Answer
$\exists$ an integer $d$, such that $6/d$ is an integer and $d \neq 3.$
Work Step by Step
Recall the form of the negation of a universal conditional statement:
$~(\forall x, P(x) \rightarrow Q(x)) \equiv \exists x$ such that $(P(x) \land $ ~$Q(x))$.
In this case P(x) is: 6/d is an integer.
Q(x) is: d=3.
