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."

No vote yet

1 vote

×

Problem Loading...

Note Loading...

Set Loading...

Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

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

Log in to reply

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

Log in to reply

The Collatz Conjecture? or The Ulam's Conjecture?

Log in to reply

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

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

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

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

Log in to reply

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

Log in to reply

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

Log in to reply

That wouldn't be helpful in any way.

Log in to reply

Yep, just taking the lazy way out.

Log in to reply

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

Log in to reply

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

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

Le problème de Syracuse ;)

Log in to reply