Let $^*$ be the binary operation on $N$ given by $a ^* b = \text{L.C.M. of } a \text{ and } b$. Find which elements of $N$ are invertible for the operation $^*$?

(A) An element $a$ in $N$ is invertible with respect to the operation $^*$ if there exists an element $b$ in $N$ such that $a ^* b = e = b ^* a$,where $e$ is the identity element.
For the operation of $L.C.M.$ on the set of natural numbers $N$,the identity element $e$ must satisfy $L.C.M.(a, e) = a$ for all $a \in N$. This holds true for $e = 1$.
Thus,we need to find $b \in N$ such that $L.C.M.(a, b) = 1$.
Since $L.C.M.(a, b) \geq a$ and $L.C.M.(a, b) \geq b$ for all $a, b \in N$,the condition $L.C.M.(a, b) = 1$ implies that $a = 1$ and $b = 1$.
Therefore,$1$ is the only invertible element of $N$ with respect to the operation $^*$.