The following operation is allowed on the positive integers: if a number is even, we can divide it by $2$, otherwise we can multiply it by a power of $3$ (different from $3^0$) and add $1$. Prove that we can reach $1$ from any starting positive integer $n$.
Problem
Source: Serbia TST 2021, P5
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26.10.2021 03:43
Beautiful problem . One of the hardest problems I've done!
26.10.2021 03:48
tigerzhang wrote: Beautiful problem . One of the hardest problems I've done!
Wait... confusion?
26.10.2021 03:52
Isn't this a more general version of collatz
26.10.2021 03:55
presumably this is just bad translation (or "powers" have a different definition in serbia i dunno)
26.10.2021 03:58
pog wrote: Arrowhead575 wrote: Isn't this a more general version of collatz it's less general. also you can literally multiply by $1$ and add $1$ as $3^0 = 1$, so thus you just add to a power of $2$ and divide until you get to $1$ Maybe powers of $3$ are supposed to mean $3^n$, where $n$ is a positive integer.
26.10.2021 04:06
26.10.2021 14:23
Bump this.