How many residues?

We will consider each of the 2 terms on the LHS separately. Firstly, $\dbinom{xp-1}{p} = \dfrac{(xp-1)(xp-2)...(p(x-1))}{p!}$ . Evidently $p(x-1) \equiv 0 \pmod{p}$ , $p(x-1) + 1\equiv 1 \pmod{p}$ , ... , $px-1 \equiv p-1 \pmod{p}$ . Therefore the terms on the numerator form the full set of residues modulo $p$ . Cancelling through gives us $\dfrac{p(x-1) \times (p-1)!}{p!} \equiv \dfrac{p(x-1)}{p} \equiv x-1 \pmod{p}$ Now consider the second term. $x(p-1) \le xp-1 < xp$ . Therefore $\lfloor \dfrac{xp-1}{x} \rfloor = p-1$ . So now plugging these into the congruence we get that $\dbinom{xp-1}{p} - \lfloor \dfrac{xp-1}{x} \rfloor \equiv x-1-(p-1) \equiv x \pmod{p}$ But we have that $x$ is ranging over the positive multiples of $p$ , so $x \equiv 0 \pmod{p}$ . Therefore the only valid value of $r$ is 0, so there is $\fbox{1}$ value