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."

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## Comments

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TopNewestWhen do you predict that the Collatz conjecture will be solved? @Sharky Kesa @Daniel Liu @Daniel Chiu @Trevor B.

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

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

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The Collatz Conjecture? or The Ulam's Conjecture?

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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 ?

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

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

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That wouldn't be helpful in any way.

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Yep, just taking the lazy way out.

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Le problème de Syracuse ;)

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

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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.

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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.

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