Let R be a relation on the set N of natural numbers denoted by nRm ⇔ n is a factor of m (i.e. n | m). Then, R is

The correct option is "Reflexive, transitive but not symmetric".

Let us analyze the given relation $R$ defined on the set of natural numbers $\mathrm{\&DoubleStruckCapitalN;}$ as:
$nRm⟹n\text{ is a factor of }m\text{ (i.e., }n|m\text{)}$
We need to check the properties of reflexivity, transitivity, and symmetry for this relation.

1. Reflexivity:
A relation $R$ on a set $\mathrm{\&DoubleStruckCapitalN;}$ is reflexive if every element is related to itself, i.e., $nRn$ for all $n\in \mathrm{\&DoubleStruckCapitalN;}$.
Here, $nRn$ means $n$ is a factor of $n$ (or $n|n$). Since any natural number divides itself (i.e., $\frac{n}{n}=1$, which is an integer), $n$ is always a factor of $n$.
Therefore, the relation $R$ is reflexive.

2. Transitivity:
A relation $R$ is transitive if whenever $nRm$ and $mRp$, then $nRp$ for all $n,m,p\in \mathrm{\&DoubleStruckCapitalN;}$.
Let $nRm$ and $mRp$. This implies:
$n|m\Rightarrow m=kn$ for some integer $k$.
$m|p\Rightarrow p=jm$ for some integer $j$.
Substituting the expression for $m$ into the equation for $p$, we get:
$p=j(kn)=(jk)n$
Since $j$ and $k$ are integers, their product $jk$ is also an integer. This shows that $n$ is a factor of $p$ (i.e., $n|p$), which means $nRp$.
Therefore, the relation $R$ is transitive.

3. Symmetry:
A relation $R$ is symmetric if $nRm$ implies $mRn$ for all $n,m\in \mathrm{\&DoubleStruckCapitalN;}$.
Let us test this with a counterexample. Take $n=2$ and $m=4$.
Here, $2R4$ is true because $2$ is a factor of $4$ (i.e., $2|4$).
However, $4R2$ is false because $4$ is not a factor of $2$ (i.e., $4\parallel 2$).
Since $nRm$ does not necessarily imply $mRn$, the relation $R$ is not symmetric.

Conclusion:
Since the relation $R$ is reflexive and transitive but not symmetric, the correct option is indeed "Reflexive, transitive but not symmetric".