 months ago

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.Log in to reply

– Kushagra Sahni · 3 years, 2 months ago

I did not understand what were you saying about odd numbers. Even one is true but can you explain me the odd process ?Log in to reply

– Shourya Pandey · 3 years, 2 months ago

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.Log in to reply

– Kushagra Sahni · 3 years, 2 months ago

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.Log in to reply

– حكيم الفيلسوف الضائع · 3 years, 2 months ago

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!Log in to reply

– Kushagra Sahni · 3 years, 2 months ago

OK. Now I got it very well. ThanksLog in to reply

– حكيم الفيلسوف الضائع · 3 years, 1 month ago

You're welcome!Log in to reply

\(C(5) = 16, C(16) = 8, C(8) = 4, C(4) = 2, C(2) = 1\). And thus the operation ends. – Sharky Kesa · 3 years, 2 months ago

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couldn't we just consider it true until proven wrong? =) – Gmasha Hussain · 3 years, 1 month ago

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– حكيم الفيلسوف الضائع · 3 years, 1 month ago

That wouldn't be helpful in any way.Log in to reply

– Gmasha Hussain · 3 years, 1 month ago

Yep, just taking the lazy way out.Log in to reply

Le problème de Syracuse ;) – Antony Diaz · 3 years, 2 months ago

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Wouldn't it be correct if I say it will also reach every time reach 2 . – Utkarsh Dwivedi · 3 years, 2 months ago

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– Yuxuan Seah · 3 years, 1 month ago

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.Log in to reply

– Utkarsh Dwivedi · 3 years, 1 month ago

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.Log in to reply