Decidability of language L = { n : ∃x∈N, n = 3x+2 } I've recently came across this language and I don't know if my hint for proving its class is correct or not. 

$$L = \{ n : \exists x \in N, \,n=3x+2 \} $$$n$ is in binary format. 

I think $L$ is decidable and as proof I was considering using a canonical enumerator of $\{0,1\}^*$. 

For each word $x$ printed by the enumerator, my algorithm would compute $3x+2$ and halt if the result was equal or greater to the provided input.
If equal, then accept, if greater, then reject. 

I would like someone to confirm if I'm correct, pointing me in the right direction if I'm not.
Yes, your approach is correct. 

The description of your algorithm would be clearer if it also mentions the following.


The algorithm rejects the empty word immediately.
That canonical enumerator of $\{0,1\}^∗$ enumerates all positive integers in increasing order.
If $3x+2$ is smaller than the input then the result will be ignored and the enumerator will keep going.


Another algorithm that decides the language would be subtracting 3 from the input number and checking the result repeatedly. If the result is 2, it accepts. If the result is negative, it rejects. It also rejects the the empty word immediately.